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Difference between revisions of "2022 AMC 8 Problems/Problem 20"
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== Solution ==
 
== Solution ==
−
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 first two numbers must add up to 13:
+
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 first two numbers must add up to <math>13</math>:
 
<asy>
 
<asy>
 
unitsize(0.5cm);
 
unitsize(0.5cm);
Line 62: Line 62:
 
label((0,0),"$7$");
 
label((0,0),"$7$");
 
</asy>
 
</asy>
−
So, our answer is <math>\textbf{(D) } 8</math>.
+
So, our answer is <math>\boxed{\textbf{(D) } 8}</math>.
 +
 
 +
~ihatemath123
 +
 
 +
==Solution 2==
 +
 
 +
The sum of the numbers in each row is <math>-2+9+5=12,</math> and the sum of the numbers in each column is <math>5+(-1)+8=12.</math>
 +
 
 +
Let <math>y</math> be the number in the lower middle. It follows that <math>x+y+8=12,</math> or <math>x+y=4.</math> We express the other two missing numbers in terms of <math>x</math> and <math>y,</math> as shown below:
  
−
~ ihatemath123
 
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=19|num-a=21}}
 
{{AMC8 box|year=2022|num-b=19|num-a=21}}

Revision as of 18:28, 28 January 2022

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

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 first two numbers must add up to $13$: [asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,lightgray); 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:

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
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