id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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3,601 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | 1 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | First, let $R(n)$ be the sum of the $n$ th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$
Now we try to prove it by induction,
$R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case)
$R(k) = 2^k - k$
$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$
By definition from the question, the n... | 5 |
3,602 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | 2 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | Let the sum of the numbers in row $2022$ be $S_{2022}$ . Let each number in row $2022$ be $x_i$ where $1 \leq i \leq 2022$
Then \begin{align*} S_{2023}&=1+(x_1+x_2+1)+(x_2+x_3+1)+...+(x_{2021}+x_{2022}+ 1)+1 \\ S_{2023}&=x_1+2(S_{2022}-x_1-x_{2022})+2023+x_{2022} \\ S_{2023} &= 2S_{2022} + 2021 \end{align*} From this w... | 5 |
3,603 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | 3 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | Let the sum of the $n^{th}$ row be $S_n$
For each of the $n-2$ non-1 entries in the $n^{th}$ row, they are equal to the sum of the $2$ numbers diagonally above it in the $n-1^{th}$ row plus $1$ . Therefore all $n-3$ non-1 entries in the $n-1^{th}$ row appear twice in the sum of the $n-2$ non-1 entries in the $n^{th}$ r... | 5 |
3,604 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | 4 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the $n$ th row is $2^{(n-1)}$ . For the 2023rd row in the Pascal's triangle, the sum of numbers is $2^{2022}$
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); ... | 5 |
3,605 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 | 5 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label(... | Observe that: $S_n = 2{S_{n-1}} + (n-2) \\ = 2{S_{n-1}} + D_{n} \\ = 2{S_{n-1}} + D_{n-1} + 1$ where $D_1=-1$ , and $D_n=D_{n-1}+1$ for $n>1$
Make a table of values, $(S_n, D_n) \mod 2$ and $(S_n, D_n) \mod 5$ , a until the cycle completes for each. ( $\mod 2$ has period $1$ $\mod 5$ has period $20$ ).
( In both cases... | 5 |
3,606 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_21 | 1 | If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\... | Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and
S basically the same, we can first count the probability that $d(Q,R) = d(R,S)$
$\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1$
There are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \times 4 = \b... | 20 |
3,607 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | 1 | Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | First, we note that $f(1) = 1$ , since the only divisor of $1$ is itself.
Then, let's look at $f(p)$ for $p$ a prime. We see that \[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\] \[1 \cdot f(p) + p \cdot f(1) = 1\] \[f(p) = 1 - p \cdot f(1)\] \[f(p) = 1-p\] Nice.
Now consider $f(p^k)$ , for $k \in \mathbb{N}$ ... | 96 |
3,608 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | 2 | Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | First, change the problem into an easier form. \[\sum_{d\mid n}d\cdot f(\frac{n}{d} )=\sum_{d\mid n}\frac{n}{d}f(d)=1\] So now we get \[\frac{1}{n}= \sum_{d\mid n}\frac{f(d)}{d}\] Also, notice that both $\frac{f(d)}{d}$ and $\frac{1}{n}$ are arithmetic functions. Applying Möbius inversion formula, we get \[\frac{f(n)}{... | 96 |
3,609 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | 3 | Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | From the problem, we want to find $f(2023)$ . Using the problem, we get $f(2023)+7f(289)+17f(119)+119f(17)+289f(7)+2023f(1)=1$ . By plugging in factors of $2023$ , we get \begin{align} f(7)+7f(1)=1\\ f(17)+17f(1)=1\\ f(119)+7f(17)+17f(7)+119f(1)=1\\ f(289)+17f(17)+289f(1)=1 \end{align} Notice that $(4)-17(2)=f(289)$ , ... | 96 |
3,610 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 | 4 | Let $f$ be the unique function defined on the positive integers such that \[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\] for all positive integers $n$ . What is $f(2023)$
$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$ | Consider any $n \in \Bbb N$ with prime factorization $n = \Pi_{i=1}^k p_i^{\alpha_i}$ .
Thus, the equation given in this problem can be equivalently written as \[ \sum_{\beta_1 = 0}^{\alpha_1} \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_k = 0}^{\alpha_k} \Pi_{i=1}^k p_i^{\alpha_i - \beta_i} \cdot f \left( \Pi_{i=1... | 96 |
3,611 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_23 | 1 | How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$ | Using AM-GM on the two terms in each factor on the left, we get \[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\] meaning the equality condition must be satisfied. This means $1 = 2a = b$ , so we only have $\boxed{1}$ solution. | 1 |
3,612 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_23 | 2 | How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation \[(1+2a)(2+2b)(2a+b) = 32ab?\]
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{an infinite number}$ | Equation $(1+2a)(2+2b)(2a+b)=32ab$ is equivalent to \[b(2a-1)^2+2a(b-1)^2+(2a-b)^2=0,\] where $a$ $b>0$ . Therefore $2a-1=b-1=2a-b=0$ , so $(a,b)=\left(\tfrac12,1\right)$ . Hence the answer is $\boxed{1}$ | 1 |
3,613 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 1 | Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $... | Consider any sequence with $n$ terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$ th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence... | 5 |
3,614 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 3 | Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $... | Seeing that all the answers are different modulus 5, and that 10 is divisible by 5, we cheese this problem.
Let $A_1, A_2, \cdots, A_n$ be one sequence satisfying the constraints of the problem. Let $b_1, b_2, \cdots, b_n, b_{n+1}$ be the sequence of nonnegative integers such that $A_k$ has $b_k$ elements for all $1\le... | 5 |
3,615 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24 | 4 | Let $K$ be the number of sequences $A_1$ $A_2$ $\dots$ $A_n$ such that $n$ is a positive integer less than or equal to $10$ , each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$ , and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$ , inclusive. For example, $\{\}$ $\{5, 7\}$ $\{2, 5, 7\}$ $\{2, 5, 7\}$ $... | We observe that in each sequence, if element $e \in A_i$ , then $e \in A_j$ for all $j \geq i$ .
Therefore, to determine a sequence with a fixed length $n$ , we only need to determine the first set $A_i$ that each element in $\left\{ 1, 2, \cdots , 10 \right\}$ is inserted into, or an element is never inserted into any... | 5 |
3,616 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 1 | There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qqua... | \begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2023} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023}... | 1 |
3,617 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 2 | There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qqua... | Note that $\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}$ , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test,... | 1 |
3,618 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25 | 3 | There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qqua... | For odd $n$ , we have \begin{align*} \tan nx & = \frac{\sin nx}{\cos nx} \\ & = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)} {\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \\ & = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\ & = - i \frac{\left( \cos x + i \sin x \right)^n - \left( \co... | 1 |
3,619 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1 | 1 | Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four... | Given that the first three glasses are full and the fourth is only $\frac{1}{3}$ full, let's represent their contents with a common denominator, which we'll set as 6. This makes the first three glasses $\dfrac{6}{6}$ full, and the fourth glass $\frac{2}{6}$ full.
To equalize the amounts, Mrs. Jones needs to pour juice ... | 16 |
3,620 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1 | 2 | Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four... | We let $x$ denote how much juice we take from each of the first $3$ children and give to the $4$ th child.
We can write the following equation: $1-x=\dfrac13+3x$ , since each value represents how much juice each child (equally) has in the end. (Each of the first three children now have $1-x$ juice, and the fourth child... | 16 |
3,621 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | 1 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | We can create the equation: \[0.8x \cdot 1.075 = 43\] using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get \[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\] \[\frac{4}{5} \c... | 50 |
3,622 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | 2 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | The discounted shoe is $20\%$ off the original price. So that means $1 - 0.2 = 0.8$ . There is also a $7.5\%$ sales tax charge, so $0.8 * 1.075 = 0.86$ . Now we can set up the equation $0.86x = 43$ , and solving that we get $x=\boxed{50}$ ~ kabbybear | 50 |
3,623 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | 3 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | Let the original price be $x$ dollars.
After the discount, the price becomes $80\%x$ dollars.
After tax, the price becomes $80\% \times (1+7.5\%) = 86\% x$ dollars.
So, $43=86\%x$ $x=\boxed{50}.$ | 50 |
3,624 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | 4 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | We can assign a variable $c$ to represent the original cost of the shoes. Next, we set up the equation $80\%\cdot107.5\%\cdot c=43$ . We can solve this equation for $c$ and get $\boxed{50}$ | 50 |
3,625 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2 | 5 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | We know the discount price will be 5/4, and 0.075 is equal to 3/40. So we look at answer choice $\textbf{(B) }$ , see that the discount price will be 40, and with sales tax applied it will be 43, so the answer choice is $\boxed{50}$ | 50 |
3,626 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | 1 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | First, note that since the rectangle covers 2 squares, we only need to guess squares that are not adjacent to any of our other guesses. To minimize the amount of guesses, each of our guessed squares should try to touch another guess on one vertex and one vertex only. There are only two ways to do this: one with $5$ gue... | 4 |
3,627 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5 | 2 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a ... | Since the hidden rectangle can only hide two adjacent squares, we may think that we eliminate 8 squares and we're done, but think again. This is the AMC 10, so there must be a better solution (also note that every other solution choice is below 8 so we're probably not done) So, we think again, we notice that we haven't... | 4 |
3,628 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | 1 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$ . We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 throu... | 6 |
3,629 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | 2 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$ $x-3$ $x-5$ ... | 6 |
3,630 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | 3 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
First, we evaluate any value on the interval $(-\infty, 1)$ . Since the degree of $P(x)$ is $1+2+...+9+10$ $\frac{10\times11}{2}$ $55$ , and every term in $P(x)$ is negative, multiplying $55$ ... | 6 |
3,631 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6 | 4 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\q... | Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$
Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1... | 6 |
3,632 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_7 | 1 | For how many integers $n$ does the expression \[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\] represent a real number, where log denotes the base $10$ logarithm?
$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$ | We have \begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}
Because $n$ is an integer and $\log n$ is well defined, $n$ m... | 901 |
3,633 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_7 | 2 | For how many integers $n$ does the expression \[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\] represent a real number, where log denotes the base $10$ logarithm?
$\textbf{(A) }900 \qquad \textbf{(B) }2\qquad \textbf{(C) }902 \qquad \textbf{(D) } 2 \qquad \textbf{(E) }901$ | Notice $\log(n^2)$ can be written as $2\log(n)$ . Setting $a=\log(n)$ , the equation becomes $\sqrt{\frac{2a-a^2}{a-3}}$ which can be written as $\sqrt{\frac{a(2-a)}{a-3}}$
Case 1: $a \ge 3$ The expression is undefined when $a=3$ . For $a>3$ , it is trivial to see that the denominator is positive and the numerator is n... | 901 |
3,634 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_8 | 1 | How many nonempty subsets $B$ of ${0, 1, 2, 3, \cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$ ? For example, $B = {4, 6, 8, 11}$ satisfies the condition.
$\textbf{(A) } 256 \qquad\textbf{(B) } 136 \qquad\textbf{(C) } 108 \qquad\textbf{(D) } 144 \qquad\textbf{(E) }... | There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$ . If the set... | 144 |
3,635 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | 1 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\sqrt{2}$ ). The area of the square is $\sqrt{2}^2 = 2.$
Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.
So now we have $2$ squa... | 8 |
3,636 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | 2 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations
are $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ . Grap... | 8 |
3,637 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | 3 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \times \frac{2^2}{2}$ which is $\boxed{8}.$ | 8 |
3,638 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9 | 4 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$
$\text{(A)}\ 2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 12$ | We start by considering the graph of $|x|+|y|\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \leq 1$ we have to translate it by $\pm 1$ on the $x$ axis and $\pm 1$ on the $y$ axis.
Graphing $|x|+|y|\leq 1$ we get a square with side length of $\sqrt{2}$ , so the area of one of these squares is just $2$
We have to m... | 8 |
3,639 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14 | 1 | For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 4$ | Denote three roots as $r_1 < r_2 < r_3$ .
Following from Vieta's formula, $r_1r_2r_3 = -6$
Case 1: All roots are negative.
We have the following solution: $\left( -3, -2, -1 \right)$
Case 2: One root is negative and two roots are positive.
We have the following solutions: $\left( -3, 1, 2 \right)$ $\left( -2, 1, 3 \rig... | 5 |
3,640 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | 1 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\q... | This problem asks to find largest $x$ that cannot be written as \[ 6 a + 10 b + 15 c = x, \hspace{1cm} (1) \] where $a, b, c \in \Bbb Z_+$
Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2.
Modulo 2 on Equation (1), we get
By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, ... | 11 |
3,641 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | 2 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\q... | Let the number of $6$ cent coins be $a$ , the number of $10$ cent coins be $b$ , and the number of $15$ cent coins be $c$ . We get the Diophantine equation
\[6a + 10b + 15c = k\]
and we wish to find the largest possible value of $k$
Construct the following $\mod 6$ table of $6$ $10$ , and $15$
\[\begin{array}{c|ccc} &... | 11 |
3,642 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 | 3 | In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
$\textbf{(A) }8\qquad\textbf{(B) }10\qquad\textbf{(C) }7\qquad\textbf{(D) }11\q... | We claim that the largest number that cannot be obtained using $6$ $10$ , and $15$ cent coins is $29$
Let's first focus on the combination of $6$ $10$ . As both of them are even numbers, we cannot obtain any odd numbers from these two but requires $15$ to sum up to an odd number. Notice that by Chicken McNugget Theorem... | 11 |
3,643 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | 1 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first sem... | Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$ $Y_2 = Y_1 + 18$ $Y_2 = Z_2 + 3$
Denote by $A_{12}$ the average of person with initial $A$ in the full year.
Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number... | 22 |
3,644 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | 2 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first sem... | We can use process of elimination by finding possible solutions to answer choices. Let $y_1$ and $y_2$ be the number of quizzes Yolanda took in the first and second semesters, respectively. Define $z_1$ and $z_2$ similarly for Zelda.
Answer choice B is satisfied by $(y_1,y_2,z_1,z_2) = (289,1,1,289)$
Answer choice C an... | 22 |
3,645 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | 3 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first sem... | Let Yolanda's average for semester $1$ be $y_1$ , the number of quizzes Yolanda took in semester $1$ be $n_1$ , Zelda's average for semester $1$ be $z_1$ , the number of quizzes Zelda took in semester $1$ be $k_1$ , Yolanda's average for semester $2$ be $y_2$ , the number of quizzes Yolanda took in semester $2$ be $n_2... | 22 |
3,646 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18 | 4 | Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first sem... | Denote $y_1$ and $y_2$ as the quiz averages of Yolanda in the $1$ st and $2$ nd semesters, respectively. Similarly, denote $z_1$ and $z_2$ as the quiz averages of Zelda in the $1$ st and $2$ nd semesters.
We have $y_1 = z_1 + 3$ , so $y_1 - 3 = z_1$ . We also know that $y_2 = 18 +y_1 = 3 + z_2$ , implying $z_2 = 15 + y... | 22 |
3,647 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | 1 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | Substituting $a = b$ we get \[f(2a) + f(0) = 2f(a)^2\] Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$ , so answer choice $\boxed{2}$ is impossible. | 2 |
3,648 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | 2 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | First, we set $a \leftarrow 0$ and $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ f(0) + f(0) = 2 f(0) \times f(0) . ]$
Thus, $f(0) = 0$ or 1.
Case 1: $f(0) = 0$
We set $b \leftarrow 0$ .
Thus, the equation given in the problem becomes $[ 2 f(a) = 0 . ]$
Thus, $f(a) = 0$ for all $a$
Case 2: $f(0)... | 2 |
3,649 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22 | 3 | A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | The relationship looks suspiciously like a product-to-sum identity. In fact, \[\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))\] which is basically the relation. So we know that $f(x) = \cos(x)$ is a valid solution to the function. However, if we define $x=ay,$ where $a$ is arbitrary, the a... | 2 |
3,650 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | 1 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | We start by trying to prove a function of $n$ , and then we can apply the function and equate it to $936$ to find the value of $n$
It is helpful to think of this problem in the format $(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots$ . Note that if we represent the scenario in this manner, we can think of picking a $1$ for one... | 11 |
3,651 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | 2 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | The product can be written as \begin{align*} 2^a 3^b 4^c 5^d 6^e & = 2^{a + 2c + e} 3^{b + e} 5^d . \end{align*}
Therefore, we need to find the number of ordered tuples $\left( a + 2c + e, b+e, d \right)$ where $a$ $b$ $c$ $d$ $e$ are non-negative integers satisfying $a+b+c+d+e \leq n$ .
We denote this number as $f(n)$... | 11 |
3,652 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 | 3 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$
$\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9$ | The product can be written as \begin{align*} 2^x 3^y 5^z \end{align*}
Letting $n=1$ , we get $(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)$ , 6 possible values. But if the only restriction of the product if that $2x\le n,y\le n,z\le n$ , we can get $(2+1)(1+1)(1+1)=12$ possible values. We calculate the ratio... | 11 |
3,653 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_24 | 1 | Suppose that $a$ $b$ $c$ and $d$ are positive integers satisfying all of the following relations.
\[abcd=2^6\cdot 3^9\cdot 5^7\] \[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\] \[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\] \[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\] \[\text{lcm}(b,d)=2^2\c... | Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$
We index Equations given in this problem from (1) to (7).
First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$
Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$ .
Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (... | 3 |
3,654 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | 1 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \tex... | Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$ | 5 |
3,655 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | 2 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \tex... | Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+4... | 5 |
3,656 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2 | 3 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \tex... | In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] vladimir.shelomovskii@gmail.com, vvsss | 5 |
3,657 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4 | 1 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$ | 6 |
3,658 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4 | 2 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$ | 6 |
3,659 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | 2 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \q... | Since the second point is the origin, this is equivalent to finding all points $(x, y)$ such that $|x| + |y| \leq 20$ . Due to the absolute values, the set of all such points will be symmetric about the origin meaning we can focus on the first quadrant and multiply by $4$
To avoid overcounts, ignore points on the axes.... | 841 |
3,660 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | 3 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \q... | This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 1; --y) { for (int x = 0; x <= 20-y; ++x) { dot((x,y),green+linewidth(4)); } } for (int y = 0; y >= -20; --y) { for (int x = 1; x <= y+20; ++x) { dot((x,y),blue+linewidth(4)); } } for (... | 841 |
3,661 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | 4 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \q... | This solution refers to the Diagram section.
As shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd. [asy] /* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= 20-y; x+=2) { ... | 841 |
3,662 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | 5 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \q... | Let $P = (x, y)$ . Since the problem asks for taxicab distances from the origin, we want $|x| + |y| \le 20$ . The graph of all solutions to this equation on the $xy$ -plane is a square with vertices at $(0, \pm 20)$ and $(\pm 20, 0)$ (In order to prove this, one can divide the sections of this graph into casework on th... | 841 |
3,663 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5 | 6 | The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \q... | Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \le 20$ where $x$ and $y$ are nonnegative integers. We can introduce a third variable, say $z$ , such that $z = 20 - (x + y)$ . Note that counting the number of ways to have $x + y +... | 841 |
3,664 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_6 | 1 | A data set consists of $6$ (not distinct) positive integers: $1$ $7$ $5$ $2$ $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \te... | First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$ | 36 |
3,665 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_7 | 1 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)-... | The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of ... | 540 |
3,666 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_7 | 2 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)-... | Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.
Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left an... | 540 |
3,667 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | 1 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$ . Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$
By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... | 10 |
3,668 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | 2 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | We can write this infinite product as $L$ (we know from the answer choices that the product must converge): \[L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] If we raise everything to the third power, we get: \[L^3 = 10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[... | 10 |
3,669 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8 | 3 | The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$ | Move the first term inside the second radical. We get \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] Do this for the third radical as well: \[\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{1... | 10 |
3,670 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | 1 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent state... | Note that:
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{7}.$ | 7 |
3,671 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9 | 2 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent state... | Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. Th... | 7 |
3,672 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | 1 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$
Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:
Together, the answer is $72+72=\boxed{144}.$ | 144 |
3,673 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | 2 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.
We know that $8$ or $9$ can pair with any integer from $1$ to $4$ $10$ or $11$ can pair with any integer from $1$ to $5$ , and $12$ or $13$ can pair with any integer from $1$ to $6$ . Thus, $8$ will have $4$ choices to pair w... | 144 |
3,674 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10 | 3 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.
Assign partners in decreasing order for $x \in \{7, \dots, 1\}$
Note that $7$ must pair with $14$ $\mathbf{1} \textbf{ choice}$
For $5 \leq x \leq 7$ , the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$ . As $x... | 144 |
3,675 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_11 | 1 | What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$
$\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81$ | Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = |\log_6 \frac{9}{x}|$
$\pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1}$
$9b^1 \cdot 9b^{-1} = \boxed{81}$ | 81 |
3,676 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_11 | 2 | What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$
$\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81$ | First, notice that there must be two such numbers: one greater than $\log_69$ and one less than it. Furthermore, they both have to be the same distance away, namely $2(\log_610 - 1)$ . Let these two numbers be $\log_6a$ and $\log_6b$ . Because they are equidistant from $\log_69$ , we have $\frac{\log_6a + \log_6b}{2} =... | 81 |
3,677 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_11 | 3 | What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$
$\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81$ | \[\log_6 9 - \log_6 x_2 = -2d\] Subbing in for $-2d$ and using log rules, \[\log_6 \frac{9}{x_2} = \log_6 \frac{9}{25}\] From this we conclude that \[\frac{9}{x_2} = \frac{9}{25} \implies x_2 = 25\]
Finding the product of the distinct values, $x_1x_2 = \boxed{81}$ | 81 |
3,678 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_12 | 1 | Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$ . What is $\cos(\angle CMD)$
$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$ | Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$
Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$
In right $\triangle CMO,$ we have \[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{13}.\] ~MRENTHUSIASM | 13 |
3,679 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_12 | 3 | Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$ . What is $\cos(\angle CMD)$
$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$ | As done above, let the edge-length equal $2$ (usually better than $1$ because we can avoid fractions when dropping altitudes). Notice that the triangle stated in the question has two side-lengths that are the altitudes of two equilateral triangles. By dropping the equilateral triangles' altitude and using $30^{\circ}$ ... | 13 |
3,680 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_13 | 1 | Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$ , where $z_1$ lies on the segment with endpoints $3$ and $4i$ , and $z_2$ has magnitude at most $1$ . What integer is closest to the area of $\mathcal{R}$
$\textbf... | [asy] size(250); import TrigMacros; rr_cartesian_axes(-2,6,-2,6,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-1,5,Ticks(f, 1.0)); yaxis(-1,5,Ticks(f, 1.0)); dot((3,0)); dot((0,4)); draw((0,4)--(3,0), blue); draw((0.8, 4.6)..(-.6,4.8)..(-.8, 3.4),red); draw((-.8, 3.4)--(2.2, -0.6), red); draw((2... | 13 |
3,681 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14 | 1 | What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$ | Let $\text{log } 2 = x$ . The expression then becomes \[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\] | 2 |
3,682 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14 | 2 | What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$ | Using sum of cubes \[(\log 5)^{3}+(\log 20)^{3}\] \[= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})\] \[= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})\] Let x = $\log 5$ and y = $\log 2$ , so $x+y=1$
The entire expression becomes \[2(x^2-x(2y+x)+(2y+x)^2)-6y^2\] \[=2(x^2+2xy+... | 2 |
3,683 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14 | 3 | What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$ | We can estimate the solution. Using $\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9$ and $\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,$ we have
\[0.7^3 + 1.7^3 + .9\cdot(-0.6) = \boxed{2}\] ~kxiang | 2 |
3,684 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | 1 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \begin{alignat*}{8} a+b+c &= -\frac{B}{A... | 30 |
3,685 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | 2 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable.
The coefficients of $x$ must multiply to $10$ , so these coefficients must be $5,2,1$ or $10,1,$ in some order. We can try one at a time, and therefore write the factored form as follows: \[... | 30 |
3,686 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | 3 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form $\frac{p}{q}$ , where $p$ is a factor of the constant $(-6)$ and $q$ is a factor of the leading coefficient $(10)$ . Therefore, $p$ is $\pm (1, 2, 3, 6)$ and q is $\pm (1, 2, 5, 10... | 30 |
3,687 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | 4 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ , and let $a, b, c$ be the roots of $P(x)$ . The roots of $P(x-2)$ are then $a + 2, b + 2, c + 2,$ so the product of the roots of $P(x-2)$ is the area of the desired rectangular prism.
$P(x-2)$ has leading coefficient $10$ and constant term $P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29... | 30 |
3,688 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15 | 5 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}... | Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ . This can be factored m as $P(x) = 10(x-a)(x-b)(x-c)$ , where $a$ $b$ , and $c$ are the roots of $P(x)$ . We want $V = (a+2)(b+2)(c+2)$
"Luckily" $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10V$ $P(-2) = -300$ , giving $V = \boxed{30}$ | 30 |
3,689 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | 1 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth ... | We have $t_n = \frac{n (n+1)}{2}$ .
If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$ ,
where $k$ is a positive integer.
Thus, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}... | 18 |
3,690 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | 2 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth ... | As mentioned above, $t_n = \frac{n (n+1)}{2}$ . If $t_n$ is a perfect square, one of two things must occur when the fraction is split into a product. Either $\frac{n}{2}$ and $n+1$ must both be squares, or $n$ and $\frac{n+1}{2}$ must both be squares, and thus the search for the next perfect square triangular number ca... | 18 |
3,691 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | 3 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth ... | According to the problem, we want to find integer $p$ such $\frac{n(n+1)}{2}=p^2$ , after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$ , we call $2n+1=q$ , the equation becomes $q^2-8p^2=1$ , obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solution se... | 18 |
3,692 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | 4 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth ... | If $n \choose 2$ is a square, then ${(2n-1)^2 \choose 2}$ is also a square. We can prove this quite simply:
\[{(2n-1)^2 \choose 2}\] \[= \frac{(2n-1)^2 \cdot ((2n-1)^2 - 1)}{2}\] \[= \frac{(2n-1)^2 \cdot (2n \cdot (2n - 2))}{2}\] \[= (2n-1)^2 \cdot 4{n \choose 2}.\]
Therefore, ${(2 \cdot 9 - 1)^2 \choose 2}$ is a squar... | 18 |
3,693 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16 | 5 | $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth ... | We want to find integer $n_i$ and $m_i$ such that $t_{n_i} =\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.$
We use the formula $\sqrt{n_{i+1}} = \sqrt{2n_i} + \sqrt{n_i + 1}$ and get
\[n_1 = ( \sqrt{2n_0} + \sqrt{n_0 + 1})^2 = (0+1)^2 = 1,\] \[n_2= ( \sqrt{2n_1} + \sqrt{n_1 + 1})^2 = ( \sqrt{2} + \sqrt{1 + 1})^2 = 8,\] \[n_3=... | 18 |
3,694 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17 | 1 | Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-... | We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$
Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$ , which can be derived using sine angle addition with $\sin{(2x + x)}$ , we have \[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\] Since $\sin{x... | 4 |
3,695 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17 | 2 | Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-... | We can optimize from the step from \[a\cdot(1+2\cos{x})=4\cos^2{x}-1\] in solution 1 by writing
\[a = \frac{4\cos^2{x}-1}{1+2\cos{x}} = 2\cos x - 1\]
and then get \[\cos x = \frac{a+1}{2}.\]
Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$
Since $\cos{x} = -\frac{1}{2}$ yields ... | 4 |
3,696 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17 | 3 | Suppose $a$ is a real number such that the equation \[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\] has more than one solution in the interval $(0, \pi)$ . The set of all such $a$ that can be written
in the form \[(p,q) \cup (q,r),\] where $p, q,$ and $r$ are real numbers with $p < q< r$ . What is $p+q+r$
$\textbf{(A) } {-... | Use the sum to product formula to obtain $2a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{3x}$ . Use the double angle formula on the RHS to obtain $a\cdot\sin{\frac{3x}{2}}\cos{\frac{x}{2}}=\sin{\frac{3x}{2}}\cos{\frac{3x}{2}}$ . From here, it is obvious that $x=\frac{2\pi}{3}$ is always a solution, and thus we divide ... | 4 |
3,697 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_18 | 1 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point... | Let $P=(r,\theta)$ be a point in polar coordinates, where $\theta$ is in degrees.
Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$ -axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta)... | 359 |
3,698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_18 | 2 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point... | Note that since we're reflecting across the $y$ -axis, if the point ever makes it to $(-1,0)$ then it will flip back to the original point. Note that after $T_1$ the point will be $1$ degree clockwise from the negative $x$ -axis. Applying $T_2$ will rotate it to be $1$ degree counterclockwise from the negative $x$ -axi... | 359 |
3,699 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_18 | 3 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point... | In degrees:
Starting with $n=0$ , the sequence goes ${0}\rightarrow {179}\rightarrow {359}\rightarrow {178}\rightarrow {358}\rightarrow {177}\rightarrow {357}\rightarrow\cdots.$
We see that it takes $2$ steps to downgrade the point by $1^{\circ}$ . Since the $1$ st point in the sequence is ${179}$ , the answer is $1+2(... | 359 |
3,700 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19 | 1 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on... | For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $... | 178 |
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