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Difference between revisions of "2024 AMC 10A Problems/Problem 21"
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Assume the number in position <math>(3, 3)</math> is <math>x</math>. The integer in position <math>(2, 3)</math> will be <math>2x-16</math>, as <math>2x-16</math> and <math>16</math> average out to x. Similarly, the integer in position <math>(3, 2)</math> is <math>0.5x+6</math>. The integer in position <math>(2, 2)</math> is <math>4x-80</math>. This makes the number in position <math>(1, 2)</math>  <math>7.5x-166</math>.  
 
Assume the number in position <math>(3, 3)</math> is <math>x</math>. The integer in position <math>(2, 3)</math> will be <math>2x-16</math>, as <math>2x-16</math> and <math>16</math> average out to x. Similarly, the integer in position <math>(3, 2)</math> is <math>0.5x+6</math>. The integer in position <math>(2, 2)</math> is <math>4x-80</math>. This makes the number in position <math>(1, 2)</math>  <math>7.5x-166</math>.  
  
−
The only answer choice that makes x an integer is {(C) } 29.
+
The only answer choice that makes x an integer is \qquad \textbf{(C) } 29  
  
 
~ElaineGu
 
~ElaineGu
  
−
(Note: I'm not very good at writing solutions, so people who wish to edit this solution to be more understandable may do so.)
+
(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.)
  
  

Revision as of 17:52, 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: 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 \qquad \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.)


See also

This problem is remarkably similar to 1988 AIME Problems/Problem 6.

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

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