Difference between revisions of "2022 AMC 8 Problems/Problem 20"
MRENTHUSIASM (talk | contribs) m (→Solution 1) |
MRENTHUSIASM (talk | contribs) m (→See Also) |
||
| Line 99: | Line 99: | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=19|num-a=21}} | {{AMC8 box|year=2022|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Revision as of 00:40, 29 January 2022
Contents
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 1
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 
:
![[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
Solution 2
The sum of the numbers in each row is 
and the sum of the numbers in each column is 
Let 
be the number in the lower middle. It follows that 
or 
We express the other two missing numbers in terms of and 
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); label((-2,0),"$y+10$",red+fontsize(8)); label((0,0),"$x-1$",red+fontsize(8)); [/asy]](http://latex.artofproblemsolving.com/9/f/9/9f9718351b90e2f216c0cef04e369082f58df0a6.png)
We have 
and 
Note that the first inequality is true for all values of 
We only need to solve the second inequality so that the third inequality is true for all values of By substitution, we get 
from which 
Therefore, the smallest possible value of is 
~MRENTHUSIASM
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
