2022 AMC 8 Problems/Problem 20
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Answer Choices)
- 6 Solution 5 (Super fast! No algebra; no testing answer choices)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (🚀Super Fast. Just 1 min!🚀)
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution
- 12 Video Solution
- 13 Video Solution
- 14 Video Solution
- 15 See Also
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 two shaded 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
Solution 3
This is based on the Solution 2 above and it is perhaps a little simpler than that.
Let be the number in the lower middle. Applying summation to first two columns yields the following.
Since is greater than the other three, we have or
Therefore, the smallest possible value of is
~vetaltekdi6
Solution 4 (Answer Choices)
Note that the sum of the rows and columns must be . We proceed to test the answer choices.
Testing , when , the number above must be , which contradicts the precondition that the numbers surrounding is less than .
Testing , the number above is , which does not work.
Testing , the number above is , which does not work.
Testing , the number above is , which does work. Hence, the answer is .
We do not need to test , because the problem asks for the smallest value of .
~MrThinker
Solution 5 (Super fast! No algebra; no testing answer choices)
The sum of the numbers in each column and row should be . If we look at the column, the gray squares (shown below) sum to .
If square has to be greater than or equal to the three blank squares, then the least can be is half the sum of the value of the gray squares, which is . But square has to be greater than and not greater than or equal to the three blank squares, so the least can be is . Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest can be is indeed ; the other two squares are less than . Therefore, the answer is
~ 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!🚀)
~Education, the Study of Everything
Video Solution
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Video Solution
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~Mathematical Dexterity
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https://youtu.be/Ij9pAy6tQSg?t=1857
~Interstigation
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~STEMbreezy
Video Solution
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=FINk9LgSJpU
~David
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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