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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+ | ==Video Solution== | ||
+ | https://youtu.be/JEjib74EmiY | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 22:12, 31 January 2020
Contents
Problem 7
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 .
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 |
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