# 2022 AMC 8 Problems/Problem 20

## Problem

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$? $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"-2"); label((0,2),"9"); label((2,2),"5"); label((2,0),"-1"); label((2,-2),"8"); label((-2,-2),"x"); [/asy]$ $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

## Solution 1

The sum of the numbers in each row is $12$. Consider the second row. In order for the sum of the numbers in this row to equal $12$, the two shaded numbers must add up to $13$: $[asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"-2"); label((0,2),"9"); label((2,2),"5"); label((2,0),"-1"); label((2,-2),"8"); label((-2,-2),"x"); [/asy]$ If two numbers add up to $13$, one of them must be at least $7$: If both shaded numbers are no more than $6$, their sum can be at most $12$. Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$. We can construct a working scenario where $x=8$: $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"-2"); label((0,2),"9"); label((2,2),"5"); label((2,0),"-1"); label((2,-2),"8"); label((-2,-2),"8"); label((0,-2),"-4"); label((-2,0),"6"); label((0,0),"7"); [/asy]$ So, our answer is $\boxed{\textbf{(D) } 8}$.

~ihatemath123

## Solution 2

The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$

Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$

We express the other two missing numbers in terms of $x$ and $y,$ as shown below: $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"-2"); label((0,2),"9"); label((2,2),"5"); label((2,0),"-1"); label((2,-2),"8"); label((-2,-2),"x"); label((0,-2),"y",red+fontsize(11)); label((-2,0),"y{+}10",red+fontsize(11)); label((0,0),"x{-}1",red+fontsize(11)); [/asy]$ We have $x>x-1, x>y+10,$ and $x>y.$ Note that the first inequality is true for all values of $x.$ We only need to solve the second inequality so that the third inequality is true for all values of $x.$ By substitution, we get $x>(4-x)+10,$ from which $x>7.$

Therefore, the smallest possible value of $x$ is $\boxed{\textbf{(D) } 8}.$

~MRENTHUSIASM

## Solution 3

This is based on the Solution 2 above and it is perhaps a little simpler than that.

Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following. $[asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"-2"); label((0,2),"9"); label((2,2),"5"); label((2,0),"-1"); label((2,-2),"8"); label((-2,-2),"x"); label((0,-2),"y",red+fontsize(11)); label((-2,0),"14{-}x",red+fontsize(11)); label((0,0),"3{-}y",red+fontsize(11)); [/asy]$

Since $x$ is greater than the other three, we have $x>14-x,$ or $x>7.$

Therefore, the smallest possible value of $x$ is $\boxed{\textbf{(D) } 8}.$

~vetaltekdi6

Note that the sum of the rows and columns must be $8+5-1=12$. We proceed to test the answer choices.

Testing $\textbf{(A)}$, when $x = -1$, the number above $x$ must be $15$, which contradicts the precondition that the numbers surrounding $x$ is less than $x$.

Testing $\textbf{(B)}$, the number above $x$ is $9$, which does not work.

Testing $\textbf{(C)}$, the number above $x$ is $8$, which does not work.

Testing $\textbf{(D)}$, the number above $x$ is $6$, which does work. Hence, the answer is $\boxed{\textbf{(D) }8}$.

We do not need to test $\textbf{(E)}$, because the problem asks for the smallest value of $x$.

~MrThinker

## Video Solution

~Mathematical Dexterity

~Interstigation

~STEMbreezy

~savannahsolver

## Video Solution

~David

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