Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 3)
Line 1: Line 1:
−
==Solution 3==
 
−
 
−
This is based on the Solution 2 above and it is perhaps a little simpler than that.
 
−
 
−
Let <math>y</math> 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 <math>x</math> is greater than the other three, we have <math>x>14-x,</math> or <math>x>7.</math>
 
−
 
−
Therefore, the smallest possible value of <math>x</math> is <math>\boxed{\textbf{(D) } 8}.</math>
 
−
 
−
~vetaltekdi6
 
−
 
 
==Solution 4 (Answer Choices)==
 
==Solution 4 (Answer Choices)==
 
Note that the sum of the rows and columns must be <math>8+5-1=12</math>. We proceed to test the answer choices.
 
Note that the sum of the rows and columns must be <math>8+5-1=12</math>. We proceed to test the answer choices.

Revision as of 20:35, 7 September 2025

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_20&oldid=259967"