Difference between revisions of "2022 AMC 8 Problems/Problem 20"
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| − | ==Problem== | + | == 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 <math>x</math> in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of <math>x</math>? | 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 <math>x</math> in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of <math>x</math>? | ||
| Line 15: | Line 15: | ||
label((0,2),"$9$"); | label((0,2),"$9$"); | ||
label((2,2),"$5$"); | label((2,2),"$5$"); | ||
| − | label((2,0),"$ | + | label((2,0),"$-1$"); |
label((2,-2),"$8$"); | label((2,-2),"$8$"); | ||
label((-2,-2),"$x$"); | label((-2,-2),"$x$"); | ||
</asy> | </asy> | ||
| − | <math>\textbf{(A) } | + | <math>\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad</math> |
| − | ==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: | |
| − | + | <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 <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>: | ||
| + | <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 <math>\textbf{(D) } 8</math>. | ||
| + | ~ ihatemath123 | ||
==See Also== | ==See Also== | ||
| − | {{AMC8 box|year=2022|num-b= | + | {{AMC8 box|year=2022|num-b=18|num-a=20}} |
| − | |||
Revision as of 19:15, 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 
in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of ?
![[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]](http://latex.artofproblemsolving.com/7/1/d/71d1b8183aafd0b7db5af174a291282dccce49ec.png)
Solution
The sum of the numbers in each row is 
. Consider the second row. In order for the sum of the numbers in this row to equal , 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]](http://latex.artofproblemsolving.com/9/4/1/9412b194acb7f33563ae5102fc76384faaf9e343.png)
If two numbers add up to 
, one of them must be at least 
- if both shaded numbers are no more than 
, their sum can be at most . Therefore, for to be larger than the three missing numbers, must be at least 
. We can construct a working scenario where 
:
![[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]](http://latex.artofproblemsolving.com/6/3/4/634b125094f297b3a8e5e036f57c8b69a27a51b1.png)
So, our answer is 
.
~ ihatemath123
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
