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5ad4bf9836198b7b75cbec11530e9400 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$ | Let the two primes be $a$ and $b$ , with $a$ being the larger prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, \[a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549\] \[ab = 5183.\] Because we have $b = a+2$ $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \boxed{16}$ | E | 16 |
5ad4bf9836198b7b75cbec11530e9400 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$ | Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$
By the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$
Plugging in $p-q=2$ and $p^{3}-q^{3}=31106$
$31106=2(p^{2}+pq+q^{2})$
Through the givens, we can see that $p \approx q$
Thus, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200$
Recall that $70^2=4900$ and $80^2=6400$ . It follows that our primes must be only marginally larger than $70$ , where we conveniently find $p=73, q=71$
The least prime greater than these two primes is $79 \implies 7 + 9 \implies \boxed{16}$ | E | 16 |
5ad4bf9836198b7b75cbec11530e9400 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$ | Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \[6x^2 + 2 = 31106 \implies x^2 = 5184.\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger than $73$ is $79$ , which has a digit sum of $7 + 9 = \boxed{16}.$ | null | 16 |
7fa49879cddf0997ff774c489ced096c | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14 | Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$ | Let $M$ be the largest number in $S$ .
We categorize numbers $\left\{ 1, 2, \ldots , M-1 \right\}$ (except $\frac{M}{2}$ if $M$ is even) into $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups, such that the $i$ th group contains two numbers $i$ and $M-i$
Recall that $M \in S$ and the sum of two numbers in $S$ cannot be equal to $M$ , and the sum of numbers in each group above is equal to $S$ . Thus, each of the above $\left\lfloor \frac{M-1}{2} \right\rfloor$ groups can have at most one number in $S$ .
Therefore, \begin{align*} |S| & \leq 1 + \left\lfloor \frac{M-1}{2} \right\rfloor \\ & \leq 1 + \left\lfloor \frac{25}{2} \right\rfloor \\ & = 13. \end{align*}
Next, we construct an instance of $S$ with $|S| = 13$ .
Let $S = \left\{ 13, 14, \ldots , 25 \right\}$ .
Thus, this set is feasible.
Therefore, the most number of elements in $S$ is $\boxed{13}$ | B | 13 |
7fa49879cddf0997ff774c489ced096c | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14 | Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$ | We know that two odd numbers sum to an even number, so we can easily say that odd numbers $1-25$ can be included in the list, making for $13$ elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than $13$ , odd numbers as well as even numbers would have to be included in the list (since there are only $12$ even numbers from $1-25$ , and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, $\boxed{13}$ elements is the highest we can go. | B | 13 |
7fa49879cddf0997ff774c489ced096c | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14 | Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$ | The smallest sum of a number $a + b$ where $b \geq a$ is $a + a = 2a$ as we are using the smallest value of $b$ . Using this, we can say that if $12$ were an element of $S$ , then one of the sums (the smallest) would be $12 + 12 = 24 < 25$ . Thus $13$ must be the smallest element. So the largest amount of elements that could be in $S$ is the list of numbers from $13$ to $25$ as they all work. Because it is inclusive we have, $25 - 13 + 1 = \boxed{13}$ | B | 13 |
7fa49879cddf0997ff774c489ced096c | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14 | Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$ | We construct a possible subset $S$ with $13$ elements by including all odd integers from $1$ to $25$ , inclusive. $S=\left\{ 1, 3, 5, \cdots , 25 \right\}$ . The sum of any $2$ elements is even, and thus cannot be an element of $S$
To show that $S$ cannot have more than $13$ elements, assume for sake of contradiction that $|S| \geq 14$ . Let $S=\left\{ x_1, x_2, \cdots , x_n \right\}$ where $n \geq 14$ and $x_1 < x_2 < \cdots < x_n$ . Because the sums of any $2$ (not necessarily distinct) elements do not appear in $S$ $x_1+x_i$ is not an element of $S$ for all $1 \leq i \leq n$ . So, $x_1, x_2, \cdots , x_n , x_1+x_1, x_1+x_2, \cdots , x_1+x_n$ are all distinct integers. Let these integers be elements of the set $T$ $|T|=2n$ , and because $n \geq 14$ $|T| \geq 28$ . But all elements of $T$ must be $\geq x_1$ and $\leq x_1+x_n \leq x_1+25$ , leaving only 26 possible values for the elements in $T$ . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.
Thus, $\boxed{13}$ is the maximum possible size of $S$ | B | 13 |
7fa49879cddf0997ff774c489ced096c | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_14 | Suppose that $S$ is a subset of $\left\{ 1, 2, 3, \cdots , 25 \right\}$ such that the sum of any two (not necessarily distinct) elements of $S$ is never an element of $S.$ What is the maximum number of elements $S$ may contain?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 16$ | We can start by building a list of the elements of $S$ ,and see if we can find a pattern. Let's start with $1$ . If we include 1, that means we cannot include $2$ (which is $1$ $1$ ), so we write the next valid number, $3$ . Similarly, we cannot include $4$ or $6$ , so we write $5$ . Proceeding, our list looks like this:
$S$ = { $1$ $3$ $5$ $7$ ....
We've found a pattern! It's easy understand why this pattern of odd numbers will continue, seeing as all the even numbers can be written as sums between odds. Calculating how many odd numbers fit into the range $1$ $25$ , we conclude that the answer is $\boxed{13}$ | B | 13 |
d11b0abf2ea5cbc3944eaaac8472ea45 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_15 | Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$ | Suppose that the first number of the arithmetic sequence is $a$ . We will try to compute the value of $S_{n}$ . First, note that the sum of an arithmetic sequence is equal to the number of terms multiplied by the median of the sequence. The median of this sequence is equal to $a + n - 1$ . Thus, the value of $S_{n}$ is $n(a + n - 1) = n^2 + n(a - 1)$ . Then, \[\frac{S_{3n}}{S_{n}} = \frac{9n^2 + 3n(a - 1)}{n^2 + n(a - 1)} = 9 - \frac{6n(a-1)}{n^2 + n(a-1)}.\] Of course, for this value to be constant, $6n(a-1)$ must be $0$ for all values of $n$ , and thus $a = 1$ . Finally, we have $S_{20} = 20^2 = \boxed{400}$ | D | 400 |
d11b0abf2ea5cbc3944eaaac8472ea45 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_15 | Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence that has a common difference of $2$ . The quotient $\frac{S_{3n}}{S_n}$ does not depend on $n$ . What is $S_{20}$
$\textbf{(A) } 340 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 380 \qquad \textbf{(D) } 400 \qquad \textbf{(E) } 420$ | Recall that the sum of the first $n$ odd numbers is $n^2$
Since $\frac{S_{3n}}{S_{n}} = \frac{9n^2}{n^2} = 9$ , we have $S_n = 20^2 = \boxed{400}$ | D | 400 |
0f34ca34acec2788a9000f355f70b92e | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18 | Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$ | Let $M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix},$ and $M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.$
We wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework:
There are $9+3=12$ ordered triples $(M_1,M_2,M_3).$
Similarly, for each of $M_1+M_3=M_2$ and $M_2+M_3=M_1,$ there are $12$ ordered triples $(M_1,M_2,M_3).$
In this subcase, we have $\boldsymbol{12\cdot3=36}$ ordered triples $\boldsymbol{(M_1,M_2,M_3).}$
Together, the answer is $8+168+126+36=\boxed{338}.$ | B | 338 |
0f34ca34acec2788a9000f355f70b92e | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18 | Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$ | We will use complementary counting and do casework on the equations.
There are $8$ possible equations:
Equation 1: $0 = 0$
Equation 2: $x = 0$
Equation 3: $y = 0$
Equation 4: $z = 0$
Equation 5: $x + y = 0$
Equation 6: $x + z = 0$
Equation 7: $y + z = 0$
Equation 8: $x + y + z = 0$
We will continue to refer to the equations by their number on this list.
$8^3 = 512$ total systems. Note that no two equations by themselves can force $x = y = z = 0$ . Therefore no system with Equation 1 or with repeated equations can force $x = y = z = 0$
Case 1: Equation 8 ( $x + y + z = 0$ ) is present.
Case 1a: Equation 8, and two equations from $\{5, 6, 7\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{5, 6, 7\}$ and $3! = 6$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 3! = 18$
Case 1b: Equation 8, one equation from $\{5, 6, 7\}$ , and one equation from $\{2, 3, 4\}$
There are $\binom{3}{1} = 3$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let us choose Equation 7. Given $x + y + z = 0$ and $y + z = 0$ , we conclude that $x = 0$ . The third equation can be either $y = 0$ or $z = 0$ . There are $3!$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 2 \cdot 3! = 36$
Case 1c: Equation 8, and two equations from $\{2, 3, 4\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{2, 3, 4\}$ and $3! = 6$ ways to arrange each case. Each of these cases forces $x = y = z = 0$ $3 \cdot 3! = 18$ total options.
Case 2: Equation 8 is $\textbf{not}$ present, at least one equation from $\{5, 6, 7\}$ is present.
Case 2a: Equations $\{5, 6, 7\}$ are all present.
There are $3!$ ways to arrange the three equations. $6$ options.
Case 2b: Two equations from $\{5, 6, 7\}$ are present. One equation from $\{2, 3, 4\}$ is present.
There are $\binom{3}{2}$ ways to choose two equations from $\{5, 6, 7\}$ . WLOG let Equations 5 and 6 be in our system: $x + y = 0$ and $x + z = 0$ . Any equation from $\{2, 3, 4\}$ will force $x = y = z = 0$ . There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{2} \cdot \binom{3}{1} \cdot 3! = 54$
Case 2c: One equation from $\{5, 6, 7\}$ is present. Two equations from $\{2, 3, 4\}$ are present.
There are $\binom{3}{1}$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let Equation 5 ( $x + y = 0$ ) be present. One of the two equations from $\{2, 3, 4\}$ must be Equation 4, $z = 0$ , since it is the only equation that restricts $z$ . The last equation can be either 2 or 3. There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{1} \cdot \binom{2}{1} \cdot 3! = 36$
Case 3: Only equations $\{2, 3, 4\}$ are present.
There are $3!$ ways to arrange the three equations. $6$ options.
We add up the cases: $18 + 36 + 18 + 6 + 54 + 36 + 6 = 174$ total systems force $x = y = z = 0$ . Thus $512 - 174 = \boxed{338}$ do not. | B | 338 |
0f34ca34acec2788a9000f355f70b92e | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18 | Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$ | The total number of possible systems is $2^9 = 512$ , with $8$ possible sets of coefficients per equation. We will use complementary counting to find the number of systems which only have the solution $(0, 0, 0)$ and subtract that from the total. Similar to what is observed in Solution 2, if any equation is repeated or $0x + 0y + 0z = 0$ , there will only be two or fewer equations for three variables, making one unique solution impossible. Therefore, we must choose $3$ different equations from $7$ possible ones, giving $7 \cdot 6 \cdot 5 = 210$ systems. However, there are two exceptions to consider, which will have more than one solution. The first is of the form $x + y + z = 0$ $x + y = 0$ $z = 0$ ; the second is of the form $x = 0$ $y = 0$ $x + y = 0$ . In both cases, there are $3$ ways to choose the variables in the equations, and then $6$ ways to arrange them, giving $2 \cdot 3 \cdot 6 = 36$ exceptions. Subtracting this gives $210 - 36 = 174$ systems with only one solution, and the answer is then $512 - 174 = \boxed{338}$ | B | 338 |
0f34ca34acec2788a9000f355f70b92e | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_18 | Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$ | Denote vector $\overrightarrow{i} = \left( i_1, i_2, i_3 \right)^T$ for $i \in \left\{ a, b, c \right\}$ .
Thus, we need to count how many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly dependent.
We do complementary counting.
First, the total number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $\left( 2^3 \right)^3 = 512$
Second, we count how many many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly independent.
To meet this condition, no vector can be a zero vector $\overrightarrow{0} = \left( 0, 0, 0 \right)^T$
Next, we do the casework analysis.
Case $1^c$ : Three vectors are all on axes.
In this case, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $3!$
Case $2^c$ : Two vectors are on axes and the third vector is not.
We construct such an instance in the following steps.
Step 1: We determine which two vectors lie on axes.
The number of ways is $3$
Step 2: For two vectors selected in Step 1, we determine which two axes they lie on.
The number of ways is $3 \cdot 2$
Step 3: For the third unselected vector, we determine its value.
To make three vectors linear independent, the third vector cannot be on the plane formed by the first two vectors.
So the number of ways is $3$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 2 \cdot 3$
Case $3^c$ : One vector is on an axis and the other two are not.
We construct such an instance in the following steps.
Step 1: We determine which vector lies on an axis.
The number of ways is $3$
Step 2: For the selected vector, we determine which axis it lies on.
The number of ways is $3$
Step 3: We determine the values of the two unselected vectors.
First, to be linearly independent, these two vectors are distinct.
Second, to be linearly independent, we cannot have one vector $(1,1,1)$ and another one that is a diagonal vector on the plane that is perpendicular to the first selected vector.
Thus, the number or ways in this step is $4 \cdot 3-2 = 10$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 10$
Case $(4.4)^c$ : No vector is on any axis.
In this case, any three distinct vectors are linearly independent.
So the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $4 \cdot 3 \cdot 2$
Putting all cases together, the number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ that are linearly independent is \[ 8^3 - 3! - 3 \cdot 3 \cdot 2 \cdot 3 - 3 \cdot 3 \cdot 10 - 4 \cdot 3 \cdot 2 = \boxed{338}. \] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | B | 338 |
b6b6184a11a773b6050d5be3bce67e9f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_19 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.) [asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy] $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$ | There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$ | C | 22 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | Without loss of generality, we assume the length of each side of $ABCD$ is $2$ .
Because $E$ is the midpoint of $CD$ $CE = 1$
Because $ABCD$ is a rhombus, $\angle BCE = 180^\circ - \angle D$
In $\triangle BCE$ , following from the law of sines, \[ \frac{CE}{\sin \angle FBC} = \frac{BC}{\sin \angle BEC} . \]
We have $\angle BEC = 180^\circ - \angle FBC - \angle BCE = 46^\circ - \angle FBC$
Hence, \[ \frac{1}{\sin \angle FBC} = \frac{2}{\sin \left( 46^\circ - \angle FBC \right)} . \]
By solving this equation, we get $\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}$
Because $AF \perp BF$ \begin{align*} BF & = AB \cos \angle ABF \\ & = 2 \cos \left( 46^\circ - \angle FBC \right) . \end{align*}
In $\triangle BFC$ , following from the law of sines, \[ \frac{BF}{\sin \angle BCF} = \frac{BC}{\sin \angle BFC} . \]
Because $\angle BCF = 180^\circ - \angle BFC - \angle FBC$ , the equation above can be converted as \[ \frac{BF}{\sin \left( \angle BFC + \angle FBC \right)} = \frac{BC}{\sin \angle BFC} . \]
Therefore, \begin{align*} \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \right) - \cos \angle FBC} \\ & = \frac{1}{\sin 46^\circ - \left( 1 - \cos 46^\circ \right) \cot \angle FBC} \\ & = \frac{\sin 46^\circ}{\cos 46^\circ - 1} \\ & = - \frac{\sin 134^\circ}{1 + \cos 134^\circ} \\ & = - \tan \frac{134^\circ}{2} \\ & = - \tan 67^\circ \\ & = \tan \left( 180^\circ - 67^\circ \right) \\ & = \tan 113^\circ . \end{align*}
Therefore, $\angle BFC = \boxed{113}$ | D | 113 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$
Because $\overline{AB} \parallel \overline{ED}$ , we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$ , so $\triangle ABG \sim \triangle DEG$ by AA.
Because $ABCD$ is a rhombus, $AB = CD = 2DE$ , so $AG = 2GD$ , meaning that $D$ is a midpoint of segment $\overline{AG}$
Now, $\overline{AF} \perp \overline{BE}$ , so $\triangle GFA$ is right and median $FD = AD$
So now, because $ABCD$ is a rhombus, $FD = AD = CD$ . This means that there exists a circle from $D$ with radius $AD$ that passes through $F$ $A$ , and $C$
AG is a diameter of this circle because $\angle AFG=90^\circ$ . This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$ , so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$ , which means that $\angle BFC = \boxed{113}$ | D | 113 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | Let $\overline{AC}$ meet $\overline{BD}$ at $O$ , then $AOFB$ is cyclic and $\angle FBO = \angle FAO$ . Also, $AC \cdot BO = [ABCD] = 2 \cdot [ABE] = AF \cdot BE$ , so $\frac{AF}{BO} = \frac{AC}{BE}$ , thus $\triangle AFC \sim \triangle BOE$ by SAS, and $\angle OEB = \angle ACF$ , then $\angle CFE = \angle EOC = \angle DAC = 67^\circ$ , and $\angle BFC = \boxed{113}$ | D | 113 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | Observe that all answer choices are close to $112.5 = 90+\frac{45}{2}$ . A quick solve shows that having $\angle D = 90^\circ$ yields $\angle BFC = 135^\circ = 90 + \frac{90}{2}$ , meaning that $\angle BFC$ increases with $\angle D$ .
Substituting, $\angle BFC = 90 + \frac{46}{2} = \boxed{113}$ | D | 113 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | This solution refers to the Diagram section.
We extend $AD$ and $BE$ to point $G$ , as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); [/asy] We know that $AB=AD=2$ and $CE=DE=1$
By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$ . This implies that $2AD=AG$ and $AD \cong DG$ , so $AG=2AD=2\cdot2=4$ . That is, $D$ is the midpoint of $AG$
Now, let's redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$ , which is on the circle, as shown below: [asy] /* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NE,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); draw(Circle(D,6),dashed); [/asy] Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$
Let's call $\angle CFE = \theta$
Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$ , So $\angle CDG = 2\angle CFE$
Let $\angle CDG = 2\theta$ . Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that \begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67. \end{align*} Since $\angle BFC=180-\theta$ , we have $\angle BFC=180-67=\boxed{113}$ | D | 113 |
009d921392e8ec2e3b586aba3c76e15b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_20 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$ . Let $E$ be the midpoint of $\overline{CD}$ , and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$ . What is the degree measure of $\angle BFC$
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | If angle $ADC$ was a right angle, it would be much easier. Thus, first pretend that $ADC$ is a right angle. $ABCD$ is now a square. WLOG, let each of the side lengths be 1. We can use the Pythagorean Theorem to find the length of line $AE$ , which is $\sqrt{5}/2$ . We want the measure of angle $BFC$ , so to work closer to it, we should try finding the length of line $BF$ . Angle $FAB$ and angle $ABF$ are complementary. Angle $ABF$ and angle $FBC$ are also complementary. Thus, $\sin FAB=\cos ABF=\sin FBC$ $\sin FAB=\sin FBC=(1/2)/(\sqrt{5}/2)=1/\sqrt{5}$ . Since $\sin FAB=1\sqrt{5}$ ,and $AB=1$ $FB=\sin FAB$ . It follows now that $FE=3*\sqrt{5}/10$
Now, zoom in on triangle $BEC$ . To use the Law of Cosines on triangle $FBC$ , we need the length of $FC$ . Use the Law of Cosines on triangle $EFC$ . Cos $E=1/\sqrt{5}$ . Thus, after using the Law of Cosines, $FC=\sqrt{2/5}$
Since we now have SSS on $BEC$ , we can get use the Law of Cosines. $\cos BFC=1/-\sqrt{2}$ $\arccos 1/-\sqrt{2}$ is 45, but if the cosine is negative that means that the angle is the supplement of the positive cosine value. $180-45=135$ . Angle $BFC$ is $135^\circ$
Realize that, around point F, there will always be 3 right angles, regardless of what angle $ADC$ is. There are only two angles that change when $ADC$ changes. Break up angle $BFC$ into angle $BFB'$ , which is always 90 degrees, and angle $B'FC$ , which we have discovered to to be half of $ADC$ . Thus, when angle $ADC$ is 46 degrees, then $B'FC$ will be 23. $23+90=113$ . Angle $BFC$ is $\boxed{113}$ degrees. | D | 113 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.
Throughout this solution, we will express all polynomials in base $x$ . E.g. $x^2 + x + 1 = 111_{x}$
We are given: \[111a + 12 = 101b + 21 = P(x).\] We add $111$ and $101$ to each side and balance respectively: \[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\] We make the unit's digits equal: \[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\] We now notice that: \[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\] Therefore $a = 11_{x} = x + 1$ $b = 12_{x} = x + 2$ , and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$ $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$ ‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{23}$ | E | 23 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | Let $P(x) = Q(x)(x^2+x+1) + x + 2$ , then $P(x) = Q(x)(x^2+1) + xQ(x) + x + 2$ , therefore $xQ(x) + x + 2 \equiv 2x + 1 \pmod{x^2+1}$ , or $xQ(x) \equiv x-1 \pmod{x^2+1}$ . Clearly the minimum is when $Q(x) = x+1$ , and expanding gives $P(x) = x^3+2x^2+3x+3$ . Summing the squares of coefficients gives $\boxed{23}$ | E | 23 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | Let $P(x) = (x^2+x+1)Q_1(x) + x + 2$ ,
then $P(x) = (x^2+1)Q_1(x) + xQ_1(x) + x + 2$
Also $P(x) = (x^2+1)Q_2(x) + 2x + 1$
We infer that $Q_1(x)$ and $Q_2(x)$ have same degree, we can assume $Q_1(x) = x + a$ , and $Q_2(x) = x + b$ , since $P(x)$ has least degree. If this cannot work, we will try quadratic, etc.
Then we get: $(x^2+1)(Q_1(x) - Q_2(x)) + xQ_1(x) - x + 1 = 0$
The constant term gives us: $(Q_1(x) - Q_2(x)) + 1 = 0$
So $Q_1(x) - Q_2(x) = -1$
Substituting this in gives: $-(x^2+1) + xQ_1(x) - x + 1 = 0$
Solving this equation, we get $Q_1(x) = x + 1$
Plugging this into our original equation we get $P(x) = x^3 + 2x^2 + 3x + 3$
Verify this works with $P(x) = (x^2+1)Q_2(x) + 2x + 1$
Therefore the answer is $1^2 + 2^2 + 3^2 + 3^2 = \boxed{23}$ | E | 23 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | Notice that we cannot have the quotients equal to some constants, since the same constant will yield different constant terms for $P(x)$ (which is bad) and different constants will yield different first coefficients (also bad). Thus, we try setting the quotients equal to linear terms (for minimizing degree).
Let $P(x)=(x^2+x+1)(ax+b)+(x+2)$ and $P(x)=(x^2+1)(ax+c)+(2x+1)$ . The quotients have the same $x$ coefficient, since $P(x)$ must have the same $x^3$ coefficient in both cases. Expanding, we get \[P(x)=ax^3+(a+b)x^2+(a+b+1)x+(b+2)\] and \[P(x)=ax^3+cx^2+(a+2)x+(c+1).\]
Equating coefficients, we get $b+2=c+1$ $a+b+1=a+2$ , and $a+b=c$ . From the second equation, we get $b=1$ , then substituting into the first, $c=2$ . Finally, from $a+b=c$ , we have $a=1$ . Now, $P(x)=(x^2+x+1)(ax+b)+(x+2)=(x^2+x+1)(x+1)+(x+2)=x^3+2x^2+3x+3$ and our answer is \[1^2+2^2+3^2+3^2=\boxed{23}.\] | E | 23 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | We construct the following equations in terms of $P(x)$ and the information given by the problem: \[\textbf{(1) } P(x)=(x^2+x+1)\cdot Q(x)+x+2\] \[\textbf{(2) } P(x)=(x^2+1)\cdot R(x)+2x+1\] Upon inspection, $Q(x)$ and $R(x)$ cannot be constant, so the smallest possible degree of $P(x)$ is $3,$ and both $Q(x)$ and $R(x)$ are linear.
Let $Q(x)=x-q$ and $R(x)=x-r.$ We know there will be values for $q$ and $r$ that make the below equation hold, so we can assume that $P(x)$ has a leading coefficient of $1$
Substituting these values in, and setting $\textbf{(1)}$ and $\textbf{(2)}$ equal to each other, \[(x^2+x+1)(x-q)+x+2=(x^2+1)(x-r)+2x+1.\] We plug in $x=0$ , yielding $r+1=q.$ Substituting this value into the above equation, \[(x^2+x+1)(x-r-1)+x+2=(x^2+1)(x-r)+2x+1.\] Letting $x=1,$ we conclude that $r=-2,$ so $R(x)=x+2.$ Therefore, \[P(x)=(x^2+1)(x+2)+2x+1 = x^3+2x^3+3x+3.\] The requested sum is \[1^2+2^2+3^2+3^2=\boxed{23}\] | E | 23 |
e633e6c624fafc5d5ad99f4bae46de5a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_21 | Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$ , the remainder is $x+2$ , and when $P(x)$ is divided by the polynomial $x^2+1$ , the remainder
is $2x+1$ . There is a unique polynomial of least degree with these two properties. What is the sum of
the squares of the coefficients of that polynomial?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 13 \qquad\textbf{(C)}\ 19 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 23$ | By remainder theorem, the polynomial can be written as follows.
\[P(x) = (x^2+x+1)Q_{1}(x)+x+2 = (x^2+1)Q_{2}(x)+2x+1\] This is a timed exam, we can use the information given by answer choices. The answer choices tell us this is the polynomial with integer coefficients, and we need to find the polynomial with the least degree so we can assume both $Q_{1}(x)$ and $Q_{2}(x)$ are linear (the coefficient of x should be same).
Then we can write $P(x)$ as a cubic polynomial.
\[P(x) = (x^2+x+1)(ax+b)+x+2 = (x^2+1)(ax+c)+2x+1\] Substituting $x=0,1,-1$ to determine the value of $a$ and $b$
We have: \[P(0) = b+2 = c+1\] \[P(1) = 3a+3b+3 = 2a+2c+3\] \[P(-1) = -a+b+1 = -2a+2c-1\]
We can solve the simultaneous equations: $a=1,b=1,c=2$
Hence, $P(x)=(x^2+x+1)(x+1)+x+2=x^3+2x^2+3x+3$ . The answer is $1^2+2^2+3^2+3^2=\boxed{23}$ | E | 23 |
0c5fd1b20828ca337921505ebaf54bc3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_22 | Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$ $x^{2}+y^{2}=64$ , and $(x-5)^{2}+y^{2}=3$ . What is the sum of the areas of all circles in $S$
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$ | We denote by $C_1$ the circle that has the equation $x^2 + y^2 = 4$ .
We denote by $C_2$ the circle that has the equation $x^2 + y^2 = 64$ .
We denote by $C_3$ the circle that has the equation $(x-5)^2 + y^2 = 3$
We denote by $C_0$ a circle that is tangent to $C_1$ $C_2$ and $C_3$ .
We denote by $\left( u, v \right)$ the coordinates of circle $C_0$ , and $r$ the radius of this circle.
From the graphs of circles $C_1$ $C_2$ $C_3$ , we observe that if $C_0$ is tangent to all of them, then $C_0$ must be internally tangent to $C_2$ .
We have \[ u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1) \]
We do the following casework analysis in terms of the whether $C_0$ is externally tangent to $C_1$ and $C_3$
Case 1: $C_0$ is externally tangent to $C_1$ and $C_3$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 2: $C_1$ is internally tangent to $C_0$ and $C_3$ is externally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r + \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 3: $C_1$ is externally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r + 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r + 2 = 8 - r$ . Thus, $r = 3$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Case 4: $C_1$ is internally tangent to $C_0$ and $C_3$ is internally tangent to $C_0$
We have \[ u^2 + v^2 = \left( r - 2 \right)^2 \hspace{1cm} (2) \] and \[ (u-5)^2 + v^2 = \left( r - \sqrt{3} \right)^2 . \hspace{1cm} (3) \]
Taking $(2) - (1)$ , we get $r - 2 = 8 - r$ . Thus, $r = 5$ .
We can further compute (omitted here) that there exist feasible $(u,v)$ with this given $r$
Because the graph is symmetric with the $x$ -axis, and for each case above, the solution of $v$ is not 0. Hence, in each case, there are two congruent circles whose centers are symmetric through the $x$ -axis.
Therefore, the sum of the areas of all the circles in $S$ is $2\left( 3^2 \pi +5^2 \pi +3^2 \pi +5^2 \pi \right) = \boxed{136}$ | E | 136 |
8468580eaf3e08e9dd177b2fe005b510 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24 | Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$ | We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$
Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{50} ~MRENTHUSIASM | B | 50 |
8468580eaf3e08e9dd177b2fe005b510 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24 | Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$ | Denote $f(900)-f(600) = a$ .
Because $f(300) = f(900)$ $f(300) - f(600) = a$
Following from the Lipschitz condition given in this problem, $|a| \leq 150$ and \[ f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\} \] and \[ f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} . \] Thus, \begin{align*} f(800) - f(400) & \leq \min \left\{ a + 50 , 100 \right\} - \max \left\{ a - 50 , -100 \right\} \\ & = 100 + \min \left\{ a, 50 \right\} - \max \left\{ a , - 50 \right\} \\ & = 100 + \left\{ \begin{array}{ll} a + 50 & \mbox{ if } a \leq -50 \\ 0 & \mbox{ if } -50 < a < 50 \\ -a + 50 & \mbox{ if } a \geq 50 \end{array} \right. . \end{align*} Thus, $f(800) - f(400)$ is maximized at $a = 0$ $f(800)-f(600) = 50$ $f(400)-f(600)=-50$ , with the maximal value 100.
By symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$ $f(800)-f(600) = -50$ $f(400)-f(600)=50$ , with the minimal value $-100$
Following from the Lipschitz condition, \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*} We have already construct instances in which the second inequality above is augmented to an equality.
Now, we construct an instance in which the first inequality above is augmented to an equality.
Consider the following piecewise-linear function: \[ f(x) = \left\{ \begin{array}{ll} \frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\ -\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\ \frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\ -\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800 \end{array} \right.. \] Therefore, the maximum value of $f(f(800)) - f(f(400))$ is $\boxed{50}$ | B | 50 |
8468580eaf3e08e9dd177b2fe005b510 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24 | Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$ | Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$ . This means that when we take any two points on $f$ , the absolute value of the slope between the two points is at most $\frac{1}{2}$
Let $f(300) = f(900) = c$ , and since we want to find the maximum value of $|f(800) - f(400)|$ , we can take the most extreme case and draw a line with slope $-\frac{1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\frac{1}{2}$ up from $f(800)$ to $f(900)$ . Then $f(400) = c - 50$ and $f(800) = c + 50$ , so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$ , and this is attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value less than $\frac{1}{2}$
Therefore, $|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{50}$ | B | 50 |
8468580eaf3e08e9dd177b2fe005b510 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24 | Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$ | Consider $g(x) = f(x)-f(300)$ . Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$ . We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$ , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50 )| \leq \frac{1}{2}(100) = \boxed{50}$ as the answer. | B | 50 |
6df2ca67a518461e3c0256f0b5edf7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$ | In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{6}.\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | A | 6 |
6df2ca67a518461e3c0256f0b5edf7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$ | First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$ \[0 \equiv 2^{2019}k + 1 \pmod{7}.\] Using Fermat's Little Theorem, \[2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.\] Thus, \[0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.\] Therefore, \[7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.\]
We may repeat this same calculation with $S_{2023}$ to yield \[S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.\] Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits.
This gives rise to a simple inequality, \[0 \leqslant S_n \leqslant 2^n.\] Since the maximum possible number that can be generated with $n$ bits is \[\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.\] Looking at our calculations for $S_{2019}$ and $S_{2023}$ , we see that the only valid integers that satisfy that constraint are $j = h = 0$ \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{6}.\] ~zoomanTV | A | 6 |
6df2ca67a518461e3c0256f0b5edf7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$ | As in Solution 2, we note that \[x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.\] We also know that $7S_{2023} \equiv 1 \pmod{2^{2023}}$ and $7S_{2019} \equiv 1 \pmod{2^{2019}}$ , this implies: \[\textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1,\] \[\textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1.\] Dividing by $7$ , we can isolate the previous sums: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}.\] The maximum value of $S_n$ occurs when every $x_i$ is equal to $1$ . Even when this happens, the value of $S_n$ is less than $2^n$ . Therefore, we can construct the following inequalities: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}.\] From these two equations, we can deduce that both $x$ and $y$ are less than $7$
Reducing $\textbf{1}$ and $\textbf{2}$ $\pmod{7},$ we see that \[2^{2023}\cdot{x}\equiv 6\pmod{7},\] and \[2^{2019}\cdot{y}\equiv 6\pmod{7}.\]
The powers of $2$ repeat every $3, \pmod{7}.$
Therefore, $2^{2023}\equiv 2 \pmod 7$ and $2^{2019} \equiv 1 \pmod {7}.$ Substituing this back into the above equations, \[2x\equiv{6}\pmod{7}\] and \[y\equiv{6}\pmod{7}.\]
Since $x$ and $y$ are integers less than $7$ , the only values of $x$ and $y$ are $3$ and $6$ respectively.
The requested sum is \begin{align*} \frac{S_{2023}-S_{2019}}{2^{2019}} &= \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} \\ &= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7} \right)\right) \\ &= \frac{3\cdot{2^4}-6}{7} \\ &= \boxed{6} -Benedict T (countmath1) | A | 6 |
6df2ca67a518461e3c0256f0b5edf7d5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_25 | Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$ | Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0. A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{6}.$ | A | 6 |
99d42854570f636b874a967f1bdd3dea | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{91^2}{13^2}=\left(\frac{91}{13}\right)^2=7^2=\boxed{49}.\] ~MRENTHUSIASM | C | 49 |
99d42854570f636b874a967f1bdd3dea | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We have \[\frac{(2112-2021)^2}{169}=\frac{91^2}{169}=\frac{(10^2-3^2)^2}{13^2}=\frac{((10+3)(10-3))^2}{13^2}=\frac{(13\cdot7)^2}{13^2}=\frac{13^2 \cdot 7^2}{13^2}=7^2=\boxed{49}.\] | C | 49 |
99d42854570f636b874a967f1bdd3dea | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_1 | What is the value of $\frac{(2112-2021)^2}{169}$
$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$ | We know that $2112-2021 = 91$ . Approximate this as $100$ as it is pretty close to it. Also, approximate $169$ to $170$ . We then have \[\frac{(2112 - 2021)^2}{169} \approx \frac{100^2}{170} \approx \frac{1000}{17} \approx 58.\] Now check the answer choices. The two closest answers are $49$ and $64$ . As the numerator is actually bigger than it should be, it should be the smaller answer, or $\boxed{49}$ | C | 49 |
3962c6da4f1b20773d38e6411fe7f176 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_2 | Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?
$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$ | We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\] Therefore, the answer is $\boxed{20}.$ | E | 20 |
c08f67961535fdb986740b4acb0f3ab2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3 | What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$
Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \[\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.\] Approximating with $\pi\approx3.14,$ we have $12<4\pi<13,$ or $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor.$ We simplify to get \[6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$ | D | 6 |
c08f67961535fdb986740b4acb0f3ab2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3 | What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
By an underestimation $\pi\approx3,$ we have $4\pi>12,$ or $\frac{81}{4\pi}<6\frac34.$
By an overestimation $\pi\approx\frac{22}{7},$ we have $4\pi<\frac{88}{7},$ or $\frac{81}{4\pi}>6\frac{39}{88}.$
Together, we get \[6 < 6\frac{39}{88} < \frac{81}{4\pi} < 6\frac34 < 7,\] from which $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$ | D | 6 |
c08f67961535fdb986740b4acb0f3ab2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_3 | What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$ | As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{81}{4\pi}\right\rfloor.$
Approximating with $\pi\approx3,$ we have $\frac{81}{4\pi}\approx6\frac34.$ Since $\pi$ is about $5\%$ greater than $3,$ it is safe to claim that $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{6}.$ | D | 6 |
ed99ded39618784c8e0ce788348fe463 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | First, modulo $2$ or $5$ $\underline{20210A} \equiv A$ .
Hence, $A \neq 0, 2, 4, 5, 6, 8$
Second modulo $3$ $\underline{20210A} \equiv 2 + 0 + 2 + 1 + 0 + A \equiv 5 + A$ .
Hence, $A \neq 1, 4, 7$
Third, modulo $11$ $\underline{20210A} \equiv A + 1 + 0 - 0 - 2 - 2 \equiv A - 3$ .
Hence, $A \neq 3$
Therefore, the answer is $\boxed{9}$ | E | 9 |
ed99ded39618784c8e0ce788348fe463 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\textbf{(C)}$
If the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ . Thus we can eliminate options $\textbf{(A)}$ and $\textbf{(D)}$
So the correct option is either $\textbf{(B)}$ or $\textbf{(E)}$ . Let's try dividing the number with some integers.
$20210A/7 = 2887x$ , where $x$ is $1A/7$ . Since $13$ and $19$ are both indivisible by $7$ , this does not help us narrow the choices down.
$20210A/11 = 1837x$ , where $x$ is $3A/11$ . Since $33/11 = 3$ , option $\textbf{(B)}$ would make $20210A$ divisible by $11$ . Thus, by elimination, the correct choice must be option $\boxed{9}$ | E | 9 |
ed99ded39618784c8e0ce788348fe463 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_5 | The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$ | $202100 \implies$ divisible by $2$
$202101 \implies$ divisible by $3$
$202102 \implies$ divisible by $2$
$202103 \implies$ divisible by $11$
$202104 \implies$ divisible by $2$
$202105 \implies$ divisible by $5$
$202106 \implies$ divisible by $2$
$202107 \implies$ divisible by $3$
$202108 \implies$ divisible by $2$
This leaves only $A=\boxed{9}$ | E | 9 |
7c7f1375a007e6e2252d526ac69f2028 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_6 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$ | There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride. | B | 8 |
7c7f1375a007e6e2252d526ac69f2028 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_6 | Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$ | There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$ | B | 8 |
23fa4cd719d8194472b2b513366a4ce5 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_7 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$ | By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{170}$ degrees. | D | 170 |
23fa4cd719d8194472b2b513366a4ce5 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_7 | As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$ . Point $F$ lies on $\overline{AD}$ so that $DE=DF$ , and $ABCD$ is a square. What is the degree measure of $\angle AFE$
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$ | We can extend $\overline{AD}$ to $G$ , making $\angle CDG$ a right angle. It follows that $\angle GDE$ is $110^\circ - 90^\circ = 20^\circ$ , as shown below. [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); pair G = (15,10); label("$G$", G, E); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F^^G); draw(A--B--C--D--G--cycle); draw(D--EE--F--cycle); [/asy] Since $\angle DFE = \angle DEF$ , we see that $\angle DFE = \angle DEF = \frac{20}{2} = 10^\circ$ . Thus, $\angle AFE = 180^\circ - 10^\circ = \boxed{170}$ degrees. | D | 170 |
a805b8cb77cb41b297df52ffde3c75a9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_8 | A two-digit positive integer is said to be $\emph{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | Note that the number $\underline{xy} = 10x + y.$ By the problem statement, \[10x + y = x + y^2 \implies 9x = y^2 - y \implies 9x = y(y-1).\] From this we see that $y(y-1)$ must be divisible by $9.$ This only happens when $y=9.$ Then, $x=8.$ Thus, there is only $\boxed{1}$ cuddly number, which is $89.$ | B | 1 |
a805b8cb77cb41b297df52ffde3c75a9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_8 | A two-digit positive integer is said to be $\emph{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$ | If the tens digit is $a$ and the ones digit is $b$ then the number is $10a+b$ so we have the equation $10a + b = a + b^2$ . We can guess and check after narrowing the possible cuddly numbers down to $13,14,24,25,35,36,46,47,57,68,78,89,$ and $99$ . (We can narrow it down to these by just thinking about how $a$ 's value affects $b$ 's value and then check all the possiblities.) Checking all of these we get that there is only $\boxed{1}$ 2-digit cuddly number, and it is $89$ . Yay!!! | B | 1 |
1861eee8004af280f1e0eb084bd24822 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_10 | A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$ . Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$
$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$ | The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{13.5}.$ | B | 13.5 |
88c2c163970f842a079d34adcde8a3b2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$ | Let $x$ be the length of the ship.
Then, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps.
Also, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps.
Since the ship and Emily have the same ratio of absolute speeds in either direction, $\frac{210}{210-x} = \frac{42}{x-42}$ . Dividing both sides by $42$ and cross multiplying, we get $5(x-42) = 210-x$ , so $6x = 420$ , and $x = \boxed{70}$ | A | 70 |
88c2c163970f842a079d34adcde8a3b2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$ | Let the speed at which Emily walks be $42$ steps per hour. Let the speed at which the ship is moving be $s$ . Walking in the direction of the ship, it takes her $210$ steps, or $\frac {210}{42} = 5$ hours, to travel. We can create an equation: \[d = 5(42-s),\] where $d$ is the length of the ship. Walking in the opposite direction of the ship, it takes her $42$ steps, or $42/42 = 1$ hour. We can create a similar equation: \[d = 1(42+s).\] Now we have two variables and two equations. We can equate the expressions for $d$ and solve for $s$ \begin{align*} 210-5s &= 42 + s \\ s &= 28. \\ \end{align*} Therefore, we have $d = 42 + s = \boxed{70}$ | A | 70 |
88c2c163970f842a079d34adcde8a3b2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$ | Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.
Let $L$ be the length of the ship, $E$ be Emily's step length, and $S$ be the ship's step length. We wish to find $\frac LE.$
When Emily walks from the back of the ship to the front, she walks a distance of $210E$ and the front of the ship moves a distance of $210S.$ We have $210E=L+210S$ for this scenario, which rearranges to \[210E-210S=L. \hspace{15mm}(1)\] When Emily walks in the opposite direction, she walks a distance of $42E$ and the back of the ship moves a distance of $42S.$ We have $42E=L-42S$ for this scenario, which rearranges to \[42E+42S=L. \hspace{19.125mm}(2)\] We multiply $(2)$ by $5$ and then add $(1)$ to get $420E=6L,$ from which $\frac LE = \boxed{70}.$ | A | 70 |
88c2c163970f842a079d34adcde8a3b2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$ | Every time Emily takes a step, the boat also "takes a step". Call the length of the boats step $s$ . Call the length of the boat $x$
When Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of $s$ . This means that she travels a total distance of $x + 210 s$ to reach the other end of the boat.
When Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by $s$ (since the boat is coming towards her and moves a distance of $s$ ). This means that she travels a total distance of $x - 42 s$ to reach the other end of the boat.
Taking Emily's step as a unit of distance, we now have two equations \begin{align*} 210 &= x + 210 s, \\ 42 &= x - 42s. \end{align*} Solving for $x$ you get $\boxed{70}$ | A | 70 |
88c2c163970f842a079d34adcde8a3b2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_11 | Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$ | Call the speed of the boat $v_s$ and the speed of Emily $v_e$
Consider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is $v_e-v_s$
Consider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is $v_e+v_s$
Since Emily takes $210$ steps to walk along with the boat and $42$ steps to walk opposite the boat, that means it takes her $5$ times longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$
This means that $5(v_e-v_s)=v_e+v_s$ , so $v_s = \frac{2v_e}{3}$
As Emily takes $210$ steps to walk the length of the boat at a speed of $v_e- \frac{2v_e}{3}=\frac{v_e}{3}$ , she must take $\frac13$ of the time to walk the length of the boat at a speed of $v_e$ , so our answer is $\frac{210}{3} = \boxed{70}$ | A | 70 |
5307591cc7f7910c19f7e9a8083b54d2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_12 | The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$ | Recall that $9\equiv-1\pmod{5}.$ We expand $N$ by the definition of bases: \begin{align*} N&=27006000052_9 \\ &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ &\equiv 2-7+6-5+2 &&\pmod{5} \\ &\equiv -2 &&\pmod{5} \\ &\equiv \boxed{3} ~Aidensharp ~Kante314 ~MRENTHUSIASM | D | 3 |
5307591cc7f7910c19f7e9a8083b54d2 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_12 | The base-nine representation of the number $N$ is $27006000052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$
$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$ | We need to first convert $N$ into a regular base- $10$ number: \[N = 27006000052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]
Now, consider how the last digit of $9$ changes with changes of the power of $9:$ \begin{align*} 9^0&=1, \\ 9^1&=9, \\ 9^2&=\ldots 1, \\ 9^3&=\ldots 9, \\ 9^4&=\ldots 1, \\ & \ \vdots \end{align*} Note that if $x$ is odd, then $9^x \equiv 4\pmod{5}.$ On the other hand, if $x$ is even, then $9^x \equiv 1\pmod{5}.$
Therefore, we have \begin{align*} N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &&\pmod{5} \\ &\equiv 2+28+6+20+2 &&\pmod{5} \\ &\equiv 58 &&\pmod{5} \\ &\equiv \boxed{3}$ may simplify the process further, as given by Solution 1. | D | 3 |
e58c06af306d10b5ae4bbaa399756ecf | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_14 | How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$ | The second equation is $(|x|+|y| - 4)^2 = 1$ . We know that the graph of $|x| + |y|$ is a very simple diamond shape, so let's see if we can reduce this equation to that form: \[(|x|+|y| - 4)^2 = 1 \implies |x|+|y| - 4 = \pm 1 \implies |x|+|y| = \{3,5\}.\] We now have two separate graphs for this equation and one graph for the first equation, so let's put it on the coordinate plane: [asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 1.0,0.5)); yaxis(-8,8,Ticks(f, 1.0,0.5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return -x-3; } draw(graph(f,0,-3)); real f(real x) { return 5-x; } draw(graph(f,0,5)); real f(real x) { return 5+x; } draw(graph(f,0,-5)); real f(real x) { return x-5; } draw(graph(f,0,5)); real f(real x) { return -x-5; } draw(graph(f,0,-5)); real f(real x) { return 3-x; } draw(graph(f,0,3)); real f(real x) { return 3+x; } draw(graph(f,0,-3)); real f(real x) { return x-3; } draw(graph(f,0,3)); real f(real x) { return (-x^2)/3+3; } draw(graph(f,-5,5)); [/asy] We see from the graph that there are $5$ intersections, so the answer is $\boxed{5}$ | D | 5 |
b9d69c4d56382afc7c00e83bf2e4db48 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15 | Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$ | Let $\odot O_1$ be the circle with radius $5\sqrt2$ that is tangent to $\overleftrightarrow{AB}$ at $B$ and to $\overleftrightarrow{AC}$ at $C.$ Note that $\angle ABO_1 = \angle ACO_1 = 90^\circ.$ Since the opposite angles of quadrilateral $ABO_1C$ are supplementary, quadrilateral $ABO_1C$ is cyclic.
Let $\odot O_2$ be the circumcircle of quadrilateral $ABO_1C.$ It follows that $\odot O_2$ is also the circumcircle of $\triangle ABC,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O1, O2; A = (0,2sqrt(26)); O1 = (0,0); B = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[0]; C = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[1]; O2 = midpoint(A--O1); fill(A--B--C--cycle, yellow); dot("$A$",A,1.5*N,linewidth(4)); dot("$B$",B,1.5*W,linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$O_1$",O1,1.5*S,linewidth(4)); dot("$O_2$",O2,1.5*N,linewidth(4)); label("$3\sqrt6$",midpoint(A--B),scale(0.5)*rotate(90)*dir(midpoint(A--B)--A),red+fontsize(10)); label("$3\sqrt6$",midpoint(A--C),scale(0.5)*rotate(90)*dir(midpoint(A--C)--C),red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--B),0.5*SW,red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--C),0.5*SE,red+fontsize(10)); markscalefactor=0.05; draw(rightanglemark(A,B,O1)^^rightanglemark(A,C,O1),red); draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); [/asy] By the Inscribed Angle Theorem, we conclude that $\overline{AO_1}$ is the diameter of $\odot O_2.$ By the Pythagorean Theorem on right $\triangle ABO_1,$ we have \[AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.\] Therefore, the area of $\odot O_2$ is $\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{26}.$ | C | 26 |
b9d69c4d56382afc7c00e83bf2e4db48 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15 | Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$ | [asy] import olympiad; unitsize(50); pair A,B,C,D,E,I,F,G,O; A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,D,E); draw(A--B--C--cycle); dot(O); dot(I); dot(F); dot(G); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,D,E)); draw(I--B); draw(I--C); draw(I--A); draw(rightanglemark(A,C,I)); draw(rightanglemark(A,B,I)); draw(O--F); draw(O--G); draw(rightanglemark(A,F,O)); draw(rightanglemark(A,G,O)); label("$O$",O,W); label("$A$",A,S); label("$B$",B,N); label("$C$",C,W); label("$D$",F,S); label("$E$",G,W); label("$3\sqrt{6}$",(1.5,1.5),S); label("$3\sqrt{6}$",(-1.5,1.5),S); label("$5\sqrt{2}$",(1,3.625),N); label("$5\sqrt{2}$",(-1,3.625),N); label("$I$",I,N); label("$r$",(-0.25,1.5),E); label("$r$",(0.5,2.125),S); add(pathticks(A--F,1,0.5,0,2)); add(pathticks(F--B,1,0.5,0,2)); add(pathticks(A--G,1,0.5,0,2)); add(pathticks(G--C,1,0.5,0,2)); [/asy] Because circle $I$ is tangent to $\overline{AB}$ at $B, \angle{ABI} \cong 90^{\circ}$ . Because $O$ is the circumcenter of $\bigtriangleup ABC, \overline{OD}$ is the perpendicular bisector of $\overline{AB}$ , and $\angle{BAI} \cong \angle{DAO}$ , so therefore $\bigtriangleup ADO \sim \bigtriangleup ABI$ by AA similarity. Then we have $\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}$ . We also know that $\overline{AD} = \frac{3\sqrt{6}}{2}$ because of the perpendicular bisector, so the hypotenuse of $\bigtriangleup ADO$ is \[\sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2+\left(\frac{3\sqrt{6}}{2}\right)^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}.\] This is the radius of the circumcircle of $\bigtriangleup ABC$ , so the area of this circle is $\boxed{26}$ | C | 26 |
b9d69c4d56382afc7c00e83bf2e4db48 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_15 | Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$ , and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$ . What is the area of the circle that passes through vertices $A$ $B$ , and $C?$
$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$ | Denote by $O$ the center of the circle that is tangent to line $AB$ at $B$ and to line $AC$ at $C$
Because this circle is tangent to line $AB$ at $B$ , we have $OB \perp AB$ and $OB = 5 \sqrt{2}$
Because this circle is tangent to line $AC$ at $C$ , we have $OC \perp AC$ and $OC = 5 \sqrt{2}$
Because $AB = AC$ $OB = OC$ $AO = AO$ , we get $\triangle ABO \cong \triangle ACO$ . Hence, $\angle BAO = \angle CAO$
Let $AO$ and $BC$ meet at point $D$ .
Because $AB = AC$ $\angle BAO = \angle CAO$ $AD = AD$ , we get $\triangle ABD \cong \triangle ACD$ . Hence, $BD = CD$ and $\angle ADB = \angle ADC = 90^\circ$
Denote $\theta = \angle BAO$ . Hence, $\angle BAC = 2 \theta$
Denote by $R$ the circumradius of $\triangle ABC$ .
In $\triangle ABC$ , following from the law of sines, $2 R = \frac{BC}{\sin \angle BAC}$
Therefore, the area of the circumcircle of $\triangle ABC$ is \begin{align*} \pi R^2 & = \pi \left( \frac{BC}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{2 BD}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{BD}{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{2 \sin \theta \cos \theta} \right)^2 \\ & = \pi \left( \frac{AB }{2 \cos \theta} \right)^2 \\ & = \pi \left( \frac{AO}{2} \right)^2 \\ & = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\ & = \boxed{26} ~Steven Chen (www.professorchenedu.com) | C | 26 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\boxed{17}.$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so \begin{align*} x+3-9&=2 \cdot (x-10) \\ x-6&=2 \cdot (x-10) \\ x&=14. \end{align*} The answer is $x+3=\boxed{17}.$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\overline{GH}$ [asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); pair G = (-4*sqrt(3),-2); pair H = (4*sqrt(3),-2); label("$G, 21$", G, S); label("$H, 13$", H, S); draw(A--G, dashed); draw(C--H, dashed); dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5)); [/asy] Because of hexagon proportions, $\frac{BA}{AG} = \frac{1}{3}$ and $\frac{BC}{CH} = \frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ $B$ and $G$ lie on the same line, $\frac{12-9}{g-12} = \frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\boxed{17}$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | Denote by $h_X$ the height of any point $X$
Denote by $M$ the midpoint of $A$ and $C$ .
Hence, \[h_M = \frac{h_A + h_C}{2} = 11.\] Denote by $O$ the center of $ABCDEF$ . Because $ABCDEF$ is a regular hexagon, $O$ is the midpoint of $B$ and $E$ .
Hence, \[h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}.\] Because $ABCDEF$ is a regular hexagon, $M$ is the midpoint of $B$ and $O$ .
Hence, \[h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.\] Solving these equations, we get $h_E = \boxed{17}$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\vec{BE}=2\vec{BA}+2\vec{BC},$ so the rise from $B$ to $E$ is $2\cdot3+2\cdot1=8.$
Together, the height of the pillar at $E$ is $9+8=\boxed{17}$ meters. | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | WLOG, let the side length of the hexagon be $6$
Establish a 3D coordinate system, in which $A=(0,0,0)$ . Let the coordinates of $B$ and $C$ be $(6,0,0)$ $\left(9,-3\sqrt{3},0\right)$ , respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)$
The vector $\vec{PQ}=\langle 6,0,-3\rangle$ and $\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle$ . Computing $\vec{PQ} \times \vec{PR}$ by the matrix \[\begin{bmatrix} i & j & k \\ 6 & 0 & -3 \\ 9 & -3\sqrt{3} & -2 \end{bmatrix}\] gives the result $-9\sqrt{3}i -15j -18\sqrt{3} k$ . Therefore, a normal vector of the plane of the solar panel is $\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle$ , and the equation of the plane is $-9\sqrt{3}x-15y-18\sqrt{3}z=k$ . Substituting $(x,y,z)=(0,0,12)$ , we find that $k=-216\sqrt{3}$
Since $E=\left(0,-6\sqrt{3}\right)$ , we substitute $(x,y)=\left(0,-6\sqrt{3}\right)$ into $-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}$ , which gives $z=\boxed{17}$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | Let the pillars be $AA', BB', \ldots, FF'$ . Since solar panel $A'B'C'D'E'F'$ is a hexagon, the line $B'E'$ hits the midpoint $M$ of $A'C'$ . So, the 3D slope (change in $x$ : change in $y$ : change in $z$ ) of $BE$ is same as $BD$ . If $a$ is side of the hexagonal solar panel, \[B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a\] . So, $B'M:B'E'$ $1:4$ . Since the height of $M$ to ground $ABCDEF$ is $(10+12)/2 = 11$ , the rise (in z) from $B'$ to $M$ is 2 meaning the rise from $B'$ to $E'$ is $8$ . Thus, $EE' = 8+BB' = \boxed{17}$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | Set the midpoint of $\overline{AC}$ as $M$ [asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); pair M =(0,1.35); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); dot((0,1)); label("$M$", M); draw((-sqrt(3),1)--(sqrt(3),1),black); [/asy] We know that the height of $M$ is $11$ as it is the midpoint of $\overline{AC}$ , so the height is the average of $A$ and $C$ , which is $\frac{10 + 12}{2}= 11$ . Since $ABCDEF$ is a regular hexagon, $BE = 4\cdot BM$ . Because the increase in height is proportional to the length of the line segments, and the increase in height from $B$ to $M$ is $2$ , the increase in height from $B$ to $E$ is $2\cdot4=8.$ Adding to the height of $B$ , we get $8+9=\boxed{17}$ | D | 17 |
451978be96cf3a05246bd3e4e8c381b9 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_17 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$ | Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with $9$ or $17$ . Because both $A$ and $C$ are greater than $9$ and closer to $E$ than $B$ , we can assume that the height increases as the point gets closer to $E$ . Thus, we know the answer is greater than $9$ . The only choice that satisfies both these criteria is $\boxed{17}$ | D | 17 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | There are $4$ possibilities for the top-left section. It follows that the top-right and bottom-left sections each have $3$ possibilities, so they have $3^2=9$ combinations. We have two cases:
Together, the answer is $36+48=\boxed{84}.$ | C | 84 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | We will do casework on the type of crops in the field.
Case 1: all of a kind.
If all four sections have the same type of crop, there are simply $\underline{4}$ ways to choose crops for the sections.
Case 2: $\boldsymbol{3}$ of a kind, $\boldsymbol{1}$ of another kind.
Since the one of another kind must be adjacent to two of the other crops, when choosing the type of crops in this case, we cannot choose soybeans and potatoes, or corn and wheat. Therefore, there are $4 \cdot 3 - 2 \cdot 2 = 8$ choices for the two crops we choose for the section (notice we did not choose by $2$ , since the crop we pick first will be the unique one), and $4$ ways to choose which section the unique crop is planted on. This gives us a total of $4 \cdot 8 = \underline{32}$ ways to choose crops for the sections.
Case 3: $\boldsymbol{2}$ of a kind, $\boldsymbol{2}$ of another kind.
We cannot choose corn and wheat, or soybeans and potatoes, once again, because if we do, the two would have to be adjacent in some way, which the problem disallows. So, there are ${4 \choose 2} - 2 = 4$ ways to choose our two crops (notice that we did choose by $2$ , since there are two of both crops). There are ${4 \choose 2} = 6$ ways to choose where one of the crops go, so there are $4 \cdot 6 = \underline{24}$ ways to choose crops for the sections.
Case 4: $\boldsymbol{2}$ of a kind, $\boldsymbol{1}$ of another kind, $\boldsymbol{1}$ of another kind.
In cases 2 and 3, we excluded the possibility of choosing bad pairs for our crops (i.e. soybeans and potatoes, or corn and wheat). In this case, it is inevitable that we choose a bad pair, because we are choosing $3$ crops this time. The two sections of the same kind must contain the crop that is not part of the bad pair in the trio: for example, if we choose corn, soybeans and potatoes as our three crop types, nor soybeans and potatoes can be the type which occupies two sections in this case; corn must be the one to do so. There are $4$ ways to choose the crop that is not part of the bad pair, and then $1$ way to choose the bad pair, giving us $4 \cdot 1 = 4$ ways to choose the crops. To separate the bad pair of crops, the two of a kind must be diagonally placed. There are $2$ ways to choose where the two of a kind go, and $2$ ways to choose which of the bad pair goes where, giving us $2 \cdot 2 = 4$ ways to choose the positions for the crops. In total, there are $4 \cdot 4 = \underline{16}$ ways to choose crops for the sections.
Case 5: every single crop.
Bad pairs must be on the same diagonal, so there are $2$ ways to choose which pair gets which diagonal, and $2 \cdot 2 = 4$ ways to choose which of each pair goes where on the diagonal, giving us $2 \cdot 4 = \underline{8}$ ways to choose crops for the sections.
Adding up all our values, we get our final answer of $4+32+24+16+8 = \boxed{84}$ | C | 84 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | To lighten notation, we use C, W, S, P to denote corn, wheat, soybeans, and potatoes, respectively.
We use I, II, III, IV to denote four quadrants, respectively.
We determine an arrangement in the following steps.
Step 1: Determine the crop planted in I.
The number of ways is $4$
Step 2: Determine the crops planed in II, III, IV.
To find the number of arrangements in this step, without loss of generality, we assume that we plant C in I.
We do the following casework analysis.
Case 1: Both II and IV are planted with C.
In this case, the number of ways to plant in III is $3$
Case 2: In II and IV, only one quadrant is planted with C, and another quadrant is planted with either S or P.
In this case, we determine an arrangement in the following steps.
Step 2.1: Determine whether C is planted in II or IV.
The number of ways is $2$
Step 2.2: In II or IV not planted with C, determine whether it is planted with S or P.
The number of ways is $2$
Step 2.3: Determine the crop planted in III.
The number of ways is $2$
Following from the rule of product, the number of ways in this case is $2 \cdot 2 \cdot 2 = 8$
Case 3: II and IV are both planted with S or W.
In this case, we determine an arrangement in the following steps.
Step 2.1: Determine whether S or W is planted in II and IV.
The number of ways is $2$
Step 2.2: Determine the crop planted in III.
The number of ways is $3$
Following from the rule of product, the number of ways in this case is $2 \cdot 3 = 6$
Case 4: In II and IV, exactly one quadrant is planted with S and another one is planted with W.
Step 2.1: Determine which quadrant in II and IV is planted with S.
The number of ways is $2$
Step 2.2: Determine the crop planted in III.
The number of ways is $2$
Following from the rule of product, the number of ways in this case is $2 \cdot 2 = 4$
Putting all cases together, the total number of arrangements in Step 2 is $3 + 8 + 6 + 4 = 21$
Following from the rule of product, the total number of arrangements is $4 \cdot 21 = 84$
Therefore, the answer is $\boxed{84}$ | C | 84 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | The number of cases with at least one pair of corn and wheat adjacent is $4 \cdot 2 \cdot 4^2 - 4 \cdot 2 \cdot 4 + 0 - 2 = 94$ possible fields (You can easily see this for yourself using PIE.), and WLOG, the same goes for soybean and potatoes. Now, applying PIE on both sets (number of cases with at least one pair of corn and wheat and the number of cases with at least one pair of soybeans and potatoes) yields $2\cdot94 - 16 = 172$ . We want the number of cases without adjacent pairs of soybeans and potatoes or wheat and corn, so we subtract $256 - 172$ , yielding an answer of $\boxed{84}.$ | C | 84 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | We can create a tree representing an arbitrary box we start with and the possibilities for other boxes around the grid. We can just designate the crop we start with as $1$ , and the other crops as $2$ $3$ , and $4$ , where $1$ cannot be next to $2$ and $3$ cannot be next to $4$ in the grid. [asy] size(400); draw((0, 0)--(-10, -2)--(-13, -4)--(-14, -6)); draw((-13, -4)--(-13,-6)); draw((-13, -4)--(-12,-6)); draw((-10, -2)--(-10, -4)--(-10.5, -6)); draw((-10, -4)--(-9.5, -6)); draw((-10, -2)--(-7, -4)--(-7.5, -6)); draw((-7, -4)--(-6.5, -6)); label("$1$", (0, 0), N); label("$1$", (-10, -2), UnFill(0.5mm)); label("$1$", (-13, -4), UnFill(0.5mm)); label("$1$", (-14, -6), UnFill(0.5mm)); label("$3$", (-13, -6), UnFill(0.5mm)); label("$4$", (-12, -6), UnFill(0.5mm)); label("$3$", (-10, -4), UnFill(0.5mm)); label("$1$", (-10.5, -6), UnFill(0.5mm)); label("$3$", (-9.5, -6), UnFill(0.5mm)); label("$4$", (-7, -4), UnFill(0.5mm)); label("$1$", (-7.5, -6), UnFill(0.5mm)); label("$4$", (-6.5, -6), UnFill(0.5mm)); draw((0, 0)--(0, -2)--(-3, -4)--(-4, -6)); draw((-3, -4)--(-3, -6)); draw((-3, -4)--(-2, -6)); draw((0, -2)--(0, -4)--(-0.5, -6)); draw((0, -4)--(0.5, -6)); draw((0, -2)--(3, -4)--(2.5, -6)); draw((3, -4)--(3.5, -6)); label("$3$", (0, -2), UnFill(0.5mm)); label("$1$", (-3, -4), UnFill(0.5mm)); label("$2$", (0, -4), UnFill(0.5mm)); label("$3$", (3, -4), UnFill(0.5mm)); label("$4$", (-2, -6), UnFill(0.5mm)); label("$3$", (-3, -6), UnFill(0.5mm)); label("$1$", (-4, -6), UnFill(0.5mm)); label("$3$", (-0.5, -6), UnFill(0.5mm)); label("$4$", (0.5, -6), UnFill(0.5mm)); label("$1$", (2.5, -6), UnFill(0.5mm)); label("$3$", (3.5, -6), UnFill(0.5mm)); draw((0, 0)--(10, -2)--(13, -4)--(13.5, -6)); draw((13, -4)--(12.5,-6)); draw((10, -2)--(10, -4)--(10.5, -6)); draw((10, -4)--(9.5, -6)); draw((10, -2)--(7, -4)--(8, -6)); draw((7, -4)--(7, -6)); draw((7, -4)--(6, -6)); label("$4$", (10, -2), UnFill(0.5mm)); label("$4$", (13, -4), UnFill(0.5mm)); label("$4$", (13.5, -6), UnFill(0.5mm)); label("$1$", (12.5, -6), UnFill(0.5mm)); label("$2$", (10, -4), UnFill(0.5mm)); label("$4$", (10.5, -6), UnFill(0.5mm)); label("$3$", (9.5, -6), UnFill(0.5mm)); label("$1$", (7, -4), UnFill(0.5mm)); label("$4$", (8, -6), UnFill(0.5mm)); label("$3$", (7, -6), UnFill(0.5mm)); label("$1$", (6, -6), UnFill(0.5mm)); [/asy] As we can see, there are $21$ possibilities. We multiply this by the $4$ (the number of possibilities there are for the crop we started with), so our answer is $\boxed{84}$ | C | 84 |
e4b620b804a9b085d02a98d982141719 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_18 | A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$ | The top right box has $4$ choices and the top left box has $3$ choices. Thus, it is reasonable to assume that the answer is a multiple of $12$ . We know that the answer will not be too small or too large, so the answer is $\boxed{84}$ | C | 84 |
fab00f0a0ff8ab34b1225a2521e2a4c1 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_19 | A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$ . A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$ . The value of $s$ can be written as $a+\frac{b\pi}{c}$ , where $a,b$ , and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$
$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$ | The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^\circ$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1)^2-\pi\cdot1^2=4-\pi.$ As a result, we have \[A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.\] Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\pi,$ or \[A=4s+2\pi.\] We equate the expressions for $A,$ and then solve for $s:$ \[8s-20+\pi=4s+2\pi.\] We get $s=5+\frac{\pi}{4},$ so the answer is $5+1+4=\boxed{10}.$ | A | 10 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | A quadratic equation does not have two distinct real solutions if and only if the discriminant is nonpositive. We conclude that:
Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\] We apply casework to the value of $b:$
Together, there are $\boxed{6}$ ordered pairs $(b,c),$ namely $(1,1),(1,2),(2,1),(2,2),(3,3),$ and $(4,4).$ | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | Similar to Solution 1, use the discriminant to get $b^2\leq 4c$ and $c^2\leq 4b$ . These can be rearranged to $c\geq \frac{1}{4}b^2$ and $b\geq \frac{1}{4}c^2$ . Now, we can roughly graph these two inequalities, letting one of them be the $x$ axis and the other be $y$ .
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs: [asy] unitsize(2); Label f; f.p=fontsize(6); xaxis("$x$",0,5,Ticks(f, 1.0)); yaxis("$y$",0,5,Ticks(f, 1.0)); real f(real x) { return 0.25x^2; } real g(real x) { return 2*sqrt(x); } dot((1,1)); dot((2,1)); dot((1,2)); dot((2,2)); dot((3,3)); dot((4,4)); draw(graph(f,0,sqrt(20))); draw(graph(g,0,5)); [/asy] We are looking for lattice points (since $b$ and $c$ are positive integers), of which we can count $\boxed{6}$ | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We need to solve the following system of inequalities: \[ \left\{ \begin{array}{ll} b^2 - 4 c \leq 0 \\ c^2 - 4 b \leq 0 \end{array} \right.. \] Feasible solutions are in the region formed between two parabolas $b^2 - 4 c = 0$ and $c^2 - 4 b = 0$
Define $f \left( b \right) = \frac{b^2}{4}$ and $g \left( b \right) = 2 \sqrt{b}$ .
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For $b = 1$ , we have $f(b) = \frac{1}{4}$ and $g(b) = 2$ .
Hence, the feasible $c$ are $1, 2$
For $b = 2$ , we have $f(b) = 1$ and $g(b) = 2 \sqrt{2}$ .
Hence, the feasible $c$ are $1, 2$
For $b = 3$ , we have $f(b) = \frac{9}{4}$ and $g(b) = 2 \sqrt{3}$ .
Hence, the feasible $c$ is $3$
For $b = 4$ , we have $f(b) = 4$ and $g(b) = 4$ .
Hence, the feasible $c$ is $4$
For $b > 4$ , we have $f(b) > g(b)$ . Hence, there is no feasible $c$
Putting all cases together, the correct answer is $\boxed{6}$ | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | A quadratic equation $Ax^2+Bx+C=0$ has one real solution if and only if $\sqrt{B^2-4AC}=0.$ Similarly, it has imaginary solutions if and only if $\sqrt{B^2-4AC}<0.$ We proceed as following:
We want both $x^2+bx+c$ to be $1$ value or imaginary and $x^2+cx+b$ to be $1$ value or imaginary. $x^2+4x+4$ is one such case since $\sqrt {b^2-4ac}$ is $0.$ Also, $x^2+3x+3, x^2+2x+2, x^2+x+1$ are always imaginary for both $b$ and $c.$ We also have $x^2+x+2$ along with $x^2+2x+1$ since the latter has one solution, while the first one is imaginary. Therefore, we have $\boxed{6}$ total ordered pairs of integers. | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We see that $b^2 \leq 4c$ and $c^2 \leq 4b.$ WLOG, assume that $b \geq c.$ Then we have that $b^2 \leq 4c \leq 4b$ , so $b^2 \leq 4b$ and therefore $b \leq 4$ , also meaning that $c \leq 4.$ This means that we only need to try 16 cases. Now we can get rid of the assumption that $b \geq c$ , because we want ordered pairs. For $b = 1$ and $b = 2$ $c = 1$ and $c = 2$ work. When $b = 3$ $c$ can only be $3$ , and when $b = 4$ , only $c = 4$ works, for a total of $\boxed{6}$ ordered pairs of integers. | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | We need both $b^2\leq 4c$ and $c^2\leq 4b$
If $b=c$ then the above become $b^2\leq 4b\iff b\leq 4$ , so we have four solutions $(k,k)$ , where $k=1$ $2$ $3$ $4$
If $b<c$ then we only need $c^2\leq 4b$ since it implies $b^2< 4c$ . Now $c^2\leq 4b\leq 4(c-1) \implies (c-2)^2\leq 0 \implies c=2$ , so $b=1$ . We plug $b=1$ $c=2$ back into $c^2\leq 4b$ and it works. So there is another solution $(1,2)$
By symmetry, if $b>c$ then $(b,c)=(2,1)$
Therefore the total number of solutions is $\boxed{6}$ | B | 6 |
11bdfcdb95089785de4e92c43fdaa913 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_20 | For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?
$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$ | Since $b^{2} - 4c \le 0$ and $c^{2} - 4b \le 0$ , adding the two together yields $b^{2} + c^{2} \le 4(c+b)$ . Obviously, this is not true if either $b$ or $c$ get too large, and they are equal when $b = c = 4$ , so the greatest pair is $(4,4)$ and both numbers must be lesser for further pairs. For there to be two distinct real solutions, we can test all these pairs where $(b,c)$ are less than 4 (except for the already valid solution) on the original quadratics, and we find the working pairs are $(1,1)$ $(2,1)$ $(2,1)$ $(2,2)$ $(3,3)$ $(4,4)$ meaning there are $\boxed{6}$ pairs. | B | 6 |
bc402817dde0da17dae5abf7e23cfbbd | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$ | For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{16}.\] ~MRENTHUSIASM ~Jesshuang | E | 16 |
bc402817dde0da17dae5abf7e23cfbbd | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$ | For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$
Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{16}.$ | E | 16 |
bc402817dde0da17dae5abf7e23cfbbd | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$ | Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{16}$ | E | 16 |
bc402817dde0da17dae5abf7e23cfbbd | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_21 | Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$ | Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$
Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$
On the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$
We equate the expressions to get $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{16}$ | E | 16 |
9f17b5358359d5750ff14c8dbe3005ae | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$ , where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,18,20,28,44,45,50$ for $pq^2$ , and just $32$ for $p^5$ . Here $12$ especially catches our eyes, as this means if one of $f_i(n)=12$ , each of $f_{i+1}(n), f_{i+2}(n), ...$ will all be $12$ . This is because $f_{i+1}(n)=f(f_i(n))$ (as given in the problem statement), so were $f_i(n)=12$ , plugging this in we get $f_{i+1}(n)=f(12)=12$ , and thus the pattern repeats. Hence, as long as for a $i$ , such that $i\leq 50$ and $f_{i}(n)=12$ $f_{50}(n)=12$ must be true, which also immediately makes all our previously listed numbers, where $f(x)=12$ , possible values of $n$
We also know that if $f(x)$ were to be any of these numbers, $x$ would satisfy $f_{50}(n)$ as well. Looking through each of the possibilities aside from $12$ , we see that $f(x)$ could only possibly be equal to $20$ and $18$ , and still have $x$ less than or equal to $50$ . This would mean $x$ must have $10$ , or $9$ divisors, and testing out, we see that $x$ will then be of the form $p^4q$ , or $p^2q^2$ . The only two values less than or equal to $50$ would be $48$ and $36$ respectively. From here there are no more possible values, so tallying our possibilities we count $\boxed{10}$ values (Namely $12,18,20,28,32,36,44,45,48,50$ ). | D | 10 |
9f17b5358359d5750ff14c8dbe3005ae | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | First, take note that the maximum possible value of $f_1(n)$ for $1 \le n \le k$ increases as $k$ increases (it is a step function), i.e. it is increasing. Likewise, as $k$ decreases, the maximum possible value of $f_1(n)$ decreases as well. Also, let $f_1(n) = 2d(n)$ where $d(n)$ is the number of divisors of n.
Since $n \le 50$ $f_1(n) <= 20$ . This maximum occurs when $d(n) = 10 \implies n = 2^4 \cdot 3 = 48$ . Next, since $f_1(n) <=20$ $f_1(f_1(n)) \le 12 \implies f_2(n) \le 12$ . This maximum occurs when $d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12$ . Since $f_2(n) \le 12$ $f_1(f_2(n)) \le 12 \implies f_3(n) \le 12$ , once again. This maximum again occurs when $d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12$ . Now, suppose for the sake of contradiction that $f_2(n) < 12$ . Then, $f_3(n) < 12$ (since $f_2(n) = 12$ was the only number that would maximize $f_3(n))$ for $f_2(n) \le 12$ ). As a result, since $f_1(n)$ is increasing, and because $12$ is where $f_1$ steps down from a maximum of $6 \cdot 2 = 12$ , we must have that $f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12$ . We continue applying $f_1$ on both sides (which is possible since $f_1$ is increasing) until we reach $f_50$ , giving us that $f_50(n) < 12$ . However, $f_50(n) = 12$ , which is a contradiction. Thus, $f_2(n) = 12$
Now, let us finally solve for the solutions. $f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6$ $d(f_1(n)) = 6 \implies f_1(n) = p^2 \cdot q$ where $p$ and $q$ are primes. Since $f_1(n) \le 20$ $f_1(n)$ can only be $12$ $18$ , or $20$ . If $f_1(n) = 12$ , then $d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}$ , resulting in 8 solutions. If $f_1(n) = 18$ , then $d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36$ , giving us one more solution. Finally, $f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48$ . Thus, in total, we have $\boxed{10}$ solutions. | D | 10 |
9f17b5358359d5750ff14c8dbe3005ae | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_23 | For each positive integer $n$ , let $f_1(n)$ be twice the number of positive integer divisors of $n$ , and for $j \ge 2$ , let $f_j(n) = f_1(f_{j-1}(n))$ . For how many values of $n \le 50$ is $f_{50}(n) = 12?$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | $\textbf{Observation 1}$ $f_1 \left( 12 \right) = 12$
Hence, if $n$ has the property that $f_j \left( n \right) = 12$ for some $j$ , then $f_k \left( n \right) = 12$ for all $k > j$
$\textbf{Observation 2}$ $f_1 \left( 8 \right) = 8$
Hence, if $n$ has the property that $f_j \left( n \right) = 8$ for some $j$ , then $f_k \left( n \right) = 8$ for all $k > j$
$\textbf{Case 1}$ $n = 1$
We have $f_1 \left( n \right) = 2$ $f_2 \left( n \right) = f_1 \left( 2 \right) = 4$ $f_3 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_4 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 2}$ $n$ is prime.
We have $f_1 \left( n \right) = 4$ $f_2 \left( n \right) = f_1 \left( 4 \right) = 6$ $f_3 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 3}$ : The prime factorization of $n$ takes the form $p_1^2$
We have $f_1 \left( n \right) = 6$ $f_2 \left( n \right) = f_1 \left( 6 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 4}$ : The prime factorization of $n$ takes the form $p_1^3$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 5}$ : The prime factorization of $n$ takes the form $p_1^4$
We have $f_1 \left( n \right) = 10$ $f_2 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 6}$ : The prime factorization of $n$ takes the form $p_1^5$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case the only $n$ is $2^5 = 32$
$\textbf{Case 7}$ : The prime factorization of $n$ takes the form $p_1 p_2$
We have $f_1 \left( n \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 8}$ : The prime factorization of $n$ takes the form $p_1 p_2^2$
We have $f_1 \left( n \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, all $n$ are $12, 18, 20, 28, 44, 45,$ and $50$
$\textbf{Case 9}$ : The prime factorization of $n$ takes the form $p_1 p_2^3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_1 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
$\textbf{Case 10}$ : The prime factorization of $n$ takes the form $p_1 p_2^4$
We have $f_1 \left( n \right) = 20$ $f_2 \left( n \right) = f_1 \left( 20 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $48$
$\textbf{Case 11}$ : The prime factorization of $n$ takes the form $p_1^2 p_2^2$
We have $f_1 \left( n \right) = 18$ $f_2 \left( n \right) = f_1 \left( 18 \right) = 12$ . Hence, Observation 1 implies $f_{50} \left( n \right) = 12$
In this case, the only $n$ is $36$
$\textbf{Case 12}$ : The prime factorization of $n$ takes the form $p_1 p_2 p_3$
We have $f_1 \left( n \right) = 16$ $f_2 \left( n \right) = f_1 \left( 16 \right) = 10$ $f_3 \left( n \right) = f_2 \left( 10 \right) = 8$ . Hence, Observation 2 implies $f_{50} \left( n \right) = 8$
Putting all cases together, the number of feasible $n \leq 50$ is $\boxed{10}$ | D | 10 |
32fc75eb10d60c36d4926fda6265c517 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24 | Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$ | For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below. [asy] /* Made by MRENTHUSIASM */ size(150); pair A, B, C, D, E, F, G, H; A = (0,1); B = (1,1); C = (1,0); D = (0,0); E = (0.3,1.3); F = (1.3,1.3); G = (1.3,0.3); H = (0.3,0.3); draw(A--B--C--D--cycle^^A--E^^B--F^^C--G^^E--F--G); draw(H--D^^H--E^^H--G,dashed); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*(1,0),linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*NW,linewidth(4)); dot("$F$",F,1.5*NE,linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); [/asy] Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$
We apply casework to face $ABCD.$ Recall that there are $\binom42=6$ ways to label its edges:
Therefore, we have $4+16=\boxed{20}$ such labelings in total. | E | 20 |
32fc75eb10d60c36d4926fda6265c517 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24 | Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$ | Since we want the sum of the edges of each face to be $2$ , we need there to be two $1$ s and two $0$ s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$ s adjacent to $1$ s and $0$ s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$ . In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$ . Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$ . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.
Secondly, $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$ . Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$ . Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$ . This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$ . Therefore, we have another $8$ cases.
Summing these up, we have $8+4+8 = 20$ . Therefore, our answer is $\boxed{20}$ | E | 20 |
32fc75eb10d60c36d4926fda6265c517 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24 | Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$ | We see that if the $3$ edges connecting to $A$ has two $0$ 's, and one $1$ , it would have the same solutions as if it had two $1$ 's, and one $0$ . The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.
Similarly, if the $3$ edges connecting to $A$ has three $0$ 's, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.
Putting all the cases together, we have $4+6+4+6=\boxed{20}$ solutions. | E | 20 |
32fc75eb10d60c36d4926fda6265c517 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10A_Problems/Problem_24 | Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$ | The problem states the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ . That is, the sum of the labels on the $4$ edges of a face is equal to $2$ . The labels can only be $0$ or $1$ , meaning $2$ edges are labeled $1$ , the other $2$ are labeled $0$
This problem can be approached by Graph Coloring of Graph Theory . Note that each face of the cube connects to $4$ other faces, each with a shared edge. We use the following graph to represent the problem. Each vertex represents a face, each edge represent the cube's edge. Each vertex has $4$ edges connecting to $4$ other vertices. The edges can be colored red or blue, with red as label $0$ , and blue as label $1$ . Each vertex must have $2$ red edges and $2$ blue edges.
$\textbf{Case 1}$ $2$ adjacent red edges from vertex A. There are $4$ ways to choose $2$ red edges adjacent to each other and connect to $2$ vertices with an edge between them as shown below.
$\textbf{Case 1.1}$ $2$ adjacent red edges from vertex $A$ form a closed loop with a third red edge. There is only $1$ case as shown below.
$\textbf{Case 1.2}$ $2$ adjacent red edges from vertex $A$ does not form a closed loop with a third red edge. There are $3$ cases as shown below.
In case $1$ , there are total $4 \cdot (1 + 3) = 16$ ways.
$\textbf{Case 2}$ $2$ red edges from vertex $A$ with $1$ blue edge in between. There are $2$ ways to choose $2$ red edges with $1$ blue edge in between.
There are only $2$ cases as shown below.
In case $2$ , there are total $2 \cdot 2 = 4$ ways.
From both case $1$ and case $2$ , there are $16 + 4 = \boxed{20}$ ways in total. | E | 20 |
38b6550b85e0ad6e3d2e5d4d538f4d1a | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$ , we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{11110}.\] Note that it is equally valid to manually add all four numbers together to get the answer. | E | 11110 |
38b6550b85e0ad6e3d2e5d4d538f4d1a | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_1 | What is the value of $1234 + 2341 + 3412 + 4123$
$\textbf{(A)}\: 10000\qquad\textbf{(B)} \: 10010\qquad\textbf{(C)} \: 10110\qquad\textbf{(D)} \: 11000\qquad\textbf{(E)} \: 11110$ | We have \[1234 + 2341 + 3412 + 4123 = 1111 \left( 1 + 2 + 3 + 4 \right) = \boxed{11110}.\] ~Steven Chen (www.professorchenedu.com) | E | 11110 |
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