Difference between revisions of "2024 AMC 10A Problems/Problem 21"

Revision as of 17:59, 8 November 2024

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

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

-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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 (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 <math>0</math>. Going up, let the common difference be <math>a</math>, and going left, let the common difference be <math>b</math>. Therefore, we have <cmath> \begin{bmatrix} . & ? &.&.&4a \\ .&.&.&48&3a\\ 12&.&.&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}</cmath> ~Tacos_are_yummy_1 (I'm currently editing this, please don't interfere, thanks!)
 ==See also==

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