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

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==Solution 4 (Answer Choices)==
 
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Note that the sum of the rows and columns must be <math>8+5-1=12</math>. We proceed to test the answer choices.
 
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Testing <math>\textbf{(A)}</math>, when <math>x = -1</math>, the number above <math>x</math> must be <math>15</math>, which contradicts the precondition that the numbers surrounding <math>x</math> is less than <math>x</math>.
 
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Testing <math>\textbf{(B)}</math>, the number above <math>x</math> is <math>9</math>, which does not work.
 
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Testing <math>\textbf{(C)}</math>, the number above <math>x</math> is <math>8</math>, which does not work.
 
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Testing <math>\textbf{(D)}</math>, the number above <math>x</math> is <math>6</math>, which ''does'' work. Hence, the answer is <math>\boxed{\textbf{(D) }8}</math>.
 
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We do not need to test <math>\textbf{(E)}</math>, because the problem asks for the '''smallest''' value of <math>x</math>.
 
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~MrThinker
 
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==Solution 5 (Super fast! No algebra and no testing any of the answer choices!)==
 
==Solution 5 (Super fast! No algebra and no testing any of the answer choices!)==
  

Revision as of 20:35, 7 September 2025

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

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