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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 the units digit must be $0$ , since $4+3+2+1$ is $0$ . But every digit from there, will be a $1$ since we have that each time afterwards, we must carry the $1$ from the previous sum. The answer choice that satisfies these conditions is $\boxed{11110}$ | 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 can simply add the numbers. $1234 + 2341 + 3412 + 4123 = 11110 \implies \boxed{11110}$ | E | 11110 |
956b2a5e0eee881d65bbc50986453fce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$ | The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$
Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{6}.\] ~MRENTHUSIASM ~Wilhelm Z | B | 6 |
956b2a5e0eee881d65bbc50986453fce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$ | To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \[\frac12\cdot4\cdot5-\frac12\cdot4\cdot2=10-4=\boxed{6}.\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) | B | 6 |
956b2a5e0eee881d65bbc50986453fce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$ | The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem , the area is \[\frac12\cdot|(1\cdot2+3\cdot0+5\cdot5+3\cdot0)-(0\cdot3+2\cdot5+0\cdot3+5\cdot1)|=\frac12\cdot12=\boxed{6}.\] ~Taco12 ~I-AM-DA-KING | B | 6 |
956b2a5e0eee881d65bbc50986453fce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_2 | What is the area of the shaded figure shown below? [asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
$\textbf{(A)}\: 4\qquad\textbf{(B)} \: 6\qquad\textbf{(C)} \: 8\qquad\textbf{(D)} \: 10\qquad\textbf{(E)} \: 12$ | We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \[4+\frac{6}{2}-1 = 4+3-1 = \boxed{6}.\] ~danprathab | B | 6 |
c01a12e455ee4ff18154037331e010b0 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_3 | The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$ | We write the given expression as a single fraction: \[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}\] by cross multiplication. Then by factoring the numerator, we get \[\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.\] The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\boxed{4041}$ | E | 4041 |
c01a12e455ee4ff18154037331e010b0 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_3 | The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$ | Denote $a = 2020$ . Hence, \begin{align*} \frac{2021}{2020} - \frac{2020}{2021} & = \frac{a + 1}{a} - \frac{a}{a + 1} \\ & = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\ & = \frac{2 a + 1}{a \left( a + 1 \right)} . \end{align*}
We observe that ${\rm gcd} \left( 2a + 1 , a \right) = 1$ and ${\rm gcd} \left( 2a + 1 , a + 1 \right) = 1$
Hence, ${\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1$
Therefore, $p = 2 a + 1 = 4041$
Therefore, the answer is $\boxed{4041}$ | E | 4041 |
f575fbcfd45acd8a023c680188598826 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_4 | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$ | At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$
At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases:
Together, the product of all possible values of $N$ is $10\cdot6=\boxed{60}.$ | C | 60 |
f575fbcfd45acd8a023c680188598826 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_4 | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$
$\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120$ | At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.
At $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.
It follows that \[|N-8|=2.\] We continue with the casework in Solution 1 to get the answer $\boxed{60}.$ | C | 60 |
b54b8698c94ee1801a97bf098caf3141 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6 | The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$ | Let this positive integer be written as $p_1^{e_1}\cdot p_2^{e_2}$ . The number of factors of this number is therefore $(e_1+1) \cdot (e_2+1)$ , and this must equal 2021. The prime factorization of 2021 is $43 \cdot 47$ , so $e_1+1 = 43 \implies e_1=42$ and $e_2+1=47\implies e_2=46$ . To minimize this integer, we set $p_1 = 3$ and $p_2 = 2$ . Then this integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$ .
Now $m=16$ and $k=42$ so $m+k = 16 + 42 = \boxed{58}$ | B | 58 |
b54b8698c94ee1801a97bf098caf3141 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6 | The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$ | Recall that $6^k$ can be written as $2^k \cdot 3^k$ . Since we want the integer to have $2021$ divisors, we must have it in the form $p_1^{42} \cdot p_2^{46}$ , where $p_1$ and $p_2$ are prime numbers. Therefore, we want $p_1$ to be $3$ and $p_2$ to be $2$ . To make up the remaining $2^4$ , we multiply $2^{42} \cdot 3^{42}$ by $m$ , which is $2^4$ which is $16$ . Therefore, we have $42 + 16 = \boxed{58}$ | B | 58 |
b54b8698c94ee1801a97bf098caf3141 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_6 | The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \cdot 6^k$ , where $m$ and $k$ are integers and $6$ is not a divisor of $m$ . What is $m+k?$
$(\textbf{A})\: 47\qquad(\textbf{B}) \: 58\qquad(\textbf{C}) \: 59\qquad(\textbf{D}) \: 88\qquad(\textbf{E}) \: 90$ | If a number has prime factorization $p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ , then the number of distinct positive divisors of this number is $\left( k_1 + 1 \right) \left( k_2 + 1 \right) \cdots \left( k_m + 1 \right)$
We have $2021 = 43 \cdot 47$ .
Hence, if a number $N$ has 2021 distinct positive divisors, then $N$ takes one of the following forms: $p_1^{2020}$ $p_1^{42} p_2^{46}$
Therefore, the smallest $N$ is $3^{42} 2^{46} = 2^4 \cdot 6^{42} = 16 \cdot 6^{42}$
Therefore, the answer is $\boxed{58}$ | B | 58 |
10d915822e18482f89f720175a2110ce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7 | Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$
For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$
For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$
Together, there are $\boxed{11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\] ~KingRavi ~samrocksnature ~Wilhelm Z ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com) | C | 11 |
10d915822e18482f89f720175a2110ce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7 | Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Let $a=15-b,$ so the special fraction is \[\frac ab = \frac{15-b}{b} = \frac{15}{b}-1.\] We can ignore the $-1$ part and only focus on $\frac{15}{b}.$
The integers are $\frac{15}{1},\frac{15}{3},\frac{15}{5},$ which are $15,5,3,$ respectively. We get $30,20,18,10,8,6$ from this group of numbers.
The halves are $\frac{15}{2},\frac{15}{6},\frac{15}{10},$ which are $7\frac12,2\frac12,1\frac12,$ respectively. We get $15,10,9,5,4,3$ from this group of numbers.
The quarters are $\frac{15}{4},\frac{15}{12},$ which are $3\frac34,1\frac14,$ respectively. We get $5$ from this group of numbers.
Note that $10$ and $5$ each appear twice. Therefore, the answer is $\boxed{11}.$ | C | 11 |
10d915822e18482f89f720175a2110ce | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_7 | Call a fraction $\frac{a}{b}$ , not necessarily in the simplest form, special if $a$ and $b$ are positive integers whose sum is $15$ . How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | We split this up into two cases:
Case 1: integer + integer
The whole numbers we have are $\frac{10}{5}$ (or $2$ ), $\frac{12}{3}$ (or $4$ ), and $\frac{14}{1}$ (or $14$ ). There are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Case 2: fraction + fraction
The fractions we have are $\frac{5}{10}$ (or $\frac{1}{2}$ ), $\frac{9}{6}$ (or $\frac{3}{2}$ ), and $\frac{13}{2}$ . Similarly, there are $\dbinom{3}{2}=3$ ways to choose different-numbered pairs and $3$ ways to choose the same-numbered pairs. So, $3+3=6$
Thus, $6+6=12$
So now you would just go ahead and innocently choose $\textbf{(D) }12$ , right? No! We overcounted $8$ , as $\frac{13}{2}+\frac96=\frac{12}{3}+\frac{12}{3}=8$ . Therefore, the correct answer is actually $12-1=\boxed{11}$ | C | 11 |
c847888bcce9a3dc75ce283fc6601ff3 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_8 | The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$ | We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}
Since $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$ | C | 10 |
6185691a28eb98a434fdc2eef018f26e | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10 | Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$ | Let Alice have the number A, Bob B.
When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$ . Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$ $256$ , and $289$ . So, $A$ could equal $25$ $56$ , or $89$ , so $A+B$ is $27$ $58$ , or $91$ , of only $\boxed{27}$ is an answer choice. | null | 27 |
6185691a28eb98a434fdc2eef018f26e | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10 | Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$ | Denote by $A$ and $B$ the numbers drawn by Alice and Bob, respectively.
Alice's sentence “I can't tell who has the larger number.” implies $A \in \left\{ 2 , \cdots , 39 \right\}$
Bob's sentence “I know who has the larger number.” implies $B \in \left\{ 1 , 2 , 39, 40 \right\}$
Their subsequent conversation that $B$ is prime implies $B = 2$
Then, Alice's next sentence “In that case, if I multiply your number by 100 and add my number, the result is a perfect square.” implies $200 + A$ is a perfect square.
Hence, $A = 25$
Therefore, the answer is $\boxed{27}$ | A | 27 |
6185691a28eb98a434fdc2eef018f26e | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_10 | Forty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$ | We see that $225$ is one such square that works. Bob gets $2$ and Alice gets $25$ which is valid. Thus, $2 + 25 = 27.$ So $\boxed{27}$ is our answer. | A | 27 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | Note that $\triangle APB \cong \triangle BQC.$ Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $RQ = 4$ and $RC = 9.$ We eliminate the possibility of $x=4$ because $RC > QR.$ Thus, the side length of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the square is $(\sqrt{117})^2 = 117,$ so the answer is $\boxed{117}.$ | D | 117 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | As above, note that $\bigtriangleup BPA \cong \bigtriangleup CQB$ , which means that $QC = 13$ . In addition, note that $BR$ is the altitude of a right triangle to its hypotenuse, so $\bigtriangleup BQR \sim \bigtriangleup CBR \sim \bigtriangleup CQB$ . Let the side length of the square be $x$ ; using similarity side ratios of $\bigtriangleup BQR$ to $\bigtriangleup CQB$ , we get \[\frac{6}{x} = \frac{QB}{13} \implies QB \cdot x = 78\] Note that $QB^2 + x^2 = 13^2 = 169$ by the Pythagorean theorem, so we can use the expansion $(a+b)^2 = a^2+2ab+b^2$ to produce two equations and two variables;
\[(QB + x)^2 = QB^2 + 2QB\cdot x + x^2 \implies (QB+x)^2 = 169 + 2 \cdot 78 \implies QB+x = \sqrt{13(13)+13(12)} = \sqrt{13 \cdot 25} = 5\sqrt{13}\] \[(QB-x)^2 = QB^2 - 2QB\cdot x + x^2 \implies (QB - x)^2 = 169 - 2\cdot 78 \implies \pm(QB-x) = \sqrt{13(13) - 13(12)}\]
Since $QB-x$ is negative, it doesn't make sense in the context of this problem, so we go with \[x-QB = \sqrt{13(13) - 13(12)} = \sqrt{13 \cdot 1} = \sqrt{13}\]
We want $x^2$ , so we want to find $x$ . Adding the first equation to the second, we get \[2x = 6\sqrt{13} \implies x = 3\sqrt{13}\]
Then $x^2$ $(3\sqrt{13}^2) = 9 \cdot 13 = 117 = \boxed{117}$ | D | 117 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | We have that $\triangle CRB \sim \triangle BAP.$ Thus, $\frac{\overline{CB}}{\overline{CR}} = \frac{\overline{PB}}{\overline{AB}}$ . Now, let the side length of the square be $s.$ Then, by the Pythagorean theorem, $CR = \sqrt{x^2-36}.$ Plugging all of this information in, we get \[\frac{s}{\sqrt{s^2-36}} = \frac{13}{s}.\] Simplifying gives \[s^2=13\sqrt{s^2-36},\] Squaring both sides gives \[s^4 = 169s^2- 169\cdot 36 \implies s^4-169s^2 + 169\cdot 36 = 0.\] We now set $s^2=t,$ and get the equation $t^2-169t + 169\cdot 36 = 0.$ From here, notice we want to solve for $t$ , as it is precisely $s^2,$ or the area of the square. So we use the Quadratic formula , and though it may seem bashy, we hope for a nice cancellation of terms. \[t = \frac{169\pm\sqrt{169^2-4\cdot 36 \cdot 169}}{2}.\] It seems scary, but factoring $169$ from the square root gives us \[t = \frac{169\pm \sqrt{169 \cdot (169-144)}}{2} = \frac{169 \pm \sqrt{169 \cdot 25}}{2} = \frac{169 \pm 13\cdot 5}{2} = \frac{169\pm 65}{2},\] giving us the solutions \[t=52, 117.\] We instantly see that $t=52$ is way too small to be an area of this square ( $52$ isn't even an answer choice, so you can skip this step if out of time) because then the side length would be $2\sqrt{13}$ and then, even the largest line you can draw inside the square (the diagonal) is $2\sqrt{26},$ which is less than $13$ (line $PB$ ) And thus, $t$ must be $117$ , and our answer is $\boxed{117}.$ $\blacksquare$ | D | 117 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | Denote $a = RC$ . Now tilt your head to the right and view $R, \overrightarrow{RB}$ and $\overrightarrow{RC}$ as the origin, $x$ -axis and $y$ -axis, respectively. In particular, we have points $B(6,0), C(0,a), P(-7,0)$ . Note that side length of the square $ABCD$ is $BC = \sqrt{a^2 + 36}$ . Also equation of line $BC$ is \[\underbrace{\frac{x}{6} + \frac{y}{a} = 1}_{\text{intercepts form}} \quad \implies \quad ax + 6y - 6a = 0.\] Because the distance from $P(-7,0)$ to line $\color[rgb]{0,0.4,0.65}BC: ax + 6y - 6a = 0$ is also the side length $\sqrt{a^2 + 36}$ , we can apply the point-line distance formula to get \[\frac{|a\cdot(-7) + 6 \cdot 0 - 6a|}{\sqrt{a^2 + 36}} = {\sqrt{a^2 + 36}}\] which reduces to $|13a| = a^2 + 36$ . Since $a$ is positive, the last equations factors as $a^2 - 13a + 36 = (a-4)(a-9) = 0$ . Now judging from the figure, we learn that $a > RB = 6$ . So $a = 9$ .
Therefore, the area of the square $ABCD$ is $BC^2 = RC^2 + RB^2 = a^2 + 6^2 = \boxed{117}$ | D | 117 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | Denote $\angle PBA = \alpha$ .
Because $\angle QRB = \angle QBC = 90^\circ$ $\angle BCQ = \alpha$
Hence, $AB = BP \cos \angle PBA = 13 \cos \alpha$ $BC = \frac{BR}{\sin \angle BCQ} = \frac{6}{\sin \alpha}$
Because $ABCD$ is a square, $AB = BC$ .
Hence, $13 \cos \alpha = \frac{6}{\sin \alpha}$
Therefore, \begin{align*} \sin 2 \alpha & = 2 \sin \alpha \cos \alpha \\ & = \frac{12}{13} . \end{align*}
Thus, $\cos 2 \alpha = \pm \frac{5}{13}$
$\textbf{Case 1}$ $\cos 2 \alpha = \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{3}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 3 \sqrt{13}$
Therefore, ${\rm Area} \ ABCD = AB^2 = 117$
$\textbf{Case 2}$ $\cos 2 \alpha = - \frac{5}{13}$
Thus, $\cos \alpha = \sqrt{\frac{1 + \cos 2 \alpha}{2}} = \frac{2}{\sqrt{13}}$
Hence, $AB = 13 \cos \alpha = 2 \sqrt{13}$
However, we observe $BQ = \frac{BR}{\cos \alpha} = 3 \sqrt{13} > AB$ .
Therefore, in this case, point $Q$ is not on the segment $AB$
Therefore, this case is infeasible.
Putting all cases together, the answer is $\boxed{117}$ | D | 117 |
a6e77a212e17efcf72b5d6f7fcfce679 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_15 | In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
$\textbf{(A) }85\qquad\textbf{(B) }93\qquad\textbf{(C) }100\qquad\textbf{(D) }117\qquad\textbf{(E) }125$ | Note that if we connect points $P$ and $C$ , we get a triangle with height $RC$ and length $13$ . This triangle has an area of $\frac {1}{2}$ the square. We can now use answer choices to our advantage!
Answer choice A: If $BC$ was $\sqrt {85}$ $RC$ would be $7$ . The triangle would therefore have an area of $\frac {91}{2}$ which is not half of the area of the square. Therefore, A is wrong.
Answer choice B: If $BC$ was $\sqrt {93}$ $RC$ would be $\sqrt {57}$ . This is obviously wrong.
Answer choice C: If $BC$ was $10$ , we would have that $RC$ is $8$ . The area of the triangle would be $52$ , which is not half the area of the square. Therefore, C is wrong.
Answer choice D: If $BC$ was $\sqrt {117}$ , that would mean that $RC$ is $9$ . The area of the triangle would therefore be $\frac {117}{2}$ which IS half the area of the square. Therefore, our answer is $\boxed{117}$ | D | 117 |
4b41e175035cc7878f83d56c81917083 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_16 | Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$ | After the first swap, we do casework on the next swap.
Case 1: Silva swaps the two balls that were just swapped
There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.
Case 2: Silva swaps one ball that has just been swapped with one that hasn't been swapped
There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions.
Case 3: Silva swaps two balls that have not been swapped
There are two ways for Silva to do this, and it leaves 1 ball occupying their original position.
Our answer is the average of all 5 possible swaps, so we get \[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{2.2}.\] | D | 2.2 |
4b41e175035cc7878f83d56c81917083 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_16 | Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$ | The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: \[\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{25}.\] Multiply by 5 for 5 balls to get $\boxed{2.2}.$ | D | 2.2 |
05b365c0ddba410598dc62aac61007fc | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18 | Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$ | First note the useful fact that if $R$ is the circumradius of a dodecagon ( $12$ -gon) the area of the figure is $3R^2.$ If we connect the vertices of the $3$ squares we get a dodecagon. The radius of circumcircle of the dodecagon is simply half the diagonal of the square, which is $3\sqrt{2}.$ Thus the area of the dodecagon is $3 \cdot (3\sqrt{2})^2 = 3 \cdot 18 = 54.$ But, the problem asked for the area of the combined figure which was made from the rotated squares. This area is the area of the dodecagon, which was found, subtracting the $12$ isosceles triangles, which are formed when connecting the vertices of the squares to created the dodecagon. To find this area, we need to know the base of the isosceles triangle, call this $x.$ Then, we can use the Law of Cosines on the triangle that is formed from the two vertices of the square and the center of the square. After computing, we get that $x = 3\sqrt{3} -3.$ Realize that the $12$ isosceles are congruent with an angle measure of $120^{\circ},$ this means that we can create $4$ congruent equilateral triangles with side length of $3\sqrt3 - 3.$ The area of the equilateral triangle is $\frac{\sqrt{3}}{4} \cdot (3\sqrt{3} -3)^2 = \frac{\sqrt{3}}{4} \cdot (36 - 18\sqrt{3}) = \frac{36\sqrt{3} - 54}{4}.$ Thus, the area of all the twelve small equilateral traingles are $4 \cdot \frac{36\sqrt{3} - 54}{4} = 36\sqrt{3} - 54$ . Thus, the requested area is $54 - (36\sqrt{3} - 54) = 108 - 36\sqrt{3}.$ Thus, $a+b+c = 108 + 36 + 3 = 147,$ so the answer is $\boxed{147}.$ | E | 147 |
05b365c0ddba410598dc62aac61007fc | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18 | Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$ | To make things simpler, let's take only the original sheet and the 30 degree rotated sheet. Then the diagram is this;
[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ [/asy]
The area of this diagram is the original square plus the area of the four triangles that 'jut' out of the square. Because the square is rotated $30^{\circ}$ , each triangle is a 30-60-90 triangle. Similarly, the triangles that are bounded on the inside of the original square outside of the rotated square are also congruent 30-60-90 triangles. Noting this, we can do some labelling:
[asy] size(10cm,0); path p = box((0,0), (1,1)); draw(p, black + linewidth(2.0pt)); draw(rotate(30,(1/2,1/2))*p,black + linewidth(2.0pt)); /*Rotate 60 degrees*/ label("$y$",(0.1,-0.05)); label("$x$",(0.4,0.05)); label("$y\sqrt{3}$",(0.8,-0.05)); label("$\frac{x}{2}$",(0.22,-0.12)); label("$\frac{x\sqrt{3}}{2}$",(0.5,-0.15)); label("$2y$",(0.8,0.15)); label("$y$",(1.05,0.1)); label("$\frac{x}{2}$",(1.12,0.22)); [/asy]
Since the side lengths of the squares must be the same, and they are both 6, we have a system of equations; \[y+x+y\sqrt{3} = 6\] \[\frac{x\sqrt{3}}{2} + 2y + \frac{x}{2} = 6\]
We solve this to get $x = 6-2\sqrt{3}$ and $y = 3-\sqrt{3}$
The area of each triangle is $\frac{x}{2} \cdot \frac{x\sqrt{3}}{2} \cdot \frac{1}{2} = 6\sqrt{3} - 9$ by plugging in $x$
The rotated 60 degree square is the same thing as rotating it 30 degrees counterclockwise, so it's triangles that jut out of the square will be congruent to the triangles we have found, and therefore they will have the same area.
Unfortunately, when drawing all three squares, we see the two triangles overlap; take the very top for example.
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); [/asy]
The area of this shape is twice the area of each of the triangles that we have already found minus the area of the small triangle that is overlapped by the two by PIE. Now we only need to find the area of $\bigtriangleup BCF$
$\angle GBD \cong \angle ECA \cong 30^{\circ}$ and by symmetry $\bigtriangleup BCF$ is isosceles, so it is a 30-30-120 triangle. If we draw a perpendicular, we split it into two 30-60-90 triangles;
[asy] import olympiad; size(10cm); pair A,B,C,D,E,F,G; A = (0.211,0); B=(0.3657,0); C = (0.63397,0); D = (0.789,0); E = (0.31666,0.1823); F=(0.5,0.077); G=(0.68334,0.1823); draw((0,0)--(1,0),black+linewidth(2pt)); draw(A--E--C--cycle); draw(B--D--G--cycle); draw(F--(0.5,0)); label("$A$",A,S); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$E$",E,N); label("$F$",F,N); label("$G$",G,N); label("$H$",(0.5,0),S); [/asy]
By symmetry, the distance from A to the edge of the square is equal to the distance from D to the edge of the square is equal to $y$ . AC = BD = $x$ , and the side length of the square is 6, so we use PIE to obtain \[x+x-BC = 6-y-y \implies BC = 12 - 6\sqrt{3}\] To find the height of $\bigtriangleup BFC$ , we see that $HC = \frac{BC}{2} = 6-3\sqrt{3}$ . Then by 30-60-90 triangles, $HF = \frac{HC}{\sqrt{3}} = 2\sqrt{3} - 3$ . Finally, the area of $\bigtriangleup BFC = \frac{BC \cdot HF}{2} = 21\sqrt{3}-36$
Putting it all together, the area of the entire diagram is the area of the square plus four of these triangle-triangle intersections. The area of these intersections by PIE is $2 \cdot [ACE] - [BFC] = 12\sqrt{3}-18-(21\sqrt{3}-36) = 18-9\sqrt{3}$ .
Therefore the total area is $36 + 4 \cdot(18-9\sqrt{3}) = 36 + 72 - 36 \sqrt{3} = 108 - 36\sqrt{3}$
Thus $a + b + c = 108+36+3 = 147 = \boxed{147}$ | E | 147 |
05b365c0ddba410598dc62aac61007fc | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18 | Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$ | As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} & \ [ABCDEFGHIJKL] - 12[MAB] \\ & = 12 [OAB] - 12 [MAB] \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36\sqrt{3} \end{align*}
Therefore, the answer is $\boxed{147}$ | E | 147 |
05b365c0ddba410598dc62aac61007fc | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_18 | Three identical square sheets of paper each with side length $6{ }$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c?$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$ | Let $O$ be the center of the polygon, $A$ be the bottom right corner of the first square, $C$ be the next vertex to the left of $A$ , and $M$ be the midpoint between $A$ and $B$ , where $B$ is the bottom left corner of the first square. Note that because there are three $90^{\circ}$ squares separated by $\frac{90^{\circ}}{3} = 30^{\circ}$ , each side of the 24-sided polygon is equal in length, meaning to calculate the area of the whole polygon, we find the area of $\bigtriangleup OAC$ and multiply by 24.
To find $[\bigtriangleup OAC]$ , we already know the height $\overline{OM}$ is the sidelength of the square over $2$ , or $\frac{6}{2}=3$ , so we just need the length of the base $\overline{AC}$ . Notice that $\bigtriangleup OCM$ is a $30-60-90$ triangle since $\angle COM = \frac{360^{\circ}}{12} = 30^{\circ}$ , so $\overline{CM} = \frac{\overline{OM}}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ . Then $\overline{AC} = \overline{AM} - \overline{CM} = \frac{6}{2} - \sqrt{3} = 3 - \sqrt{3}$ , so \begin{align*} & [\bigtriangleup AOC] = \frac{1}{2} \cdot \overline{OM} \cdot \overline{AC} \\ & = \frac{1}{2} (3)(3 - \sqrt{3}) \\ & = \frac{9 - 3\sqrt{3}}{2} \end{align*}
Then the whole area of the polygon is $\frac{9 - 3\sqrt{3}}{2} \cdot 24 = 108 - 36\sqrt{3}$ . The desired solution is then $108 + 36 + 3 = 147$ , so the answer is $\boxed{147}$ | E | 147 |
31b86a0a62b7dc2970dcd1bb4f084550 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19 | Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$ | We can rewrite $N$ as $\frac{7}{9}\cdot 9999\ldots999 = \frac{7}{9}\cdot(10^{313}-1)$ .
When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it.
When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of $f(r)$ will be equal to the leading digit of $\sqrt[r]{\frac{7}{9} \cdot 10^{313(\mod r)}}$
Then $f(2)$ is the first digit of $\sqrt{\frac{7}{9}\cdot(10)} = \sqrt{\frac{70}{9}} = \sqrt{7.\ldots} \approx 2$
$f(3) - \sqrt[3]{\frac{7}{9} \cdot 10} = \sqrt[3]{\frac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1$
$f(4) - \sqrt[4]{\frac{7}{9} \cdot 10} = \sqrt[4]{\frac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1$
$f(5) - \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3$
$f(6) - \sqrt[6]{\frac{7}{9} \cdot 10} = \sqrt[6]{\frac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1$
The final answer is therefore $2+1+1+3+1 = \boxed{8}$ | A | 8 |
31b86a0a62b7dc2970dcd1bb4f084550 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19 | Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$ | For notation purposes, let $x$ be the number $777 \ldots 777$ with $313$ digits, and let $B(n)$ be the leading digit of $n$ . As an example, $B(x) = 7$ , because $x = 777 \ldots 777$ , and the first digit of that is $7$
Notice that \[B(\sqrt{\frac{n}{100}}) = B(\sqrt{n})\] for all numbers $n \geq 100$ ; this is because $\sqrt{\frac{n}{100}} = \frac{\sqrt{n}}{10}$ , and dividing by $10$ does not affect the leading digit of a number. Similarly, \[B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).\] In general, for positive integers $k$ and real numbers $n > 10^{k}$ , it is true that \[B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).\] Behind all this complex notation, all that we're really saying is that the first digit of something like $\sqrt[3]{123456789}$ has the same first digit as $\sqrt[3]{123456.789}$ and $\sqrt[3]{123.456789}$
The problem asks for \[B(\sqrt[2]{x}) + B(\sqrt[3]{x}) + B(\sqrt[4]{x}) + B(\sqrt[5]{x}) + B(\sqrt[6]{x}).\]
From our previous observation, we know that \[B(\sqrt[2]{x}) = B(\sqrt[2]{\frac{x}{100}} = B(\sqrt[2]{\frac{x}{10,000}} = B(\sqrt[2]{\frac{x}{1,000,000}} = \ldots .\] Therefore, $B(\sqrt[2]{x}) = B(\sqrt[2]{7.777 \dots})$ . We can evaluate $B(\sqrt[2]{7.777 \dots})$ , the leading digit of $\sqrt[2]{7.777 \dots}$ , to be $2$ . Therefore, $f(2) = 2$
Similarly, we have \[B(\sqrt[3]{x}) = B(\sqrt[3]{\frac{x}{1,000}} = B(\sqrt[3]{\frac{x}{1,000,000}} = B(\sqrt[3]{\frac{x}{1,000,000,000}} = \ldots .\] Therefore, $B(\sqrt[3]{x}) = B(\sqrt[3]{7.777 \ldots})$ . We know $B(\sqrt[3]{7.777 \ldots}) = 1$ , so $f(3) = 1$
Next, \[B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(4) = 1$
We also have \[B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})\] and $B(\sqrt[5]{777.777 \ldots}) = 3$ , so $f(5) = 3$
Finally, \[B(\sqrt[6]{x}) = B(\sqrt[6]{7.777 \ldots})\] and $B(\sqrt[4]{7.777 \ldots}) = 1$ , so $f(6) = 1$
We have that $f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{8}$ | A | 8 |
31b86a0a62b7dc2970dcd1bb4f084550 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19 | Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$ | Since $7777..7$ is a $313$ digit number and $\sqrt {7}$ is around $2.5$ , we have $f(2)$ is $2$ $f(3)$ is the same story, so $f(3)$ is $1$ . It is the same as $f(4)$ as well, so $f(4)$ is also $1$ . However, $313$ is $3$ mod $5$ , so we need to take the 5th root of $777$ , which is between $3$ and $4$ , and therefore, $f(5)$ is $3$ $f(6)$ is the same as $f(4)$ , since it is $1$ more than a multiple of $6$ . Therefore, we have $2+1+1+3+1$ which is $\boxed{8}$ | A | 8 |
31b86a0a62b7dc2970dcd1bb4f084550 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_19 | Let $N$ be the positive integer $7777\ldots777$ , a $313$ -digit number where each digit is a $7$ . Let $f(r)$ be the leading digit of the $r{ }$ th root of $N$ . What is \[f(2) + f(3) + f(4) + f(5)+ f(6)?\] $(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$ | First, we compute $f \left( 2 \right)$
Because $N > 4 \cdot 10^{312}$ $\sqrt{N} > 2 \cdot 10^{156}$ .
Because $N < 9 \cdot 10^{312}$ $\sqrt{N} < 3 \cdot 10^{156}$
Therefore, $f \left( 2 \right) = 2$
Second, we compute $f \left( 3 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[3]{N} > 1 \cdot 10^{104}$ .
Because $N < 8 \cdot 10^{312}$ $\sqrt[3]{N} < 2 \cdot 10^{104}$
Therefore, $f \left( 3 \right) = 1$
Third, we compute $f \left( 4 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[4]{N} > 1 \cdot 10^{78}$ .
Because $N < 16 \cdot 10^{312}$ $\sqrt[4]{N} < 2 \cdot 10^{78}$
Therefore, $f \left( 4 \right) = 1$
Fourth, we compute $f \left( 5 \right)$
Because $N > 3^5 \cdot 10^{310}$ $\sqrt[5]{N} > 3 \cdot 10^{62}$ .
Because $N < 4^5 \cdot 10^{310}$ $\sqrt[5]{N} < 4 \cdot 10^{62}$
Therefore, $f \left( 5 \right) = 3$
Fifth, we compute $f \left( 6 \right)$
Because $N > 1 \cdot 10^{312}$ $\sqrt[6]{N} > 1 \cdot 10^{52}$ .
Because $N < 2^6 \cdot 10^{312}$ $\sqrt[6]{N} < 2 \cdot 10^{52}$
Therefore, $f \left( 6 \right) = 1$
Therefore, \begin{align*} f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right) & = 2 + 1 + 1 + 3 + 1 \\ & = 8 . \end{align*}
Therefore, the answer is $\boxed{8}$ | A | 8 |
90a48e30c81187552131f5fceac1cce5 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_21 | Regular polygons with $5,6,7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?
$(\textbf{A})\: 52\qquad(\textbf{B}) \: 56\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 64\qquad(\textbf{E}) \: 68$ | Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice.
(We can see this because to have a vertex of the $m$ -gon on an arc subtended by a side of the $n$ -gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi)
This means that we will end up with $2$ times the number of sides in the polygon with fewer sides.
If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections.
Throughout $6$ of these pairs, the $5$ -sided polygon has the least number of sides $3$ times, the $6$ -sided polygon has the least number of sides $2$ times, and the $7$ -sided polygon has the least number of sides $1$ time.
Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{68}$ | E | 68 |
ea6fb3168b5d2b237af1a3e00e3d9a84 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22 | For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$ | To get from $S_n$ to $S_{n+1}$ , we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1)^2}{2}$
Now, we can look at the different values of $n$ mod $3$ . For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$ , then we have $\frac{n(n+1)^2}{2}\equiv 0\pmod{3}$ . However, for $n\equiv 1\pmod{3}$ , we have \[\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.\]
Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have $S_5\equiv 1\pmod{3}$ , and $S_8$ $S_9$ , and $S_{10}$ are all divisible by $3$ . After $3\cdot 3=9$ , we have $S_{17}$ $S_{18}$ , and $S_{19}$ all divisible by $3$ , as well as $S_{26}, S_{27}, S_{28}$ , and $S_{35}$ . Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{197} .$ -BorealBear | B | 197 |
ea6fb3168b5d2b237af1a3e00e3d9a84 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22 | For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$ | Since we have a wonky function, we start by trying out some small cases and see what happens. If $j$ is $1$ and $k$ is $2$ , then there is one case. We have $2$ mod $3$ for this case. If $N$ is $3$ , we have $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3$ which is still $2$ mod $3$ . If $N$ is $4$ , we have to add $1 \cdot 4 + 2 \cdot 4 + 3 \cdot 4$ which is a multiple of $3$ , meaning that we are still at $2$ mod $3$ . If we try a few more cases, we find that when $N$ is $8$ , we get a multiple of $3$ . When $N$ is $9$ , we are adding $0$ mod $3$ , and therefore, we are still at a multiple of $3$
When $N$ is $10$ , then we get $0$ mod $3$ $10(1+2+3+...+9)$ which is $10$ times a multiple of $3$ . Therefore, we have another multiple of $3$ . When $N$ is $11$ , so we have $2$ mod $3$ . So, every time we have $-1$ mod $9$ $0$ mod $9$ , and $1$ mod $9$ , we always have a multiple of $3$ . Think about it: When $N$ is $1$ , it will have to be $0 \cdot 1$ , so it is a multiple of $3$ . Therefore, our numbers are $8, 9, 10, 17, 18, 19, 26, 27, 28, 35$ . Adding the numbers up, we get $\boxed{197}$ | B | 197 |
ea6fb3168b5d2b237af1a3e00e3d9a84 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_22 | For each integer $n\geq 2$ , let $S_n$ be the sum of all products $jk$ , where $j$ and $k$ are integers and $1\leq j<k\leq n$ . What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$
$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$ | Denote $A_{n, <} = \left\{ \left( j , k \right) : 1 \leq j < k \leq n \right\}$ $A_{n, >} = \left\{ \left( j , k \right) : 1 \leq k < j \leq n \right\}$ and $A_{n, =} = \left\{ \left( j , k \right) : 1 \leq j = k \leq n \right\}$
Hence, $\sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk = S_n$
Therefore, \begin{align*} S_n & = \frac{1}{2} \left( \sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk \right) \\ & = \frac{1}{2} \left( \sum_{1 \leq j, k \leq n} jk - {\left( j , k \right) \in A_{n,=}} jk \right) \\ & = \frac{1}{2} \left( \sum_{j=1}^n \sum_{k=1}^n jk - \sum_{j=1}^n j^2 \right) \\ & = \frac{1}{2} \left( \frac{n^2 \left( n + 1 \right)^2}{4} - \frac{n \left( n + 1 \right) \left( 2 n + 1 \right) }{6} \right) \\ & = \frac{\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)}{24} . \end{align*}
Hence, $S_n$ is divisible by 3 if and only if $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by $24 \cdot 3 = 8 \cdot 9$
First, $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is always divisible by 8. Otherwise, $S_n$ is not even an integer.
Second, we find conditions for $n$ , such that $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by 9.
Because $3 n + 2$ is not divisible by 3, it cannot be divisible by 9.
Hence, we need to find conditions for $n$ , such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9.
This holds of $n \equiv 0, \pm 1 \pmod{9}$
Therefore, the 10 least values of $n$ such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9 (equivalently, $S_n$ is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35.
Their sum is 197.
Therefore, the answer is $\boxed{197}$ | B | 197 |
61d89ad63900ed1a23478c361a03edb6 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24 | A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$ | This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.
Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.
Topology.jpg
Here is how the $4$ blue unit cubes are arranged:
In Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).
In Figure $(2)$ $4$ blue unit cubes are in $T$ shape.
In Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.
In Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.
In Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.
In Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.
So the answer is $\boxed{7}$ | A | 7 |
61d89ad63900ed1a23478c361a03edb6 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24 | A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$ | Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.
Case 1: 4, 0
In this case, there is only one possibility for the top layer - all of the other color - $\binom{4}{4}$ . Therefore there is 1 construction from this case.
Case 2: 3, 1
In this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\binom{4}{3}$ . Therefore there are 4 constructions from this case.
Case 3: 2, 2
In this case, the top layer has six possibilities of arrangement - $\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.
The total number of constructions for the cube is thus $1+4+2=7=\boxed{7}$ | A | 7 |
61d89ad63900ed1a23478c361a03edb6 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24 | A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$ | Divide the $2 \times 2 \times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.
Therefore, our answer is $6+1+0=\boxed{7}$ | A | 7 |
61d89ad63900ed1a23478c361a03edb6 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24 | A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$ | Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.
The fact for Burnside lemma are
1. the sum of stablizer on the same orbit equals to the # of operators;
2. the sum of stablizer can be counted as $fix(g)$
3. the sum of the $fix(g)/|G|$ equals the # of orbit.
Let's start with defining the operator for a cube,
1. $\textbf{e (identity)}$
For identity, there are $\frac{8!}{4!4!} = 70$
2. ${\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,
each contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$
$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{i}) = 2+2+6 = 10$ , and $fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30$
3. ${\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.
each contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$
$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4$
therefore $fix(\bf r^{i}) = 4+4 = 8$ , and $fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32$
4. ${\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides
each contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36$
5. The total number of operators are
$|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24$
Based on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$
therefore the number of orbit $= \frac{168}{G=24} = \boxed{7}$ | null | 7 |
61d89ad63900ed1a23478c361a03edb6 | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_24 | A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$ | Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).
Once we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 "adjacent", the second 3 "cornering", and the last one "opposite."
We can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):
OAA, OAC, OCC, CCC, CAA, CCA, AAA.
This gives the answer to be $\boxed{7}$ | A | 7 |
8fac46204ecbb29ee514a534c810ca5d | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_25 | A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$ | We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.
Denote $\angle AFE = \theta$ . Thus, $\angle FIB = \angle CEF = \angle EKG = \angle KLC = \theta$
Hence, \begin{align*} AB & = AF + FB \\ & = EF \cos \angle EFA + IF \sin \angle FIB \\ & = 3 \cos \theta + \sin \theta , \end{align*} and \begin{align*} AC & = AE + EK + KC \\ & = EF \sin \angle EFA + EG \cos \angle CEG + KG \cos \angle EKG + KL \sin \angle CLK \\ & = 3 \sin \theta + \cos \theta + \cos \theta + \sin \theta \\ & = 2 \cos \theta + 4 \sin \theta . \end{align*}
Because $AB = AC$ $3 \cos \theta + \sin \theta = 2 \cos \theta + 4 \sin \theta$ .
Hence, $\tan \theta = \frac{1}{3}$ .
Hence, $\sin \theta = \frac{1}{\sqrt{10}}$ and $\cos \theta = \frac{3}{\sqrt{10}}$
Hence, $AB = AC = BD = CD = \sqrt{10}$
Now, we put the graph to a coordinate plane by setting point $A$ as the origin, putting $AB$ in the $x$ -axis and $AC$ on the $y$ -axis.
Hence, $A = \left( 0 , 0 \right)$ $B = \left( \sqrt{10} , 0 \right)$ $C = \left( 0 , \sqrt{10} \right)$ $D = \left( \sqrt{10} , \sqrt{10} \right)$ $E = \left( 0 , \frac{3}{\sqrt{10}} \right)$ $F = \left( \frac{9}{\sqrt{10}} , 0 \right)$ $G = \left( \frac{1}{\sqrt{10}} , \frac{6}{\sqrt{10}} \right)$ $H = \left( \frac{4}{\sqrt{10}} , \frac{7}{\sqrt{10}} \right)$ $I = \left( \sqrt{10} , \frac{3}{\sqrt{10}} \right)$
Denote $P = \left( \frac{10 - u}{\sqrt{10}} , \sqrt{10} \right)$ $Q = \left( \sqrt{10} , \frac{10 - v}{\sqrt{10}} \right)$
Because $HPQJ$ is a rectangle, $HP \perp PQ$ . Hence, $m_{HP} m_{PQ} = -1$ .
We have $m_{HP} = \frac{3}{6 - u}$ and $m_{PQ} = - \frac{v}{u}$ .
Hence, \[ \frac{3}{6 - u} \cdot \left( - \frac{v}{u} \right) = - 1 . \hspace{1cm} (1) \]
Because $HPQJ$ is a rectangle, $x_J + x_P = x_H + x_Q$ and $y_J + y_P = y_H + y_Q$ .
Hence, $J = \left( \frac{4 + u}{\sqrt{10}} , \frac{7 - v}{\sqrt{10}} \right)$
The equation of line $GI$ is \begin{align*} y & = \frac{\frac{3}{\sqrt{10}} - \frac{6}{\sqrt{10}}}{\sqrt{10} - \frac{1}{\sqrt{10}}} \left( x - \frac{1}{\sqrt{10}} \right) + \frac{6}{\sqrt{10}} \\ & = - \frac{x}{3} + \frac{19}{3 \sqrt{10}} . \end{align*} Because point $J$ is on line $GI$ , plugging the coordinates of $J$ into the equation of line $GI$ , we get \[ \frac{7 - v}{\sqrt{10}} = - \frac{\frac{4 + u}{\sqrt{10}}}{3} + \frac{19}{3 \sqrt{10}} . \hspace{1cm} (2) \]
By solving Equations (1) and (2), we get $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$ or $\left( 3 , 3 \right)$
Case 1: $\left( u , v \right) = \left( 2 , \frac{8}{3} \right)$
We have $P = \left( \frac{8}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{22}{3 \sqrt{10}} \right)$ .
Thus, $HP = \frac{5}{\sqrt{10}}$ and $PQ = \frac{10}{3\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{5}{3}$
Case 2: $\left( u , v \right) = \left( 3 , 3 \right)$
We have $P = \left( \frac{7}{\sqrt{10}} , \sqrt{10} \right)$ and $Q = \left( \sqrt{10} , \frac{7}{ \sqrt{10}} \right)$ .
Thus, $HP = \frac{3 \sqrt{2}}{\sqrt{10}}$ and $PQ = \frac{3 \sqrt{2}}{\sqrt{10}}$
Therefore, ${\rm Area} \ R = HP \cdot PQ = \frac{9}{5}$
Putting these two cases together, the sum of all possible values of the area of $R$ is $\frac{5}{3} + \frac{9}{5} = \frac{52}{15}$
Therefore, the answer is $\boxed{67}$ | E | 67 |
8fac46204ecbb29ee514a534c810ca5d | https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_10B_Problems/Problem_25 | A rectangle with side lengths $1{ }$ and $3,$ a square with side length $1,$ and a rectangle $R$ are inscribed inside a larger square as shown. The sum of all possible values for the area of $R$ can be written in the form $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n?$ [asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy] $(\textbf{A})\: 14\qquad(\textbf{B}) \: 23\qquad(\textbf{C}) \: 46\qquad(\textbf{D}) \: 59\qquad(\textbf{E}) \: 67$ | [asy] size(8cm); label("D",(0,0),SW); label("A",(0,10),NW); label("B",(10,10),NE); label("C",(10,0),SE); label("H",(1,5.7),SE); label("O",(0,6),W); label("I",(0,3),W); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); draw((1,6)--(0,6)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); draw((0.3,6)--(0.3,5.7)--(0,5.7)); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); label("E",(0,9),W); label("F",(3,10),N); label("P",(4,10),N); label("L",(8,10),N); draw((4,7)--(4,10)); draw((4,9.7)--(4.3,9.7)--(4.3,10)); label("G",(3.9,6.8),S); label("N",(6.2,4.1),S); label("M",(10,22/3),E); label("Q",(10,13/3),E); draw((10,13/3)--(6,13/3)); draw((9.7,13/3)--(9.7,4.033333333333333333333333)--(10,4.03333333333333333333333333)); label("K",(10,3),E); label("J",(9,0),S); [/asy] We will scale every number up by a factor of $\sqrt{10}.$ This implies our final area will be $\frac{1}{10}$ of the answer we receive.
We have \[FAE\sim EOH \sim IOH\sim JDI\sim KCJ \sim NQK\sim GPF.\] Let $AE=a$ and $FA=b.$ We have \[FP=AE=OH=JC=\frac13ID=a\] and \[PG=AF=EO=OI=KC=\frac13 DJ=b\] As $ABCD$ is a square, we have $AD=DC$ or \[a+2b+3a=3b+a \Rightarrow 3a=b.\] Since $a^2+b^2=10,$ we have \[a=1,b=3.\] We have $\triangle GPL\cong \triangle MQN$ which implies \[MQ=GP=3.\] Denote $QK=x.$ As $\triangle NQK\sim \triangle JDI,$ we have $NQ=3x.$
We have \begin{align*}BM&=BC-(CK+QK+MQ)\\ &=4-x.\end{align*}
In addition, \begin{align*}LB&=AB-(AF+FP+PL)\\&=6-3x.\end{align*}
Since $\triangle LBM\sim \triangle MQN,$ we have \[\frac{LB}{BM}=\frac{MQ}{QN}\Rightarrow \frac{6-3x}{4-x}=\frac{3}{3x}=\frac{1}{x}.\] Simplifying we have \[3x^2-7x+4=0 \Rightarrow x=\frac43, 1.\] We have \begin{align*}[GLMN]&=MN\cdot LM\\ &= 3\sqrt{x^2+1}\cdot \sqrt{10x^2-44x+52}.\end{align*}
Plugging in $x=1,$ we have $[GLMN]=18.$
Plugging in $x=\frac43,$ we have $[GLMN]=\frac{50}{3}.$
The sum of the two possible $R$ s is \[\frac{1}{10}\cdot\frac{104}{3}=\frac{52}{15}.\] Hence, $52+15=\boxed{67}.$ | E | 67 |
6f3ea48e13a731238b94c9129b49dcf8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 | What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | $(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\boxed{8}.$ | D | 8 |
6f3ea48e13a731238b94c9129b49dcf8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 | What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | We have \begin{align*} \left(2^2-2\right)-\left(3^2-3\right)+\left(4^2-4\right) &= 2(2-1)-3(3-1)+4(4-1) \\ &= 2(1)-3(2)+4(3) \\ &= 2-6+12 \\ &= \boxed{8} ~MRENTHUSIASM | D | 8 |
6f3ea48e13a731238b94c9129b49dcf8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 | What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | We have \begin{align*} (2^2-2)-(3^2-3)+(4^2-4) &= 2^2+4^2-3^2-2+3-4 \\ &=2^2+(4-3)(4+3)-3 \\ &=2^2+7-3=2^2+4 \\ &=4\cdot 2 \\ &=\boxed{8} PureSwag | D | 8 |
6f3ea48e13a731238b94c9129b49dcf8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 | What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | Using the difference of squares, we have \[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2).\] Knowing that $\sqrt{2} \approx 1.41$ and $\sqrt{3} \approx 1.73,$ we get \[(2-\sqrt{2})(2+\sqrt{2}) - (3-\sqrt{3})(3+\sqrt{3}) + (4-2)(4+2) \approx 0.59\cdot 3.41 -1.26\cdot 4.73 + 2 \cdot 6 =8.0521,\] which is closest to $\boxed{8}.$ | D | 8 |
fcae6db6d5f6061f8a72653db6fba503 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 | Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$ | The following system of equations can be formed with $P$ representing the number of students in Portia's high school and $L$ representing the number of students in Lara's high school: \begin{align*} P&=3L, \\ P+L&=2600. \end{align*} Substituting $P=3L$ gives $4L=2600.$ Solving for $L$ gives $L=650.$ Since we need to find $P,$ we multiply $650$ by $3$ to get $P=\boxed{1950}.$ | C | 1950 |
fcae6db6d5f6061f8a72653db6fba503 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 | Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$ | Suppose Lara's high school has $x$ students, so Portia's high school has $3x$ students. We have $x+3x=2600,$ or $4x=2600.$ The answer is \[3x=2600\cdot\frac 34=650\cdot3=\boxed{1950}.\] | C | 1950 |
fcae6db6d5f6061f8a72653db6fba503 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 | Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$ | Clearly, $2600$ is $4$ times the number of students in Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{1950}$ students. | C | 1950 |
fcae6db6d5f6061f8a72653db6fba503 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 | Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?
$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$ | The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)},\textbf{(D)},$ and $\textbf{(E)}.$ Since $\textbf{(A)}$ is too small (as it is clear that $600+\frac{600}{3}<2600$ ), we are left with $\boxed{1950}.$ | C | 1950 |
8f2c4ae7ed3a6c03f08bcbb09bac69d6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$ | The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$
Let the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$
We know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \boxed{14238}$ | D | 14238 |
8f2c4ae7ed3a6c03f08bcbb09bac69d6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$ | Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$ | D | 14238 |
8f2c4ae7ed3a6c03f08bcbb09bac69d6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$ | Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc} & A & B & C & D & 0 \\ +\quad & & A & B & C & D \\ \hline & & & & & \\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$ | D | 14238 |
8f2c4ae7ed3a6c03f08bcbb09bac69d6 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 | The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$ | We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$ | D | 14238 |
4df7e995252c8f5946ec855b51892801 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 | A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$ | Since \[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\] we seek the sum \[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\] in which there are $30$ terms.
The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{3195}.\] ~MRENTHUSIASM | D | 3195 |
4df7e995252c8f5946ec855b51892801 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 | A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$ | The distance (in inches) traveled within each $1$ -second interval is: \[5,5+1(7),5+2(7), \dots , 5+29(7).\] This is an arithmetic sequence so the total distance travelled, found by summing them up is: \[\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.\] Or, \[30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{3195}.\] ~BakedPotato66 | D | 3195 |
4df7e995252c8f5946ec855b51892801 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 | A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$ | From the $30$ -term sum \[5+12+19+26+\cdots\] in Solution 1, taking modulo $10$ gives \[5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.\] The only answer choices congruent to $5$ modulo $10$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimation, $\textbf{(A)}$ is too small, leaving us with $\boxed{3195}.$ | D | 3195 |
4df7e995252c8f5946ec855b51892801 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 | A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$ | This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases $7 \ \text{in/s}$ per second. So, the acceleration $a=7 \ \text{in/s\textsuperscript{2}}.$ The average speed of the first second is $5 \ \text{in/s}.$ We can know the initial velocity $v_0=5-0.5\cdot7=1.5.$ The displacement at $t=30$ is \[s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{3195}.\] ~Bran_Qin | D | 3195 |
09f982d83272a8ff7390516849306bb3 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$ | We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which \begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{75} ~MRENTHUSIASM | E | 75 |
09f982d83272a8ff7390516849306bb3 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$ | It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$
Let $x=\underline{a} \ \underline{b}.$ We have \[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\] Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{75}.$ | E | 75 |
09f982d83272a8ff7390516849306bb3 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_8 | When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$ | We have \[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\] Expanding both sides, we have \[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\] Subtracting $66$ from both sides, we have \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] Multiplying both sides by $50 \cdot 3 = 150,$ we have \[99(10a+b) + 75 = 100(10a+b).\] Thus, the answer is $10a+b = \boxed{75}.$ | E | 75 |
8f9efbcd233cf8820f31e5a1b37fe2ea | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9 | What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$ | Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{1}$ , which can be achieved at $x=y=0$ | D | 1 |
8f9efbcd233cf8820f31e5a1b37fe2ea | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9 | What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$ | We expand the original expression, then factor the result by grouping: \begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right). \end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{1}.\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$ | D | 1 |
8f9efbcd233cf8820f31e5a1b37fe2ea | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9 | What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$ | Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$ | D | 1 |
46b4938c8f91b8bb50e4605986c7e479 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11 | For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$ | We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{8}.$ | E | 8 |
46b4938c8f91b8bb50e4605986c7e479 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11 | For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$ | Vertically subtracting $2021_b - 221_b,$ we see that the ones place becomes $0,$ and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$ ). Let $b-2 = A.$ Then, we have our final number as \[1A00_b.\] Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2.$
Now, notice that the final number will only be congruent to \[b^3-(b-2)^2\equiv0\pmod{3}.\] If either $b\equiv0\pmod{3},$ or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3},$ and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}.$ Therefore, $b$ must be $\equiv2\pmod{3}.$ Among the answers, only $8$ is $\equiv2\pmod{3},$ and therefore our answer is $\boxed{8}.$ | E | 8 |
46b4938c8f91b8bb50e4605986c7e479 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11 | For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$ | By the definition of bases, we have \[2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).\] For values $b_1$ and $b_2$ such that $b_1\equiv b_2\pmod{3},$ we get \[\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.\] Note that answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. So, $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves us with $\boxed{8},$ the answer choice with a unique residue modulo $3.$ | E | 8 |
6f45f81a6fea686b8179800df852e4dd | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12 | Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$ | The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much. | E | 4 |
ffcb1af09cba9e9bebcafd465a78a1ee | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_13 | What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ $AC = 3$ $AD = 4$ $BC = \sqrt{13}$ $BD = 2\sqrt{5}$ , and $CD = 5$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$ | Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC$ as the base, then $\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{4}.$ | C | 4 |
ffcb1af09cba9e9bebcafd465a78a1ee | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_13 | What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ $AC = 3$ $AD = 4$ $BC = \sqrt{13}$ $BD = 2\sqrt{5}$ , and $CD = 5$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$ | We will place tetrahedron $ABCD$ in the $xyz$ -plane. By the Converse of the Pythagorean Theorem, we know that $\triangle ACD$ is a right triangle. Without the loss of generality, let $A=(0,0,0), C=(3,0,0), D=(0,4,0),$ and $B=(x,y,z).$
We apply the Distance Formula to $\overline{BA},\overline{BC},$ and $\overline{BD},$ respectively: \begin{align*} x^2+y^2+z^2&=2^2, &(1) \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. &\hspace{1mm} (3) \end{align*} Subtracting $(1)$ from $(2)$ gives $-6x+9=9,$ from which $x=0.$
Subtracting $(1)$ from $(3)$ gives $-8y+16=16,$ from which $y=0.$
Substituting $(x,y)=(0,0)$ into $(1)$ produces $z^2=4,$ or $|z|=2.$
Let the brackets denote areas. Finally, we find the volume of tetrahedron $ABCD$ using $\triangle ACD$ as the base: \begin{align*} V_{ABCD}&=\frac13\cdot[ACD]\cdot h_B \\ &=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\ &=\boxed{4} ~MRENTHUSIASM | C | 4 |
3d2419b00c582f984e8c960eff7065fb | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_14 | All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$ | By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{88}$ | A | 88 |
3d2419b00c582f984e8c960eff7065fb | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_14 | All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$ | Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{88}.\] ~ike.chen | A | 88 |
bba59d30fd76667bf71c9a33bb8984e8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_15 | Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$ | Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$ , and the answer is $\boxed{90}$ | C | 90 |
bba59d30fd76667bf71c9a33bb8984e8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_15 | Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$ | Setting $y = Ax^2+B = Cx^2+D$ , we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$ , so $x^2 = \frac {D-B}{A-C} \ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get \[\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{90}.\] ~ike.chen | C | 90 |
bba59d30fd76667bf71c9a33bb8984e8 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_15 | Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$ | Like in Solution 2 , we find $\frac {D-B}{A-C} \ge 0$ . Notice that, since $D \ne B$ , this expression can never equal $0$ , and since $A \ne C$ , there won't be a divide-by- $0$ . This means that every choice results in either a positive or a negative value.
For every choice of $(A,B,C,D)$ that results in a positive value, we can flip $B$ and $D$ to obtain a corresponding negative value. This is a bijection (we could flip $B$ and $D$ again to obtain the original choice (injectivity) and we could flip $B$ and $D$ from any negative choice to obtain the corresponding positive choice (surjectivity)), so half of the choices are positive (where the curves intersect) and half are negative (where they don't).
This means that of the $\frac{6\cdot5\cdot4\cdot3}{2} = 180$ total choices (dividing by $2$ because the order of the curves does not matter), half of them, or $\frac{180}{2} = \boxed{90}$ , lead to intersecting curves. | C | 90 |
e39e52fb7d100aa594cd7ee235129b29 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_16 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$ | The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{142}$ ~Lopkiloinm | C | 142 |
e39e52fb7d100aa594cd7ee235129b29 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_16 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$ | We can look at answer choice $\textbf{(C)}$ , which is $142$ first. That means that the number of numbers from $1$ to $142$ is roughly the number of numbers from $143$ to $200$
The number of numbers from $1$ to $142$ is $\frac{142(142+1)}{2}$ which is approximately $10000.$ The number of numbers from $143$ to $200$ is $\frac{200(200+1)}{2}-\frac{142(142+1)}{2}$ which is approximately $10000$ as well. Therefore, we can be relatively sure the answer choice is $\boxed{142}.$ | C | 142 |
e39e52fb7d100aa594cd7ee235129b29 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_16 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$ | We can arrange the numbers in the following pattern: \[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length $200.$ [asy]draw((0,0)--(200,200)--(200,0)--cycle); draw((142,0)--(142,142)); label("$x$",(142,0)--(142,142),E); label("$x$",(0,0)--(142,0),S); label("$200$",(200,0)--(200,200),E); [/asy]
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields: \begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*} We see that $\boxed{142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer. | C | 142 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$ , therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$
Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$ , therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$
Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{194}$ | D | 194 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Since $\triangle BCD$ is isosceles with base $\overline{BD},$ it follows that median $\overline{CP}$ is also an altitude. Let $OD=x$ and $CP=h,$ so $PB=x+11.$
Since $\angle AOD=\angle COP$ by vertical angles, we conclude that $\triangle AOD\sim\triangle COP$ by AA, from which $\frac{AD}{CP}=\frac{OD}{OP},$ or \[AD=CP\cdot\frac{OD}{OP}=h\cdot\frac{x}{11}.\] Let the brackets denote areas. Notice that $[AOD]=[BOC]$ (By the same base and height, we deduce that $[ACD]=[BDC].$ Subtracting $[OCD]$ from both sides gives $[AOD]=[BOC].$ ). Doubling both sides produces \begin{align*} 2[AOD]&=2[BOC] \\ OD\cdot AD&=OB\cdot CP \\ x\left(\frac{hx}{11}\right)&=(x+22)h \\ x^2&=11(x+22). \end{align*} Rearranging and factoring result in $(x-22)(x+11)=0,$ from which $x=22.$
Applying the Pythagorean Theorem to right $\triangle CPB,$ we have \[h=\sqrt{43^2-33^2}=\sqrt{(43+33)(43-33)}=\sqrt{760}=2\sqrt{190}.\] Finally, we get \[AD=h\cdot\frac{x}{11}=4\sqrt{190},\] so the answer is $4+190=\boxed{194}.$ | D | 194 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Let $CP = y$ $CP$ a is perpendicular bisector of $DB.$ Then, let $DO = x,$ thus $DP = PB = 11+x.$
(1) $\triangle CPO \sim \triangle ADO,$ so we get $\frac{AD}{x} = \frac{y}{11},$ or $AD = \frac{xy}{11}.$
(2) Applying Pythagorean Theorem on $\triangle CDP$ gives $(11+x)^2 + y^2 = 43^2.$
(3) $\triangle BPC \sim \triangle BDA$ with ratio $1:2,$ so $AD = 2y$ using the fact that $P$ is the midpoint of $BD$
Thus, $\frac{xy}{11} = 2y,$ or $x = 22.$ And $y = \sqrt{43^2 - 33^2} = 2 \sqrt{190},$ so $AD = 4 \sqrt{190}$ and the answer is $4+190=\boxed{194}.$ | D | 194 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Observe that $\triangle BPC$ is congruent to $\triangle DPC$ ; both are similar to $\triangle BDA$ . Let's extend $\overline{AD}$ and $\overline{BC}$ past points $D$ and $C$ respectively, such that they intersect at a point $E$ . Observe that $\angle BDE$ is $90$ degrees, and that $\angle DBE \cong \angle PBC \cong \angle DBA \implies \angle DBE \cong \angle DBA$ . Thus, by ASA, we know that $\triangle ABD \cong \triangle EBD$ , thus, $AD = ED$ , meaning $D$ is the midpoint of $AE$ .
Let $M$ be the midpoint of $\overline{DE}$ . Note that $\triangle CME$ is congruent to $\triangle BPC$ , thus $BC = CE$ , meaning $C$ is the midpoint of $\overline{BE}.$
Therefore, $\overline{AC}$ and $\overline{BD}$ are both medians of $\triangle ABE$ . This means that $O$ is the centroid of $\triangle ABE$ ; therefore, because the centroid divides the median in a 2:1 ratio, $\frac{BO}{2} = DO = \frac{BD}{3}$ . Recall that $P$ is the midpoint of $BD$ $DP = \frac{BD}{2}$ . The question tells us that $OP = 11$ $DP-DO=11$ ; we can write this in terms of $DB$ $\frac{DB}{2}-\frac{DB}{3} = \frac{DB}{6} = 11 \implies DB = 66$
We are almost finished. Each side length of $\triangle ABD$ is twice as long as the corresponding side length $\triangle CBP$ or $\triangle CPD$ , since those triangles are similar; this means that $AB = 2 \cdot 43 = 86$ . Now, by Pythagorean theorem on $\triangle ABD$ $AB^{2} - BD^{2} = AD^{2} \implies 86^{2}-66^{2} = AD^{2} \implies AD = \sqrt{3040} \implies AD = 4 \sqrt{190}$
The answer is $4+190 = \boxed{194}$ | D | 194 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Since $P$ is the midpoint of isosceles triangle $BCD$ , it would be pretty easy to see that $CP\perp BD$ . Since $AD\perp BD$ as well, $AD\parallel CP$ . Connecting $AP$ , it’s obvious that $[ADC]=[ADP]$ . Since $DP=BP$ $[APB]=[ADC]$
Since $P$ is the midpoint of $BD$ , the height of $\triangle APB$ on side $AB$ is half that of $\triangle ADC$ on $CD$ . Since $[APB]=[ADC]$ $AB=2CD$
As a basic property of a trapezoid, $\triangle AOB \sim \triangle COD$ , so $\frac{OB}{OD}=\frac{AB}{CD}=2$ , or $OB=2OD$ . Letting $OD=x$ , then $PB=DP=11+x$ , and $OB=22+x$ . Hence $22+x=2x$ and $x=22$
Since $\triangle AOD \sim \triangle COP$ $\frac{AD}{PC}=\frac{OD}{OP}=2$ . Since $PD=11+22=33$ $PC=\sqrt{43^2-33^2}=\sqrt{760}$
So, $AD=2\sqrt{760}=4\sqrt{190}$ . The correct answer is $\boxed{194}$ | D | 194 |
4e2ddae52b64e4c35028c9275c6b89c9 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 | Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$ , and $\overline{AD}\perp\overline{BD}$ . Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$ , and let $P$ be the midpoint of $\overline{BD}$ . Given that $OP=11$ , the length of $AD$ can be written in the form $m\sqrt{n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$
$\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215$ | Let $D$ be the origin of the cartesian coordinate plane, $B$ lie on the positive $x$ -axis, and $A$ lie on the negative $y$ -axis. Then let the coordinates of $B = (2a,0), A = (0, -2b).$ Then the slope of $AB$ is $\frac{b}{a}.$ Since $AB \parallel CD$ the slope of $CD$ is the same. Note that as $\triangle DCB$ is isosceles $C$ lies on $x = a.$ Thus since $CD$ has equation $y = \frac{b}{a}x$ $D$ is the origin), $C = (a,b).$ Therefore $AC$ has equation $y = \frac{3b}{a}x - 2b$ and intersects $BD$ $x$ -axis) at $O =\left(\frac{2}{3}a, 0\right).$ The midpoint of $BD$ is $P = (a,0),$ so $OP = \frac{a}{3} = 11,$ from which $a = 33.$ Then by Pythagorean theorem on $\triangle DPC$ $\triangle DBC$ is isosceles), we have $b = \sqrt{43^2 - 33^2} = 2\sqrt{190},$ so $2b=4\sqrt{190}.$
Finally, the answer is $4+190=\boxed{194}.$ | D | 194 |
b12a7d87dde75e2dbc226447d9fa57d7 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_19 | The area of the region bounded by the graph of \[x^2+y^2 = 3|x-y| + 3|x+y|\] is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$ | In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$
Case 1: $|x-y|=x-y, |x+y|=x+y$
Substituting and simplifying, we have $x^2-6x+y^2=0$ , i.e. $(x-3)^2+y^2=3^2$ , which gives us a circle of radius $3$ centered at $(3,0)$
Case 2: $|x-y|=y-x, |x+y|=x+y$
Substituting and simplifying again, we have $x^2+y^2-6y=0$ , i.e. $x^2+(y-3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,3)$
Case 3: $|x-y|=x-y, |x+y|=-x-y$
Doing the same process as before, we have $x^2+y^2+6y=0$ , i.e. $x^2+(y+3)^2=3^2$ . This gives us a circle of radius $3$ centered at $(0,-3)$
Case 4: $|x-y|=y-x, |x+y|=-x-y$
One last time: we have $x^2+y^2+6x=0$ , i.e. $(x+3)^2+y^2=3^2$ . This gives us a circle of radius $3$ centered at $(-3,0)$
After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:
[asy] size(10cm); Label f; f.p=fontsize(7); xaxis(-8,8,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 1.0)); draw(arc((-3,0),3,90,270) -- cycle, gray); draw(arc((0,3),3,0,180) -- cycle, gray); draw(arc((3,0),3,-90,90) -- cycle, gray); draw(arc((0,-3),3,-180,0) -- cycle, gray); draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, grey); [/asy] Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$ .
The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$ . The answer is $36+18$ which is $\boxed{54}$ | E | 54 |
b12a7d87dde75e2dbc226447d9fa57d7 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_19 | The area of the region bounded by the graph of \[x^2+y^2 = 3|x-y| + 3|x+y|\] is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$ | A somewhat faster variant of solution 1 is to use a bit of symmetry in order to show that the remaining three cases are identical to Case 1 in the above solution, up to rotations by $90^{\circ}$ about the origin. This allows us to quickly sketch the region after solving Case 1.
Upon simplifying Case 1, we obtain $(x-3)^2 + y^2 = 3^2$ which is a circle of radius 3 centered at $(3,0)$ . We remark that only the points on the semicircle where $x \ge 3$ work here, since Case 1 assumes $x-y \ge 0$ and $x+y \ge 0$ . Next, we observe that an ordered pair is a solution to the given equation if and only if any of its $90^{\circ}$ rotations about the origin is a solution. This follows as the value of $x^2+y^2-3(|x-y|+|x+y|)$ is invariant to $90^{\circ}$ rotations, since $x^2+y^2$ simply represents the square of the distance to the origin (which is unchanged upon rotation), and $|x-y|+|x+y|$ is the sum of the distances to the lines $y=x$ and $y=-x$ , multiplied by $\sqrt{2}$ (also unchanged upon $90^{\circ}$ rotation).
By the above observation, we can quickly sketch the remainder of the region, and the area is $\boxed{54}$ as above. | E | 54 |
b12a7d87dde75e2dbc226447d9fa57d7 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_19 | The area of the region bounded by the graph of \[x^2+y^2 = 3|x-y| + 3|x+y|\] is $m+n\pi$ , where $m$ and $n$ are integers. What is $m + n$
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$ | Assume $y$ $0$ . We get that $x$ $6$ . That means that this figure must contain the points $(0,6), (6,0), (0, -6), (-6, 0)$ . Now, assume that $x$ $y$ . We get that $x$ $3 \sqrt 3$ . We get the points $(3,3), (3,-3), (-3, 3), (-3, -3)$
Since this contains $x^2 + y^2$ , assume that there are circles. Therefore, we can guess that there is a center square with area $6 \cdot 6$ $36$ and $4$ semicircles with radius $3$ . We get $4$ semicircles with area $4.5 \pi$ , and therefore the answer is $36+18$ $\boxed{54}$ | E | 54 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | We write out the $5!=120$ cases, then filter out the valid ones:
$13254,14253,14352,15243,15342,21435,21534,23154,24153,24351,25143,25341,\linebreak 31425,31524,32415,32514,34152,34251,35142,35241,41325,41523,42315,42513,\linebreak 43512,45132,45231,51324,51423,52314,52413,53412.$
We count these out and get $\boxed{32}$ permutations that work. | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | By symmetry with respect to $3,$ note that $(x_1,x_2,x_3,x_4,x_5)$ is a valid sequence if and only if $(6-x_1,6-x_2,6-x_3,6-x_4,6-x_5)$ is a valid sequence. We enumerate the valid sequences that start with $1,2,31,$ or $32,$ as shown below:
[asy] /* Made by MRENTHUSIASM */ size(16cm); draw((0.25,0)--(1.75,3),red,EndArrow); draw((0.25,0)--(1.75,0),red,EndArrow); draw((0.25,0)--(1.75,-3),red,EndArrow); draw((2.25,3)--(3.75,3),red,EndArrow); draw((2.25,0)--(3.75,0.75),red,EndArrow); draw((2.25,0)--(3.75,-0.75),red,EndArrow); draw((2.25,-3)--(3.75,-2.25),red,EndArrow); draw((2.25,-3)--(3.75,-3.75),red,EndArrow); draw((4.25,3)--(5.75,3),red,EndArrow); draw((4.25,0.75)--(5.75,0.75),red,EndArrow); draw((4.25,-0.75)--(5.75,-0.75),red,EndArrow); draw((4.25,-2.25)--(5.75,-2.25),red,EndArrow); draw((4.25,-3.75)--(5.75,-3.75),red,EndArrow); draw((6.25,3)--(7.75,3),red,EndArrow); draw((6.25,0.75)--(7.75,0.75),red,EndArrow); draw((6.25,-0.75)--(7.75,-0.75),red,EndArrow); draw((6.25,-2.25)--(7.75,-2.25),red,EndArrow); draw((6.25,-3.75)--(7.75,-3.75),red,EndArrow); label("$1$",(0,0)); label("$3$",(2,3)); label("$2$",(4,3)); label("$5$",(6,3)); label("$4$",(8,3)); label("$4$",(2,0)); label("$2$",(4,0.75)); label("$5$",(6,0.75)); label("$3$",(8,0.75)); label("$3$",(4,-0.75)); label("$5$",(6,-0.75)); label("$2$",(8,-0.75)); label("$5$",(2,-3)); label("$2$",(4,-2.25)); label("$4$",(6,-2.25)); label("$3$",(8,-2.25)); label("$3$",(4,-3.75)); label("$4$",(6,-3.75)); label("$2$",(8,-3.75)); draw((12.75,0)--(14.25,4.5),red,EndArrow); draw((12.75,0)--(14.25,1.5),red,EndArrow); draw((12.75,0)--(14.25,-1.5),red,EndArrow); draw((12.75,0)--(14.25,-4.5),red,EndArrow); draw((14.75,4.5)--(16.25,5.25),red,EndArrow); draw((14.75,4.5)--(16.25,3.75),red,EndArrow); draw((14.75,1.5)--(16.25,1.5),red,EndArrow); draw((14.75,-1.5)--(16.25,-0.75),red,EndArrow); draw((14.75,-1.5)--(16.25,-2.25),red,EndArrow); draw((14.75,-4.5)--(16.25,-3.75),red,EndArrow); draw((14.75,-4.5)--(16.25,-5.25),red,EndArrow); draw((16.75,5.25)--(18.25,5.25),red,EndArrow); draw((16.75,3.75)--(18.25,3.75),red,EndArrow); draw((16.75,1.5)--(18.25,1.5),red,EndArrow); draw((16.75,-0.75)--(18.25,-0.75),red,EndArrow); draw((16.75,-2.25)--(18.25,-2.25),red,EndArrow); draw((16.75,-3.75)--(18.25,-3.75),red,EndArrow); draw((16.75,-5.25)--(18.25,-5.25),red,EndArrow); draw((18.75,5.25)--(20.25,5.25),red,EndArrow); draw((18.75,3.75)--(20.25,3.75),red,EndArrow); draw((18.75,1.5)--(20.25,1.5),red,EndArrow); draw((18.75,-0.75)--(20.25,-0.75),red,EndArrow); draw((18.75,-2.25)--(20.25,-2.25),red,EndArrow); draw((18.75,-3.75)--(20.25,-3.75),red,EndArrow); draw((18.75,-5.25)--(20.25,-5.25),red,EndArrow); label("$2$",(12.5,0)); label("$1$",(14.5,4.5)); label("$3$",(14.5,1.5)); label("$4$",(14.5,-1.5)); label("$5$",(14.5,-4.5)); label("$4$",(16.5,5.25)); label("$5$",(16.5,3.75)); label("$1$",(16.5,1.5)); label("$1$",(16.5,-0.75)); label("$3$",(16.5,-2.25)); label("$1$",(16.5,-3.75)); label("$3$",(16.5,-5.25)); label("$3$",(18.5,5.25)); label("$3$",(18.5,3.75)); label("$5$",(18.5,1.5)); label("$5$",(18.5,-0.75)); label("$5$",(18.5,-2.25)); label("$4$",(18.5,-3.75)); label("$4$",(18.5,-5.25)); label("$5$",(20.5,5.25)); label("$4$",(20.5,3.75)); label("$4$",(20.5,1.5)); label("$3$",(20.5,-0.75)); label("$1$",(20.5,-2.25)); label("$3$",(20.5,-3.75)); label("$1$",(20.5,-5.25)); draw((25.25,0)--(26.75,1.5),red,EndArrow); draw((25.25,0)--(26.75,-1.5),red,EndArrow); draw((27.25,1.5)--(28.75,2.25),red,EndArrow); draw((27.25,1.5)--(28.75,0.75),red,EndArrow); draw((27.25,-1.5)--(28.75,-0.75),red,EndArrow); draw((27.25,-1.5)--(28.75,-2.25),red,EndArrow); draw((29.25,2.25)--(30.75,2.25),red,EndArrow); draw((29.25,0.75)--(30.75,0.75),red,EndArrow); draw((29.25,-0.75)--(30.75,-0.75),red,EndArrow); draw((29.25,-2.25)--(30.75,-2.25),red,EndArrow); draw((31.25,2.25)--(32.75,2.25),red,EndArrow); draw((31.25,0.75)--(32.75,0.75),red,EndArrow); draw((31.25,-0.75)--(32.75,-0.75),red,EndArrow); draw((31.25,-2.25)--(32.75,-2.25),red,EndArrow); label("$3$",(25,0)); label("$1$",(27,1.5)); label("$2$",(27,-1.5)); label("$4$",(29,2.25)); label("$5$",(29,0.75)); label("$4$",(29,-0.75)); label("$5$",(29,-2.25)); label("$2$",(31,2.25)); label("$2$",(31,0.75)); label("$1$",(31,-0.75)); label("$1$",(31,-2.25)); label("$5$",(33,2.25)); label("$4$",(33,0.75)); label("$5$",(33,-0.75)); label("$4$",(33,-2.25)); [/asy]
There are $16$ valid sequences that start with $1,2,31,$ or $32.$ By symmetry, there are $16$ valid sequences that start with $5,4,35,$ or $34.$ So, the answer is $16+16=\boxed{32}.$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | Reading the terms from left to right, we have two cases for the consecutive digits, where $+$ means increase and $-$ means decrease:
$\textbf{Case \#1: }\boldsymbol{+,-,+,-}$
$\textbf{Case \#2: }\boldsymbol{-,+,-,+}$
For $\text{Case \#1},$ note that for the second and fourth terms, one term must be $5,$ and the other term must be either $3$ or $4.$ We have four subcases:
$(1) \ \underline{\hspace{3mm}}3\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$
$(2) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}3\underline{\hspace{3mm}}$
$(3) \ \underline{\hspace{3mm}}4\underline{\hspace{3mm}}5\underline{\hspace{3mm}}$
$(4) \ \underline{\hspace{3mm}}5\underline{\hspace{3mm}}4\underline{\hspace{3mm}}$
For $(1),$ the first two blanks must be $1$ and $2$ in some order, and the last blank must be $4.$ So, we get $2$ possibilities. Similarly, $(2)$ also has $2$ possibilities.
For $(3),$ there are no restrictions for the numbers $1, 2,$ and $3.$ So, we get $3!=6$ possibilities. Similarly, $(4)$ also has $6$ possibilities.
Together, $\text{Case \#1}$ has $2+2+6+6=16$ possibilities. By symmetry, $\text{Case \#2}$ also has $16$ possibilities.
Finally, the answer is $16+16=\boxed{32}.$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | Like Solution 3, we have two cases. Due to symmetry, we just need to count one of the cases. For the purpose of this solution, we will be doing $-,+,-,+$ . Instead of starting with 5, we start with 1.
There are two ways to place it:
_1_ _ _
_ _ _1_
Now we place 2, it can either be next to 1 and on the outside, or is place in where 1 would go in the other case. So now we have another two "sub case":
_1_2_(case 1)
21_ _ _(case 2)
There are 3! ways to arrange the rest for case 1, since there is no restriction.
For case 2, we need to consider how many ways to arrange 3,4,5 in a a>b<c fashion. It should seem pretty obvious that b has to be 3, so there will be 2! way to put 4 and 5.
Now we find our result, times 2 for symmetry, times 2 for placement of 1 and times (3!+2!) for the two different cases for placement of 2. This give us $2*2*(3!+2!)=4*(6+2)=\boxed{32}$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | We only need to find the # of rearrangements when 5 is the 4th digit and 5th digit. Find the total, and multiply by 2. Then we can get the answer by adding the case when 5 is the third digit.
Case $1$ 5 is the 5th digit. __ __ __ __ 5
Then $4$ can only be either 1st digit or the 3rd digit.
4 __ __ __ 5, then the only way is that $3$ is the 3rd digit, so it can be either $231$ or $132$ , give us $2$ results.
__ __ 4 __ 5, then the 1st digit must be $2$ or $3$ $2$ gives us $1$ way, and $3$ gives us $2$ ways. (Can't be $1$ because the first digit would increasing). Therefore, $4$ in the middle and $5$ in the last would result in $3$ ways.
Case $2$ $5$ is the fourth digit. __ __ __ 5 __
Then the last digit can be all of the 4 numbers $1$ $2$ $3$ , and $4$ . Let's say if the last digit is $4$ , then the 2nd digit would be the largest for the remaining digits to prevent increasing order or decreasing order. Then the remaining two are interchangeable, give us $2!$ ways. All of the $4$ can work, so case $2$ would result in $2!+2!+2!+2!=8$ ways.
Case $3$ $5$ is in the middle. __ __ 5 __ __
Then there are only two cases:
1. $42513$ , then 4 and 3 are interchangeable, which results in $2!*2!$ . Or it can be $43512$ , then 4 and 2 are interchangeable, but it can not be $23514$ , so there can only be 2 possible ways: $43512$ $21534$
Therefore, case 3 would result in $4+2=6$ ways.
$8+3+2=13$ , so the total ways for case 1 and case 2 with both increasing and decreasing would be $13*2=26.$
Finally, we have $26+6=\boxed{32}.$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | First, we list the triples that are invalid:
543, 542, 541, 532, 531, 521, 432, 431, 321
By symmetry, there are the same amount of increasing triplets as there are decreasing ones. This yields 18 invalid 3 digit permutations in total.
Suppose the triplet is ABC and the other 2 digits are X and Y. We then have 3 ways to arrange a triplet with 2 other digits.
ABCXY, XABCY, XYABC
X and Y can be arranged 2 ways.
XY, YX
This produces 18*3*2=108 permutations of invalid results. We have 5!
ways to arrange 5 numbers so 120-108=12.
Now, we must account for overcounting. For example, when 543 is counted,
it only registers as one invalid permutation but in fact, it is 3 whole
invalid permutations. We then complete this for the rest of the list:
54321 has 543, 432, and 321
54213 has 542 and 421
54123 has 541 and 123
53214 has 532 and 321
53124 has 531 and 124
52134 has 521 and 134
43215 has 432 and 321
43125 has 431 and 125
32145 has 321 and 145
This produces 19 values that we have overcounted but this value itself is also overcounted. We already counted 9 of the terms. This brings the final value of overcounted terms down to 10 for the decreasing triplets. By symmetry, 10 increasing triplets were overcounted.
This gives us $120-108+20=\boxed{32}.$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | It is easier to consider the complement of the desired cases, so try to find the cases that DO have three integers in increasing order. First, write down the sets of three numbers that feature the numbers in increasing order. They are 123, 124, 125, 134, 135, 145, 234, 245, 345. Each of these can be in three positions: the three are in the front with two more numbers in the back, the three are in the middle with two on each side, and two in the front and the set of three in the back. Now, count the number of combinations of 5 numbers that each of the set of three can form that have no been previously accounted for. Also, if the set features both 3 increasing and 3 decreasing, then do not count it because we will separately count them.
\[\begin{tabular}[t]{|l|c|c|c|} \hline Three Increasing Numbers&Front&Middle&Back\\\hline 123&2&2&2\\\hline 124&2&2&2\\\hline 125&1&2&2\\\hline 134&2&2&2\\\hline 135&1&2&2\\\hline 145&1&2&2\\\hline 234&2&1&1\\\hline 245&1&1&1\\\hline 345&1&0&1\\\hline \end{tabular}\]
This gives us a total of 41 possibilities. These account for the possibilities with ONLY increasing numbers. Mirror over to the other side to get the set of combinations with either at least 1 set of 3 increasing numbers in a row or only at least 1 set of 3 decreasing numbers, but not both. We can count the ones with both separately. 14532, 12543, 13542, 54123, 53124, 52134. Total of 6.
$41*2+6=82+6=88$ . This is the compliment. There are a total of $5!=120$ $120-88=32$
Thus, the answer is $\boxed{32}$ | D | 32 |
983a1250302ccf5244e7b36d6e52fb5e | https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | First, note that there is a symmetry from $+,-,+,-$ to $-,+,-,+$ as follows: $a, b, c, d, e \leftrightarrow 6-a, 6-b, 6-c, 6-d, 6-e$ . Now, consider the placement of 5 in the $+,-,+,-$ case. Clearly, 5 is the maximum value, so it must be placed in the 2nd position or the 4th position, but we also have symmetry $a, b, c, d, e \leftrightarrow e, d, c, b, a$ . We will find the number of sequences with 5 in the second position by using casework on the position of the 4.
Case 1:
4 5 __ __ __: 3 is the maximum value, so it must go into the 4th position, leaving us 2 ways to fill in 1 and 2.
Case 2:
__ 5 __ 4 __: 1, 2, and 3 are all less than 4 or 5, so they can be filled in any manner: 3! = 6.
Overall, we have 8 ways to fill in the sequences. By our two types of symmetry, there are $8 \cdot 2 \cdot 2 = \boxed{32}$ | D | 32 |
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