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Difference between revisions of "2022 AMC 8 Problems/Problem 20"

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== Solution 1 ==
 
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The sum of the numbers in each row is <math>12</math>. Consider the second row. In order for the sum of the numbers in this row to equal <math>12</math>, the two shaded numbers must add up to <math>13</math>:
 
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<asy>
 
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unitsize(0.5cm);
 
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fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray);
 
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draw((3,3)--(-3,3));
 
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draw((3,1)--(-3,1));
 
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draw((3,-3)--(-3,-3));
 
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draw((3,-1)--(-3,-1));
 
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draw((3,3)--(3,-3));
 
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draw((1,3)--(1,-3));
 
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draw((-3,3)--(-3,-3));
 
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draw((-1,3)--(-1,-3));
 
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label((-2,2),"$-2$");
 
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label((0,2),"$9$");
 
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label((2,2),"$5$");
 
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label((2,0),"$-1$");
 
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label((2,-2),"$8$");
 
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label((-2,-2),"$x$");
 
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</asy>
 
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If two numbers add up to <math>13</math>, one of them must be at least <math>7</math>: If both shaded numbers are no more than <math>6</math>, their sum can be at most <math>12</math>. Therefore, for <math>x</math> to be larger than the three missing numbers, <math>x</math> must be at least <math>8</math>. We can construct a working scenario where <math>x=8</math>:
 
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<asy>
 
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unitsize(0.5cm);
 
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draw((3,3)--(-3,3));
 
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draw((3,1)--(-3,1));
 
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draw((3,-3)--(-3,-3));
 
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draw((3,-1)--(-3,-1));
 
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draw((3,3)--(3,-3));
 
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draw((1,3)--(1,-3));
 
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draw((-3,3)--(-3,-3));
 
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draw((-1,3)--(-1,-3));
 
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label((-2,2),"$-2$");
 
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label((0,2),"$9$");
 
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label((2,2),"$5$");
 
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label((2,0),"$-1$");
 
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label((2,-2),"$8$");
 
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label((-2,-2),"$8$");
 
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label((0,-2),"$-4$");
 
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label((-2,0),"$6$");
 
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label((0,0),"$7$");
 
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</asy>
 
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So, our answer is <math>\boxed{\textbf{(D) } 8}</math>.
 
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~ihatemath123
 
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==Solution 2==
 
==Solution 2==
  

Revision as of 20:34, 7 September 2025

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

Solution 4 (Answer Choices)

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

Solution 5 (Super fast! No algebra and no testing any of the answer choices!)

The sum of the numbers in each column and row should be $5+(-1)+8=12$. If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$.

[asy] 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),"$x$"); filldraw((-3,-3)--(-1,-3)--(-1,-1)--(-3,-1)--cycle, lightgray, black+linewidth(1)); filldraw((-1,-1)--(-3,-1)--(-3,1)--(-1,1)--cycle, lightgray, black+linewidth(1)); label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (-0,-5), S); [/asy]

If square $x$ has to be greater than or equal to the three blank squares, then the least $x$ can be is half the sum of the value of the gray squares, which is $14\div2=7$. But square $x$ has to be greater than and not greater than or equal to the three blank squares, so the least $x$ can be is $7+1=8$. Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest $x$ can be is indeed $8$; the other two squares are less than $8$. Therefore, the answer is $\boxed{\text{(D) }8}$.

~ JoyfulSapling

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=Bbea8RWE2sMWN6xl&t=3643

~Math-X

Video Solution (🚀Super Fast. Just 1 min!🚀)

https://youtu.be/7J4EGPaB29Y

~Education, the Study of Everything

Video Solution

https://youtu.be/0hHlpIVeFjg

Video Solution

https://www.youtube.com/watch?v=xnGQffaxYAA

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1857

~Interstigation

Video Solution

https://youtu.be/hs6y4PWnoWg?t=369

~STEMbreezy

Video Solution

https://youtu.be/DXFwzrOF4b4

~savannahsolver

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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

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