2020 AMC 10A Problems/Problem 7
- The following problem is from both the 2020 AMC 12A #5 and 2020 AMC 10A #7, so both problems redirect to this page.
Contents
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 1
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
Solution 3
Taking the average of the first and last terms, and , we have that the mean of the set is . There are 5 values in each row, column or diagonal, so the value of the common sum is , or . ~Arctic_Bunny, edited by KINGLOGIC
Solution 4
Let us consider the horizontal rows. Since there are five of them, each with constant sum , we can add up the 25 numbers in 5 rows for a sum of . Since the sum of the 25 numbers used is , and . ~cw357
Solution 5
The mean of the set of numbers is . The numbers around it must be equal (i.e. if the mean of , , , , and is , then .) One row of the square would be
Adding the numbers would be
with a sum of .
Video Solution 1
Education, the Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~savannahsolver
Video Solution 4 by OmegaLearn
https://youtu.be/mgEZOXgIZXs?t=1
~ pi_is_3.14
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.