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 
array of integers form an arithmetic progression of length 
The numbers in positions 
and 
are 
and 
respectively. What number is in position 
![\[\begin{bmatrix} . & ? &.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ .&.&.&.&0\end{bmatrix}\]](http://latex.artofproblemsolving.com/2/c/a/2caa51c2881f478a1e28bf45df93c383dff268f6.png)
Solution
![\[\begin{bmatrix} 12 & 29 &46&63&80 \\ 12&24&36&48&60\\ 12&19&26&33&40\\ 12&14&16&18&20\\ 12&9&6&3&0\end{bmatrix}\]](http://latex.artofproblemsolving.com/1/3/5/13554e8c867fc72395411ed4e1b75544e34cc2f0.png)
-submitted by Astingo
Solution 2: Some Basic Algebra and Answer Choices
Assume the number in position 
is 
. The integer in position 
will be 
, as and 
average out to x. Similarly, the integer in position 
is 
. The integer in position 
is 
. This makes the number in position 
.
The only answer choice that makes x an integer is 
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
