Difference between revisions of "2020 AMC 10A Problems/Problem 7"
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Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>. | Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>. | ||
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| + | ===Solution 2=== | ||
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| + | Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get <math>0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math> as our answer. | ||
| + | ~Baolan | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 14:01, 2 February 2020
Problem
The 
integers from 
to 
inclusive, can be arranged to form a 
-by- square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
Solution
Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by is the total value per row. The sum of the integers is 
, and the common sum is 
.
Solution 2
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get 
as our answer.
~Baolan
Video Solution
~IceMatrix
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AMC 10 Problems and Solutions | ||
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
