Difference between revisions of "2022 AMC 8 Problems/Problem 20"
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label((0,-2),"$y$",red); label((-2,0),"$y+10$",red+fontsize(8)); label((0,0),"$x-1$",red+fontsize(8)); | label((0,-2),"$y$",red); label((-2,0),"$y+10$",red+fontsize(8)); label((0,0),"$x-1$",red+fontsize(8)); | ||
</asy> | </asy> | ||
+ | We have <math>x>x-1, x>y+10,</math> and <math>x>y.</math> Note that the first inequality is true for all values of <math>x.</math> We only need to solve the second inequality so that the third inequality is true for all values of <math>x.</math> By substitution, we get <math>x>(4-x)+10,</math> from which <math>x>7.</math> | ||
+ | |||
+ | Therefore, the smallest possible value of <math>x</math> is <math>\boxed{\textbf{(D) } 8}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==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 19:51, 28 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
?
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
:
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
:
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:
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 |