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72d524e23511fbea35c4efd7b67a23c6 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_9 | Find the value of $x$ that satisfies the equation $25^{-2} = \frac{5^{48/x}}{5^{26/x} \cdot 25^{17/x}}.$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 9$ | Manipulate the powers of $5$ in order to get a clean expression.
\[\frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 25^{\frac{17}{x}}} = \frac{5^{\frac{48}{x}}}{5^{\frac{26}{x}} \cdot 5^{\frac{34}{x}}} = 5^{\frac{48}{x}-(\frac{26}{x}+\frac{34}{x})} = 5^{-\frac{12}{x}}\] \[25^{-2} = (5^2)^{-2} = 5^{-4}\] \[5^{-4} = 5^{-\frac{12}{x}}\]
If two numbers are equal, and their bases are equal, then their exponents are equal as well. Set the two exponents equal to each other.
\begin{align*}-4&=\frac{-12}{x}\\ -4x&=-12\\ x&=\boxed{3} | B | 3 |
c77a815a9771a937a243afe42d0ae71e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_11 | A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 2 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 20$ | Using the point-slope form, the equation of each line is
\[y-15=3(x-10) \longrightarrow y=3x-15\] \[y-15=5(x-10) \longrightarrow y=5x-35\]
Substitute in $y=0$ to find the $x$ -intercepts.
\[0=3x-15\longrightarrow x=5\] \[0=5x-35\longrightarrow x=7\] The difference between them is $7-5=\boxed{2}$ | A | 2 |
c77a815a9771a937a243afe42d0ae71e | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_11 | A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 2 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 20$ | The $x$ -intercepts of a line is where the line's $y$ -coordinate is $=0$ . The slope is defined as $\frac{\text{change in } y\text{-coordinate}}{\text{change in } x\text{-coordinate}}$ . Therefore, the line with slope $3$ can be expressed as
\[\frac{\text{change in }y\text{-coordinate}}{\text{change in }x\text{-coordinate}}=3\]
From $(10,15)$ to the $x$ -intercept, the line must move $-15$ units to the $x$ -axis. Therefore, \[\frac{\text{change in }y\text{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_1}=3\implies x_1=-5\]
Therefore, the $x$ -intercept of the line with slope $3$ is $(10-5,0)=(5,0)$ .Note that some sources may state "the $x$ -intercept is $5$ " instead of "the $x$ -intercept is $(5,0)$
If an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope $5$ , we find \[\frac{\text{change in }y\text{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_2}=5\implies x_2=-3\] .
Therefore, the $x$ -intercept of the line with slope $5$ is $(10-3,0)=(7,0)$ . The distance between $\boxed{2}$ | A | 2 |
f5e57ce362f32e58b271ceed83e7002a | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_12 | Al, Betty, and Clare split $\textdollar 1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of $\textdollar 1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\textdollar100$ dollars. What was Al's original portion?
$\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500$ | For this problem, we will have to write a three-variable equation, but not necessarily solve it. Let $a, b,$ and $c$ represent the original portions of Al, Betty, and Clare, respectively. At the end of one year, they each have $a-100, 2b,$ and $2c$ . From this, we can write two equations, marked by (1) and (2).
\[a+b+c=1000\] \[(1). \text{ }2a+2b+2c=2000\] \[a-100+2b+2c=1500\] \[(2). \text{ }a+2b+2c=1600\]
(Equations (1) and (2) are derived from each equation above them.)
Since all we need to find is $a,$ subtract the second equation from the first equation to get $a=400.$
Al's original portion was $\boxed{400}$ | C | 400 |
f5e57ce362f32e58b271ceed83e7002a | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_12 | Al, Betty, and Clare split $\textdollar 1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of $\textdollar 1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\textdollar100$ dollars. What was Al's original portion?
$\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500$ | Suppose the total amount of money Betty and Clare has is $1000-x$ and Al has $x$ dollars. Then, $(x-100)+2(1000-x)=1500$ , so Al has $\boxed{400}$ dollars. | C | 400 |
f5e57ce362f32e58b271ceed83e7002a | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_12 | Al, Betty, and Clare split $\textdollar 1000$ among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of $\textdollar 1500$ dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose $\textdollar100$ dollars. What was Al's original portion?
$\textbf{(A)}\ \textdollar 250 \qquad \textbf{(B)}\ \textdollar 350 \qquad \textbf{(C)}\ \textdollar 400 \qquad \textbf{(D)}\ \textdollar 450 \qquad \textbf{(E)}\ \textdollar 500$ | Suppose if Al had not lost $\textdollar 100$ . The total amount would be $\textdollar 1600$ . As he has not gained any amount. So, Betty and Clare have collectively gained $\textdollar 600$ and as they have doubled their collective fortune,
they must have $\textdollar 600$ with them at the beginning. This leaves $\boxed{400}$ for Al.
~Anshulb | C | 400 |
9b62009686edb7388a87ce2f5b3b0770 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_13 | Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$ . For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$ . For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$ | Let $y=\clubsuit (x)$ . Since $x \leq 99$ , we have $y \leq 18$ . Thus if $\clubsuit (y)=3$ , then $y=3$ or $y=12$ . The 3 values of $x$ for which $\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{10}$ values in all. | E | 10 |
67a43318c7db7425bb04fa13c4d1c538 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_14 | Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$
$\textbf{(A) } 25 \qquad\textbf{(B) } 34 \qquad\textbf{(C) } 351 \qquad\textbf{(D) } 407 \qquad\textbf{(E) } 900$ | \[3^8\cdot5^2 = (3^4)^2\cdot5^2 = (3^4\cdot5)^2 = 405^2\]
$405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=\boxed{407}$ | D | 407 |
da3e7c81d3d9a66c26ee521ad2dc0d4b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_16 | A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$ | Let $m$ be the number of main courses the restaurant serves, so $2m$ is the number of appetizers. Then the number of dinner combinations is $2m\times m\times3=6m^2$ . Since the customer wants to eat a different dinner in all $365$ days of $2003$ , we must have
\begin{align*} 6m^2 &\geq 365\\ m^2 &\geq 60.83\ldots.\end{align*} Also, year 2003 is not a leap year, because 2003 divided by 4 does not equal an integer.
The smallest integer value that satisfies this is $\boxed{8}$ | E | 8 |
da3e7c81d3d9a66c26ee521ad2dc0d4b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_16 | A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$ | Let $m$ denote the number of main courses needed to meet the requirement. Then the number of dinners available is $3\cdot m \cdot 2m = 6m^2$ . Thus $m^2$ must be at least $365/6 \approx 61$ . Since $7^2 = 49<61<64 = 8^2$ $\boxed{8}$ | E | 8 |
da3e7c81d3d9a66c26ee521ad2dc0d4b | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_16 | A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that a restaurant should offer so that a customer could have a different dinner each night in the year $2003$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 7 \qquad\textbf{(E) } 8$ | Let $x$ be the number of main courses and let $2x$ be the number of appetizers.
Since there are 3 desserts, the number of possible dinner choices would be $2x \cdot x \cdot 3 = 6x^2$ for any number $x$ . Since a year has $365$ days, we can assume that: \begin{align*} 6x^2 \ge 365 \end{align*} \begin{align*} x^2 \ge 61 \end{align*} \begin{align*} x \ge 7.8 \end{align*} The least option that is greater than $7.8$ is $8$ , so the answer is $\boxed{8}$ | E | 8 |
0c07a5ca8005605495a7794df53fef07 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_18 | What is the largest integer that is a divisor of
\[(n+1)(n+3)(n+5)(n+7)(n+9)\]
for all positive even integers $n$
$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$ | For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \cdot 5 = 15$ , so ${\boxed{15}$ is the correct answer. | D | 15 |
0c07a5ca8005605495a7794df53fef07 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_18 | What is the largest integer that is a divisor of
\[(n+1)(n+3)(n+5)(n+7)(n+9)\]
for all positive even integers $n$
$\text {(A) } 3 \qquad \text {(B) } 5 \qquad \text {(C) } 11 \qquad \text {(D) } 15 \qquad \text {(E) } 165$ | We'll just test all the answer choices.
Note that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$
Let's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)
To support this further, list the first few consecutive lists of 3 consecutive odd integers.
We have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \ldots$
So if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.
Therefore for five consecutive odd integers, there must be at least one multiple of three.
In the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$
Therefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.
We can skip 7 since none of the answer choices are a multiple of 7.
Now we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$
Since any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$ | D | 15 |
ff4cda11ceff299e43d179d6fe936ff7 | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_22 | A clock chimes once at $30$ minutes past each hour and chimes on the hour according to the hour. For example, at $1 \text{PM}$ there is one chime and at noon and midnight there are twelve chimes. Starting at $11:15 \text{AM}$ on $\text{February 26, 2003},$ on what date will the $2003^{\text{rd}}$ chime occur?
$\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}$ | First, find how many chimes will have already happened before midnight (the beginning of the day) of $\text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+\cdots+12.$ The total number of chimes is $13+78=91.$
Every day, there will be $24$ half-hours and $2(1+2+3+\cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.
On $\text{February 26},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912 \div 180=10 \text{R}112.$ Rounding up, it is $11$ days past $\text{February 26},$ which is $\boxed{9}$ | B | 9 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are $23$ , the sum of the digits is $2+3 = 5$ (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.
$5+1 = 6$
$5+4 = 9$ , and so on.
However since the largest four-digit number ending with $23$ is $9923$ , the maximum sum is
$5+18 = 23$
Using that process we can fairly quickly compile a list of the sum of the first two digits of the number.
\[\{1, 4, 7, 10, 13, 16\}\]
Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers $xy$ in separate cases.
\[I. x+y = 1, \{10\} = 1\] \[II. x+y = 4, \{13, 22, 31, 40\} = 4\] \[III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7\] \[IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9\] \[V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6\] \[VI. x+y = 16, \{79, 88, 97\} = 3\]
And finally, we add the number of elements in each set.
\[1+4+7+9+6+3 = \boxed{30}\] | B | 30 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | A number divisible by $3$ has all its digits add to a multiple of $3.$ The last two digits are $2$ and $3$ and add up to $5 \equiv 2\ (\text{mod}\ 3).$ Therefore the first two digits must add up to $1\ (\text{mod}\ 3).$ $4$ digits (including $0$ ) are $0\ (\text{mod}\ 3),$ $3$ are $1\ (\text{mod}\ 3),$ and $3$ are $2\ (\text{mod}\ 3).$ The following combinations are equivalent to $1\ (\text{mod}\ 3)$
\[0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)\]
Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.
\[3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{30}\] | B | 30 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | We have the following: $n \equiv 0 \pmod{3}$ and $n \equiv 23\pmod{100}$ . Then $n = 3a = 100b+23$ for some integers $a$ and $b$ . Taking mod 3 gives: $0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3$ so $b=3c+1$ for some integer $c$ . But $N = 100b+23 = 100(3c+1)+23$ so $n = 300c+123$ . Bounding this gives us: $999 < 300c+123 < 10000$ so $876 < 300c < 9877$ . Dividing by $300$ gives $2.92 < c < 32.9222$ so $3\le c \le 32$ . This gives $32-3+1 = \boxed{30}$ | B | 30 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | There are $9 \cdot 10 = 90$ possible values for the first two digits. $\frac{1}{3}$ of them yield a multiple of $3$ , so the answer is $\frac{90}{3} = \boxed{30}$ | B | 30 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | Denote the four digit number by $ab23_{10}$ . Then it is well known the sum of the digits of an integer $N$ determine whether or not they are divisible by $3$ . Thus we want $a + b + 2 + 3 \equiv 0 \pmod 3$
$\underline{a = 1}: \implies$ $b \equiv 0 \pmod 3 \implies b = 0, 3, 6, 9$
$\underline{a = 2}: \implies$ $b \equiv 2 \pmod 3 \implies b = 2, 5, 8$
$\underline{a = 3}: \implies$ $b \equiv 1 \pmod 3 \implies b = 1, 4, 7$
Now we have that when $a = 4$ this is the same case as $a = 1$ as $4 \equiv 1 \pmod 3$ . Hence if we count the above number of solutions and multiply by $3$ (as $1 \leq a \leq 9$ ), we are home free. There are $4 + 3 + 3 = 10$ solutions. Multiplying this by $3$ yields $\boxed{30}$ $\blacksquare$ | null | 30 |
bd123b1a66fb6f7ffd59f664c588e52d | https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_25 | How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$ | Don't do this on a test unless you have lots of time left.
Pounding out all of the possible numbers and counting them, we get $\boxed{30}.$ | B | 30 |
7080ca56465758abe9157e656918df2a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_1 | The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$ | We factor $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}$ . As $\frac{101}{20}=5.05$ , our answer is $\boxed{5}$ | D | 5 |
cd3b483fdf38d3b6f5f19df236942974 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_2 | Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ , find $(2, 12, 9)$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | $(2, 12, 9)=\frac{2}{12}+\frac{12}{9}+\frac{9}{2}=\frac{1}{6}+\frac{4}{3}+\frac{9}{2}=\frac{1}{6}+\frac{8}{6}+\frac{27}{6}=\frac{36}{6}=6$ . Our answer is then $\boxed{6}$ | C | 6 |
cd3b483fdf38d3b6f5f19df236942974 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_2 | Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ , find $(2, 12, 9)$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | Without computing the answer exactly, we see that $2/12=\text{a little}$ $12/9=\text{more than }1$ , and $9/2=4.5$ .
The sum is $4.5 + (\text{more than }1) + (\text{a little}) = (\text{more than }5.5) + (\text{a little})$ , and as all the options are integers, the correct one is $\boxed{6}$ | C | 6 |
1f2c907bdd7353805fcf044529c2d1ca | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_3 | According to the standard convention for exponentiation, \[2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.\]
If the order in which the exponentiations are performed is changed, how many other values are possible?
$\textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 2\qquad \textbf{(D) } 3\qquad \textbf{(E) } 4$ | The best way to solve this problem is by simple brute force.
It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\uparrow 2\uparrow 2\uparrow 2$ , where $\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:
We can note that $2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal.
Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
$((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256$
$(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256$
Thus the only other result is $256$ , and our answer is $\boxed{1}$ | B | 1 |
5224566ce602732d3c1883654e518b72 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_5 | Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]
$\textbf{(A)}\ \pi \qquad \textbf{(B)}\ 1.5\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ 3\pi \qquad \textbf{(E)}\ 3.5\pi$ | The outer circle has radius $1+1+1=3$ , and thus area $9\pi$ . The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$ . Thus, our answer is $9\pi-7\pi=\boxed{2}$ | C | 2 |
ec758495da4914cd4096507551c61067 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_6 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qquad \textbf{(D) } 51\qquad \textbf{(E) } 138$ | We work backwards; the number that Cindy started with is $3(43)+9=138$ . Now, the correct result is $\frac{138-3}{9}=\frac{135}{9}=15$ . Our answer is $\boxed{15}$ | A | 15 |
ec758495da4914cd4096507551c61067 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_6 | Cindy was asked by her teacher to subtract 3 from a certain number and then divide the result by 9. Instead, she subtracted 9 and then divided the result by 3, giving an answer of 43. What would her answer have been had she worked the problem correctly?
$\textbf{(A) } 15\qquad \textbf{(B) } 34\qquad \textbf{(C) } 43\qquad \textbf{(D) } 51\qquad \textbf{(E) } 138$ | Let the number be $x$ . We transform the problem into an equation: $\frac{x-9}{3}=43$ . Solve for $x$ gives us $x=138$ . Therefore, the correct result is $\frac{138-3}{9}=\frac{135}{9}=\boxed{15}$ | A | 15 |
831af2fa9afa6e09446adc837f410f7e | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_9 | There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$ | Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$ , we will be done.
Adding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$ . Our answer is $\boxed{3}$ | B | 3 |
831af2fa9afa6e09446adc837f410f7e | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_9 | There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$ | Start by isolating $B$ and $C$ in both of the equations, in order to represent the variables $C$ and $B$ in terms of A. Ending up with the two equations $C = 2A +4$ and $B = -3A + 5$ , we have to calculate the value of the expression $\frac{A+B+C}{3}$ . Plugging in $2A + 4$ for $C$ and $-3A + 5$ for $B$ , we add them up and end up with a value of 9. Dividing 9 by 3 to compute the average, we get our answer of $\boxed{3}$ | B | 3 |
0c29b978688281fcd37e0b151cca03b1 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_11 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$ | $0.8$ MB file can either be on its own disk, or share it with a $0.4$ MB. Clearly it is better to pick the second possibility. Thus we will have $3$ disks, each with one $0.8$ MB file and one $0.4$ MB file.
We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ MB, hence we need at least $10$ more disks. And we can easily verify that $10$ disks are indeed enough: six of them will carry two $0.7$ MB files each, and four will carry three $0.4$ MB files each.
Thus our answer is $3+10 = \boxed{13}$ | B | 13 |
0c29b978688281fcd37e0b151cca03b1 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_11 | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$ | Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with $12$ files( $0.7$ MB), where the optimal number of files per disks is $2$ , so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$ | null | 13 |
be26086d4a31b115475a81a1ebc711c7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_12 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Let the time he needs to get there in be $t$ and the distance he travels be $d$ . From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$ . Setting the two equal, we have $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that $r$ , our answer, is $\boxed{48}$ | B | 48 |
be26086d4a31b115475a81a1ebc711c7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_12 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Since either time he arrives at is $3$ minutes from the desired time, the answer is merely the harmonic mean of 40 and 60.
Substituting $t=\frac ds$ and dividing both sides by $d$ , we get $\frac 2s = \frac 1{40} + \frac 1{60}$ hence $s=\boxed{48}$ | B | 48 |
be26086d4a31b115475a81a1ebc711c7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_12 | Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$ | Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly.
Setting up a system of equations, $\frac x{40} -3 = y$ and $\frac x{60} +3 = y$
Solving, we get x = 720 and y = 15.
We divide x by y to get the average speed, $\frac {720}{15} = 48$ . Therefore, the answer is $\boxed{48}$ | B | 48 |
d4358c9fca34b0244ea247e0ba1b1623 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_13 | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | This is a Pythagorean triple (a $3-4-5$ actually) with legs $15$ and $20$ . The area is then $\frac{(15)(20)}{2}=150$ . Now, consider an altitude drawn to any side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be $x$ ; we have $\frac{(x)(25)}{2}=\frac{(15)(20)}{2}$ , so $25x=300$ and $x$ is 12. Our answer is then $\boxed{12}$ | B | 12 |
d4358c9fca34b0244ea247e0ba1b1623 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_13 | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | By Heron's formula , the area is $150$ , hence the shortest altitude's length is $2\cdot\frac{150}{25}=\boxed{12}$ | B | 12 |
d4358c9fca34b0244ea247e0ba1b1623 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_13 | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | The sides are in the ratio of $3-4-5$ , and are therefore a Pythagorean Triple and a part of a right triangle.
We know two altitudes, the legs $15$ and $20$ . All that is left is to check the altitude from the right angle vertex to the hypotenuse. There is a rule that states that the three triangles formed when the altitude is dropped (the big triangle and two smaller ones it gets split into) are similar! We take the triangle with hypotenuse $20$ . The ratio of sides between that triangle and the bigger one is $4:5$ . Since the side of length $15$ in the big triangle corresponds with our altitude, we take $15\cdot\frac{4}{5}$ to get our answer ${12}$ or $\boxed{12}$ | B | 12 |
d4358c9fca34b0244ea247e0ba1b1623 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_13 | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | The triangle forms a $3-4-5$ right triangle, so we can first draw a triangle with legs $3$ and $4$ and hypotenuse $5$ . The two altitudes $3$ and $4$ are known, and the area is $6$ ; therefore $5x/2 = 6$ , and $x=12/5$ , which is clearly less than $3$ or $4$ . Now scale the triangle back up by multiplying every dimension by $5$ . The smallest altitude is then $12/5 * 5$ ${12}$ or $\boxed{12}$ | B | 12 |
c573fe1fce1fba21304d227ca3726861 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_14 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas ).
We now have that the sum of the two roots is $63$ while the product is $k$ . Since both roots are primes, one must be $2$ , otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$ . Hence, our answer is $\boxed{1}$ | B | 1 |
c573fe1fce1fba21304d227ca3726861 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_14 | Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{1}$ solution. | B | 1 |
8d92a3c7436f1ca2f5f667e27851393d | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_15 | Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$ | Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{190}$ | E | 190 |
b148b51330e16497771e9cf4fc1b97b2 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_17 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$ | We will simulate the process in steps.
In the beginning, we have:
In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:
In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:
Hence at the end we have $3+2=5$ ounces of liquid in cup $1$ , and out of these $2$ ounces is cream. Thus the answer is $\boxed{25}$ | D | 25 |
b148b51330e16497771e9cf4fc1b97b2 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_17 | Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$ | Let's consider this in steps.
We have 4 ounces of coffee in the first cup.
We hace 4 ounces of cream in the second cup.
We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).
We then take half of the second cup and pour it into the first cup. $6/2=3$ , so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.
Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces.
In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{25}$ | D | 25 |
d9ee4a63c3f4b973cf36a82edd8b3e87 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_18 | A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$ | In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$ , with two faces showing is $1+2=3$ , and with one face showing is $1$ . Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$ . Our answer is thus $\boxed{90}$ | D | 90 |
eb4ca67890ae55f6a6e9f6063730dbec | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_21 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{14}$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM | D | 14 |
eb4ca67890ae55f6a6e9f6063730dbec | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_21 | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$ | We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.
Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.
There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$
We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.
Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$
$a\textsubscript{1}$ $a\textsubscript{2}$ $a\textsubscript{3}$ $a\textsubscript{4}$ $a\textsubscript{5}$ $a\textsubscript{6}$ $a\textsubscript{7}$ $a\textsubscript{8}$ $=$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $8$ $8$ $8$ $8$ $a\textsubscript{8}$
Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{14}$ . ~By SK80, mod_x for minor edits | D | 14 |
e878cbb9cfcbd86c02d832acf0f3b2cf | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_22 | A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$ | The pattern is quite simple to see after listing a couple of terms.
\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ 18&1&1\\ \hline \end{tabular}\]
Thus, the answer is $\boxed{18}$ | C | 18 |
e878cbb9cfcbd86c02d832acf0f3b2cf | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_22 | A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$ | Given $n^2$ tiles, a step removes $n$ tiles, leaving $n^2 - n$ tiles behind. Now, $(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2$ , so in the next step $n-1$ tiles are removed. This gives $(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2$ , another perfect square.
Thus each two steps we cycle down a perfect square, and in $(10-1)\times 2 = 18$ steps, we are left with $1$ tile, hence our answer is $\boxed{18}$ | C | 18 |
e878cbb9cfcbd86c02d832acf0f3b2cf | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_22 | A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$ | We start of with $100 = 10 \cdot 10$ numbers. When we use the certain operation, call if $P(x)$ , have $100 - 10 = 90 = 10 \cdot 9$ .
Then we do $P(x)$ again, to subtract $9$ numbers and get $9 \cdot 9$ . In the end, we will want $1 = 1 \cdot 1$ . We can say we have to use $P(x)$ once to make $n \cdot n$ into $n \cdot (n-1)$ . Thus we must use it twice to get from $n \cdot n$ to $(n-1)(n-1)$ . For example, it takes us $2$ of $P(x)$ to get from $10 \cdot 10$ to $9 \cdot 9$ . Then $2$ of $P(x)$ to get from $9 \cdot 9$ to $8 \cdot 8$ . You should try this with $7$ and $6$ , and see it works. This means we can have $n$ be the number we start with, and $1$ be the number we want. Then we would use $P(x)$ $2(n - 1)$ times to get $1 \cdot 1$ . Substituting $n$ for $10$ we get $2(10-1) = 2 \cdot 9 = \boxed{18}$ .
- Wiselion | null | 18 |
8038edb0d2ef9391799e0fbc2e028d7d | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_23 | Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$ | Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\cdot 15)$ this works, giving us $AB=15-6=\boxed{9}.$ | D | 9 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$
Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$ , with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$ . Thus \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of $\triangle CDE$ are $15,36,39$ , which we recognize to be a $5 - 12 - 13$ right triangle . Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), \[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210}\] | C | 210 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | Draw altitudes from points $C$ and $D$
Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$ . We get the following triangle:
The length of $A'B$ in this triangle is equal to the length of the original $AB$ , minus the length of $CD$ .
Thus $A'B = 52 - 39 = 13$
Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$ , and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$
Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}$ | C | 210 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | Draw altitudes from points $C$ and $D$
Call the length of $AD'$ to be $y$ , the length of $BC'$ to be $z$ , and the height of the trapezoid to be $x$ .
By the Pythagorean Theorem, we have: \[z^2 + x^2 = 144\] \[y^2 + x^2 = 25\]
Subtracting these two equation yields: \[z^2-y^2=119 \implies (z+y)(z-y)=119\]
We also have: $z+y=52-39=13$
We can substitute the value of $z+y$ into the equation we just obtained, so we now have:
\[(13) (z-y)=119 \implies z-y=\frac{119}{13}\]
We can add the $z+y$ and the $z-y$ equation to find the value of $z$ , which simplifies down to be $\frac{144}{13}$ . Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.
\[\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}\]
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$ , and multiplying this and the height of the trapezoid gets us:
\[\frac{60 \cdot 91}{13 \cdot 2} = \boxed{210}\] | C | 210 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$
By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{210}.$ | C | 210 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$ .
Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$ .
The numbers multiply nicely to get $180+30=\boxed{210}$ -harsha12345 | C | 210 |
e564253d48a45f0623a55852032179ad | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10A_Problems/Problem_25 | In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | First note how the answer choices are all integers.
The area of the trapezoid is $\frac{39+52}{2} \cdot h = \frac{91}{2} h$ . So h divides 2. Let $x$ be $2h$ . The area is now $91x$ .
Trying $x=1$ and $x=2$ can easily be seen to not work. Those make the only integers possible so now you know x is a fraction.
Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13\cdot7$ .
Seeing how $39 = 3\cdot13$ and $52 = 4\cdot13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$ .
Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean $h$ is $4$ which we ruled out.
So the answer is $\boxed{210}$ . - megateleportingrobots | C | 210 |
48d9b2f212fa7ee52c3f21ed0b7fb05f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_3 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | We wish to find $\frac{9+99+\cdots +999999999}{9}$ , or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$ . This doesn't have the digit 0, so the answer is $\boxed{0}$ | A | 0 |
48d9b2f212fa7ee52c3f21ed0b7fb05f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_3 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | Notice that the final number is guaranteed to have the digits $\{1, 3, 5, 7, 9\}$ and that each of these digits can be paired with an even number adding up to 9. $\boxed{0}$ can be taken out, with the other digits fulfilling divisibility by 9. | A | 0 |
48d9b2f212fa7ee52c3f21ed0b7fb05f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_3 | The arithmetic mean of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$ -digit number $M$ , all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 4 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 8$ | The arithmetic mean is $\frac{(10^1-1)+(10^2-1)+\ldots+(10^9-1)}{9}=\frac{1111111101}{9}=123456789$ . So select $\boxed{0}$ .
~hastapasta | A | 0 |
85101534483243090c77e53eebc09051 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_4 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | By the distributive property,
\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{11}\] | D | 11 |
85101534483243090c77e53eebc09051 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_4 | What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | Inputting 4 into $x$ in the original equation,
\[[3(4)-2][4(4)+1]-[3(4)-2][4(4)]+1 = 170-160+1 = \boxed{11}\] | D | 11 |
7ddb599a96fc64e481b8f7c0d95f01a5 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_5 | Circles of radius $2$ and $3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
$\mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi$ | A line going through the centers of the two smaller circles also goes through the diameter. The length of this line within the circle is $3+3+2+2=10.$ Because this is the length of the larger circle's diameter, the length of its radius is $5.$
The area of the large circle is $25\pi$ , and the area of the two smaller circles is $9\pi + 4\pi = 13\pi.$ To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. $\longrightarrow 25\pi - 13\pi = \boxed{12}$ | E | 12 |
cb12bffe4e0299ad59e711ff25772930 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_7 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$ | Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$ $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$
From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $\boxed{84}$ | E | 84 |
cb12bffe4e0299ad59e711ff25772930 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_7 | Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$ | Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$ , it is very clear that $n=42$ makes the expression an integer. Because $n$ is a positive integer, $\frac{1}{n}$ must be less than or equal to $1$ . Thus, the only integer the expression can take is $1$ , which means the only value for $n$ is $42$ . Thus $\boxed{84}$ | E | 84 |
7114850dae4a1af7e04feb89f632bca7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_9 | Using the letters $A$ $M$ $O$ $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position
$\mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116$ | There are $4!\cdot 4$ "words" beginning with each of the first four letters alphabetically. From there, there are $3!\cdot 3$ with $U$ as the first letter and each of the first three letters alphabetically. After that, the next "word" is $USAMO$ , hence our answer is $4\cdot 4!+3\cdot 3!+1=\boxed{115}$ | D | 115 |
7114850dae4a1af7e04feb89f632bca7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_9 | Using the letters $A$ $M$ $O$ $S$ , and $U$ , we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" $USAMO$ occupies position
$\mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116$ | Let $A = 1$ $M = 2$ $O = 3$ $S = 4$ , and $U = 5$ . Then counting backwards, $54321, 54312, 54231, 54213, 54132, 54123$ , so the answer is $\boxed{115}$ | D | 115 |
860b154a3c2c59591d61c6c5a682305f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_11 | The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$ | Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$ $24=4\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{77}$ | B | 77 |
25b048bf4375a0866f5a3589cb756061 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_12 | For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | The domain over which we solve the equation is $\mathbb{R} \setminus \{2,6\}$
We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$
Simplifying that, we get $7x-6 = (k+2)x - 2k$ . Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $\boxed{5}$ | E | 5 |
8dd4b0aaafed039260c6e61dbb0fb90f | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_13 | Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$
$\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14$ | We have $8xy - 12y + 2x - 3 = 4y(2x - 3) + (2x - 3) = (4y + 1)(2x - 3)$
As $(4y + 1)(2x - 3) = 0$ must be true for all $y$ , we must have $2x - 3 = 0$ , hence $\boxed{32}$ | D | 32 |
c834173a57dbc5800025827050723917 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_14 | The number $25^{64}\cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is
$\mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35$ | Taking the root, we get $N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}$
Now, we have $N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}$
Combining the $2$ 's and $5$ 's gives us $(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}$
This is the number $2048$ with a string of sixty-four $0$ 's at the end. Thus, the sum of the digits of $N$ is $2+4+8=14\Longrightarrow\boxed{14}$ | B | 14 |
719e81f78e3bd224d1dcb2cabb806565 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_16 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | Let $x^2 = \frac{n}{20-n}$ , with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$ ). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$ . Since $\text{gcd}(x^2, x^2 + 1) = 1$ , it follows that $x^2 + 1$ divides $20$ . Listing the factors of $20$ , we find that $x = 0, 1, 2 , 3$ are the only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ). | D | 4 |
719e81f78e3bd224d1dcb2cabb806565 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_16 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | For $n<0$ and $n>20$ the fraction is negative, for $n=20$ it is not defined, and for $n\in\{1,\dots,9\}$ it is between 0 and 1.
Thus we only need to examine $n=0$ and $n\in\{10,\dots,19\}$
For $n=0$ and $n=10$ we obviously get the squares $0$ and $1$ respectively.
For prime $n$ the fraction will not be an integer, as the denominator will not contain the prime in the numerator.
This leaves $n\in\{12,14,15,16,18\}$ , and a quick substitution shows that out of these only $n=16$ and $n=18$ yield a square. Therefore, there are only $\boxed{4}$ solutions (respectively yielding $n = 0, 10, 16, 18$ ). | D | 4 |
719e81f78e3bd224d1dcb2cabb806565 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_16 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | If $\frac{n}{20-n} = k^2 \ge 0$ , then $n \ge 0$ and $20-n > 0$ , otherwise $\frac{n}{20-n}$ will be negative. Thus $0 \le n \le 19$ and \[0 = \frac{0}{20-(0)} \le \frac{n}{20-n} \le \frac{19}{20-(19)} = 19\] Checking all $k$ for which $0 \le k^2 \le 19$ , we have $0$ $1$ $2$ $3$ as the possibilities. $\boxed{4}$ | D | 4 |
719e81f78e3bd224d1dcb2cabb806565 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_16 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | For all integers x, $x^2$ is always a positive integer. So solve for $\frac{n}{20-n} = 0$ , getting $n=0$ and $\frac{n}{20-n} = 1$ , getting $n =10$ . For all values n less than 0 and greater than 20, the value $\frac{n}{20-n}$ is negative, so now try values of n between 10 and 20. Quick substitution finds $0$ $10$ $16$ , and $18$ which yields $x=0$ $x=1$ $x=2$ , and $x=3$ respectively. 4 values, or $\boxed{4}$ | D | 4 |
719e81f78e3bd224d1dcb2cabb806565 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_16 | For how many integers $n$ is $\dfrac n{20-n}$ the square of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | Simon's Favourite Factoring Trick
Since \frac{n}{20-n} is an integer k, w multiple both sides by 20-n. This gives us $n$ $20k^2$ $nk^2$ . We subtract $20k^2$ on both sides, then add $nk^2$ on both sides as a prerequisite for using Simon's Favorite Factoring Trick. We have (k^2+1)(n-20)=20. We then consider the different factors of 20 that k^2+1 can be. It could be $1$ $2$ $4$ $5$ $10$ , and $20$ . After checking case by case, we then are able to identify that there are 4 such k values that also yield an integer n value, meaning that there are 4 values, so the correct answer is $\boxed{4}$ ~CharmaineMa07292010 | D | 4 |
15bb34fe69f2f1bd9fb6c9cf0e38db42 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_18 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$ . We can construct such a situation as below, so the answer is $\boxed{12}$ | D | 12 |
15bb34fe69f2f1bd9fb6c9cf0e38db42 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_18 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | Because a pair or circles can intersect at most $2$ times, the first circle can intersect the second at $2$ points, the third can intersect the first two at $4$ points, and the fourth can intersect the first three at $6$ points. This means that our answer is $2+4+6=\boxed{12}.$ | D | 12 |
15bb34fe69f2f1bd9fb6c9cf0e38db42 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_18 | Four distinct circles are drawn in a plane . What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 16$ | Pick a circle any circle- $4$ ways. Then, pick any other circle- $3$ ways. For each of these circles, there will be $2$ intersections for a total of $4*3*2$ $24$ intersections. However, we have counted each intersection twice, so we divide for overcounting. Therefore, we reach a total of $\frac{24}{2}=\boxed{12}$ | D | 12 |
00b21382518be765d0268d61c0654b31 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_19 | Suppose that $\{a_n\}$ is an arithmetic sequence with \[a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.\] What is the value of $a_2 - a_1 ?$
$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$ | We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
\[(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100\]
...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...
$\frac{100}{100} = 1$
...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...
$\frac{1}{100} =\boxed{0.01}$ | C | 0.01 |
00b21382518be765d0268d61c0654b31 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_19 | Suppose that $\{a_n\}$ is an arithmetic sequence with \[a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.\] What is the value of $a_2 - a_1 ?$
$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$ | Adding the two given equations together gives
$a_1+a_2+...+a_{200}=300$
Now, let the common difference be $d$ . Notice that $a_2-a_1=d$ , so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is
$\frac{n}{2}(2a_1+d(n-1))$
where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$ . Therefore, we have
$50(2a_1+99d)=100$
or
$2a_1+99d=2$ . * (1)
For the sum of the equations (shown at the beginning of the solution) we have $n=200$ , so
$100(2a_1+199d)=300$
or
$2a_1+199d=3$ (2)
Now we have a system of equations in terms of $a_1$ and $d$ .
Subtracting (1) from (2) eliminates $a_1$ , yielding $100d=1$ , and $d=a_2-a_1=\frac{1}{100}=\boxed{0.01}$ | C | 0.01 |
00b21382518be765d0268d61c0654b31 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_19 | Suppose that $\{a_n\}$ is an arithmetic sequence with \[a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.\] What is the value of $a_2 - a_1 ?$
$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$ | Subtracting the 2 given equations yields
$(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$
Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms
$((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$
Simplifying and canceling $a_1$ and $x$ terms gives
$100x+100x+100x+...+100x=100$
$100x\times100=100$
$100x=1$
$x=0.01=\boxed{0.01}$ | C | 0.01 |
aea60e49f072267662890a9936082b8d | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_20 | Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$
Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.
Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=\boxed{1}$ | B | 1 |
aea60e49f072267662890a9936082b8d | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_20 | Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | The easiest way is to assume a value for $a$ and then solve the system of equations. For $a = 1$ , we get the equations $-7b + 8c = 3$ and $4b - c = -1$ Multiplying the second equation by $8$ , we have $32b - 8c = -8$ Adding up the two equations yields $25b = -5$ , so $b = -\frac{1}{5}$ We obtain $c = \frac{1}{5}$ after plugging in the value for $b$ .
Therefore, $a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}$ .
This time-saving trick works only because we know that for any value of $a$ $a^2-b^2+c^2$ will always be constant (it's a contest), so any value of $a$ will work. This is also called without loss of generality or WLOG. | B | 1 |
aea60e49f072267662890a9936082b8d | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_20 | Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | Notice that the coefficients of $a$ and $c$ are pretty similar (15s for reading and noticing), so let $b=0$ gives $a+8c=4$ , and $8a-c=7$ (10s writing). Since the desired quantity simplifies to $a^2+c^2$ , the $ac$ term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to $65(a^2+c^2)=16+49$ , or $a^2-b^2+c^2=\frac{65}{65}=\boxed{1}$ (15s writing and addition and fraction simplification and (B) circling and submission) | null | 1 |
74335f83b73b4c979c3171d969d999ef | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_22 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | 2002 12B AMC-20.png
Let $OM = x$ $ON = y$ . By the Pythagorean Theorem on $\triangle XON, MOY$ respectively, \begin{align*} (2x)^2 + y^2 &= 19^2\\ x^2 + (2y)^2 &= 22^2\end{align*}
Summing these gives $5x^2 + 5y^2 = 845 \Longrightarrow x^2 + y^2 = 169$
By the Pythagorean Theorem again, we have
\[(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{26}\] | B | 26 |
74335f83b73b4c979c3171d969d999ef | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_22 | Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$
$\mathrm{(A)}\ 24 \qquad\mathrm{(B)}\ 26 \qquad\mathrm{(C)}\ 28 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 32$ | Let $XO=x$ and $YO=y.$ Then, $XY=\sqrt{x^2+y^2}.$
Since $XN=19$ and $YM=22,$ \[XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2\] \[YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.\]
Adding these up: \[19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}\] \[845=\dfrac{5x^2+5y^2}{4}\] \[3380=5x^2+5y^2\] \[676=x^2+y^2.\]
Then, we substitute: $XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.$ | null | 26 |
0a69407007a93b53c8ff9957d4e05d5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | When $m=1$ $a_{n+1}=1+a_n+n$ . Hence, \[a_{2}=1+a_1+1\] \[a_{3}=1+a_2+2\] \[a_{4}=1+a_3+3\] \[\dots\] \[a_{12}=1+a_{11}+11\] Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\boxed{78}$ | D | 78 |
0a69407007a93b53c8ff9957d4e05d5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ $a_{m+1}=a_m+a_{1}+m$
Since $a_1 = 1$ $a_{m+1}=a_m+m+1$
Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$ , and so on until $a_2 = a_1 + 2$
Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$
Adding the Right Hand Sides of these equations gives
$(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$
These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)$
Substituting $a_1 = 1$ $a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$
Thus we have a general formula for $a_m$ and substituting $m=12$ $a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{78}$ | D | 78 |
0a69407007a93b53c8ff9957d4e05d5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since $a_{m+n} = a_m+a_n +mn$ , we know $a_2=a_1+a_1+1\cdot1=1+1+1=3$ . After this, we can use $a_2$ to find $a_4$ $a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10$ . Now, we can use $a_2$ and $a_4$ to find $a_6$ , or $a_6=a_4+a_2+4\cdot 2 = 10+3+8=21$ . Lastly, we can use $a_6$ to find $a_{12}$ $a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{78}$ | D | 78 |
0a69407007a93b53c8ff9957d4e05d5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | We can set $n$ equal to $m$ , so we can say that \[a_{m + m} = a_m + a_m + m*m\] \[a_{2m} = 2a_m + m^2\]
We set $2m = 12$ , we get $m = 6$ \[a_{12} = 2a_6 + 36\]
We set $2m = 6$ m, we get $m = 3$ \[a_6 = 2a_3 + 9\]
Solving for $a_3$ is easy, just direct substitution. \[a_2 = 1 + 1 + 1 = 3\] \[a_3 = a_{2 + 1} = 3 + 1 + 2 = 6\]
Substituting, we get \[a_6 = 2(6) + 9 = 21\] \[a_{12} = 2(21) + 36 = 78\]
Thus, the answer is $\boxed{78}$ | D | 78 |
0a69407007a93b53c8ff9957d4e05d5a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_23 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | Note that the sequence of triangular numbers $T_n=1+2+3+...+n$ satisfies these conditions. It is immediately obvious that it satisfies $a_1=1$ , and $a_{m+n}=a_m+a_n+mn$ can be visually proven with the diagram below.
[asy] for(int i=5; i > 0; --i) { for(int j=0; j < i; ++j) { draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2)); }; }; path m1 = brace((2,-.3),(0,-.3),.2); draw(m1); label("$m$",m1,S); path n1 = brace((4,-.3),(3,-.3),.2); draw(n1); label("$n$",n1,S); draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle); label("$T_m$",(1,1/3*sqrt(3))); draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle); label("$T_n$",(3.5,.5/3*sqrt(3))); path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2); draw(m2); label("$m$",m2,(.5*sqrt(3),.5)); path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2); draw(n2); label("$n$",n2,(-.5*sqrt(3),.5)); draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle); label("$mn$",(2.25,1.25*sqrt(3))); [/asy]
This means that we can use the triangular number formula $T_n = \frac{n(n+1)}{2}$ , so the answer is $T_{12} = \frac{12(12+1)}{2} = \boxed{78}$ | D | 78 |
95bebbccb9a41784a02967f5034519b7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_24 | Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$ | We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{10}$ . Alternatively, we could also say that $\triangle ABC$ is congruent to $\triangle OBC$ by SAS, so $AC$ is 20, and $\triangle AOC$ is equilateral, and $\angle BOC = 60^\circ$ | D | 10 |
95bebbccb9a41784a02967f5034519b7 | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_24 | Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$ | The path that the rider takes along the Ferris wheel can be represented by a sinusoidal graph, where $x$ represents the time in seconds. Since $x=0$ is at the crest of the graph and not at the midline, we will use a cosine graph. Therefore, we will use the form: \[f(x) = A\cos(Bx - C) + D.\]
The graph starts at the lowest point at $0$ feet, then goes up to reach the highest point at $40$ feet, then comes back down. Therefore, the amplitude is $20$ (and negative since it starts at the bottom, not the top), and the vertical shift is $20$ . There is no horizontal shift since the lowest point is at $x=0$ . It takes $60$ seconds to make one full revolution (the period), so $B = \frac{2\pi}{60} = \frac{\pi}{30}.$ Now we have all the parts we need for the equation of our graph, and we can set it equal to the height we want, $10.$
\[10 = -20\cos\left(\frac{\pi}{30}x\right) + 20.\]
We get to $\frac{1}{2} = \cos\left(\frac{\pi}{30}x\right)$ and remember that the problem is looking for the first instance of $f(x) = 10$ , so $\frac{\pi}{30}x = \frac{\pi}{3}.$ Solving, we get that $x = \boxed{10}$ | D | 10 |
afa111e916fc093ea3e555516db31f4a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_25 | When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $\dfrac{x+15}{y+1}=\dfrac{x}{y}+2$ and $\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1$ . With a lot of algebra, the solution is found to be $y= \boxed{4}$ | A | 4 |
afa111e916fc093ea3e555516db31f4a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_25 | When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | We let $n$ be the original number of elements in the set and we let $m$ be the original average of the terms of the original list. Then we have $mn$ is the sum of all the elements of the list. So we have two equations: \[mn+15=(m+2)(n+1)=mn+m+2n+2\] and \[mn+16=(m+1)(n+2)=mn+2m+n+2.\] Simplifying both equations and we get, \[13=m+2n\] \[14=2m+n\] Solving for $m$ and $n$ , we get $m=5$ and $n=\boxed{4}$ | A | 4 |
afa111e916fc093ea3e555516db31f4a | https://artofproblemsolving.com/wiki/index.php/2002_AMC_10B_Problems/Problem_25 | When $15$ is appended to a list of integers, the mean is increased by $2$ . When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$ . How many integers were in the original list?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$ | Warning: This solution will rarely ever work in any other case. However, seeing that you can so easily plug and chug in problem 25 it is funny to see this.
Plug and chug random numbers with the answer choices, starting with the choice of $4$ numbers. You see that if you have 4 5s and you add 15 to the set, the resulting mean will be 7; we can verify this with math \[\frac{5+5+5+5+15}{5}=7\] adding in 1 to the set you result in the mean to be 6. \[\frac{5+5+5+5+15+1}{6}=6\] Thus we conclude that 4 is the correct choice or $\boxed{4}$ | A | 4 |
99f6758b23f2ba5bdead09e7701ada7d | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_1 | The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$
We substitute the median for $10$ and the equation becomes $n+6=10$
Subtract both sides by 6 and we get $n=4$
$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$
The mean of those numbers is $\frac{9n+63}{9}$ which is $n+7$
Substitute $n$ for $4$ and $4+7=\boxed{11}$ | E | 11 |
5f85fc521487321633a77c83f6b9d526 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_4 | What is the maximum number of possible points of intersection of a circle and a triangle?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$ | Circle-triangle problem.PNG
We can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.
You would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.
Therefore, $2 \times 3 = \boxed{6}$ | E | 6 |
5f85fc521487321633a77c83f6b9d526 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_4 | What is the maximum number of possible points of intersection of a circle and a triangle?
$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$ | We know that the maximum amount of points that a circle and a line segment can intersect is $2$ . Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $2 \times 3 = \boxed{6}$ | E | 6 |
ad9132a9527dfe3eb903e4a6cb284999 | https://artofproblemsolving.com/wiki/index.php/2001_AMC_10_Problems/Problem_5 | How many of the twelve pentominoes pictured below have at least one line of reflectional symmetry?
[asy] unitsize(5mm); defaultpen(linewidth(1pt)); draw(shift(2,0)*unitsquare); draw(shift(2,1)*unitsquare); draw(shift(2,2)*unitsquare); draw(shift(1,2)*unitsquare); draw(shift(0,2)*unitsquare); draw(shift(2,4)*unitsquare); draw(shift(2,5)*unitsquare); draw(shift(2,6)*unitsquare); draw(shift(1,5)*unitsquare); draw(shift(0,5)*unitsquare); draw(shift(4,8)*unitsquare); draw(shift(3,8)*unitsquare); draw(shift(2,8)*unitsquare); draw(shift(1,8)*unitsquare); draw(shift(0,8)*unitsquare); draw(shift(6,8)*unitsquare); draw(shift(7,8)*unitsquare); draw(shift(8,8)*unitsquare); draw(shift(9,8)*unitsquare); draw(shift(9,9)*unitsquare); draw(shift(6,5)*unitsquare); draw(shift(7,5)*unitsquare); draw(shift(8,5)*unitsquare); draw(shift(7,6)*unitsquare); draw(shift(7,4)*unitsquare); draw(shift(6,1)*unitsquare); draw(shift(7,1)*unitsquare); draw(shift(8,1)*unitsquare); draw(shift(6,0)*unitsquare); draw(shift(7,2)*unitsquare); draw(shift(11,8)*unitsquare); draw(shift(12,8)*unitsquare); draw(shift(13,8)*unitsquare); draw(shift(14,8)*unitsquare); draw(shift(13,9)*unitsquare); draw(shift(11,5)*unitsquare); draw(shift(12,5)*unitsquare); draw(shift(13,5)*unitsquare); draw(shift(11,6)*unitsquare); draw(shift(13,4)*unitsquare); draw(shift(11,1)*unitsquare); draw(shift(12,1)*unitsquare); draw(shift(13,1)*unitsquare); draw(shift(13,2)*unitsquare); draw(shift(14,2)*unitsquare); draw(shift(16,8)*unitsquare); draw(shift(17,8)*unitsquare); draw(shift(18,8)*unitsquare); draw(shift(17,9)*unitsquare); draw(shift(18,9)*unitsquare); draw(shift(16,5)*unitsquare); draw(shift(17,6)*unitsquare); draw(shift(18,5)*unitsquare); draw(shift(16,6)*unitsquare); draw(shift(18,6)*unitsquare); draw(shift(16,0)*unitsquare); draw(shift(17,0)*unitsquare); draw(shift(17,1)*unitsquare); draw(shift(18,1)*unitsquare); draw(shift(18,2)*unitsquare);[/asy]
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | Pentonimoes.gif
The ones with lines over the shapes have at least one line of symmetry. Counting the number of shapes that have line(s) on them,
we find $\boxed{6}$ pentominoes. | D | 6 |
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