2024 AMC 10A Problems/Problem 21

The following problem is from both the 2024 AMC 10A #21 and 2024 AMC 12A #14, so both problems redirect to this page.

Problem

The numbers, in order, of each row and the numbers, in order, of each column of a $5 \times 5$ array of integers form an arithmetic progression of length $5{.}$ The numbers in positions $(5, 5), \,(2,4),\,(4,3),$ and $(3, 1)$ are $0, 48, 16,$ and $12{,}$ respectively. What number is in position $(1, 2)?$ $\begin{bmatrix} . & ? &.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ .&.&.&.&0\end{bmatrix}$ $\textbf{(A) } 19 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 29 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 39$

Solution 1 (Bashy)

Let's look at the bottom-left corner. There can be $3$ possible cases for the bottom row. Either the numbers are negative, all $0$ , or positive.

Let's assume all the numbers are negative. We know that the numbers in between the ? will be large, so we want to minimize it, because the largest answer choice is only $39$ .

$\begin{bmatrix} .&?&.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ -4&-3&-2&-1&0\end{bmatrix}$

To fill in the $1$ st and $3$ rd column, we simply find the common differences. We get:

$\begin{bmatrix} 28&?&70&.&. \\ 20&.&52&48&.\\ 12&.&34&.&.\\ 4&.&16&.&.\\ -4&-3&-2&-1&0\end{bmatrix}$

Finding the ? , we notice that it is $49$ , which is bigger than any of the other answer choices, so the bottom row can't be negative. (As the common difference for the bottom row gets larger, so does the common difference for the $1$ st and $3$ rd column, making the ? bigger, and because the smallest answer is $39$ while the minimum value for negative numbers is $49$ , it can't be negative.)

Now let's assume that all the numbers are 0. By just filling in the $1$ st and $3$ rd column, we find that it is:

$\begin{bmatrix} 24 & ? &64&.&. \\ 18&.&48&48&.\\ 12&.&32&.&.\\ 6&.&16&.&.\\ 0&0&0&0&0\end{bmatrix}$

which makes the $?=44$ . There are $2$ problems here. First of all, $44$ is not an answer choice, and the second row will not have a valid arithmetic progression, so the bottom row cannot be all $0$ 's.

$\begin{bmatrix} 12 & 29 &46&63&80 \\ 12&24&36&48&60\\ 12&19&26&33&40\\ 12&14&16&18&20\\ 12&9&6&3&0\end{bmatrix}$

-submitted by Astingo

Solution 2: Some Basic Algebra and Answer Choices

Assume the number in position $(3, 3)$ is $x$ . The integer in position $(2, 3)$ will be $2x-16$ , as $2x-16$ and $16$ average out to x. Similarly, the integer in position $(3, 2)$ is $0.5x+6$ . The integer in position $(2, 2)$ is $4x-80$ . This makes the number in position $(1, 2)$ $7.5x-166$ .

The only answer choice that makes x an integer is $\boxed{\textbf{(C) }29}$

~ElaineGu

(Note: I'm not very good at writing solutions or latex, so people who wish to edit this solution to be more understandable may do so.)

Solution 3 (Arithmetic Sequences)

Start from the $0$ . Going up, let the common difference be $a$ , and going left, let the common difference be $b$ . Therefore, we have $\begin{bmatrix} . & ? &.&.&4a \\ .&.&.&48&3a\\ 12&.&.&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}$ ~Tacos_are_yummy_1 (I'm currently editing this, please don't interfere, thanks!)

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2024 AMC 10A Problems/Problem 21

Contents

Problem

Solution 1 (Bashy)

Solution 2: Some Basic Algebra and Answer Choices

Solution 3 (Arithmetic Sequences)

See also