2013 AMC 8 Problems/Problem 17
Contents
Problem
The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
Solution 1
The arithmetic mean of these numbers is . Therefore the numbers are , , , , , , so the answer is
Solution 2
Let the number be . Then our desired number is .
Our integers are , so we have that .
Solution 3
Let the first term be . Our integers are . We have,
Solution 4
Since there are numbers, we divide by to find the mean of the numbers. . Then, (the fourth number). Fifth: ; Sixth: .
Solution 5
Let the number be . Then our list is: . Simplifying this gets you , which means that
Video Solution
https://youtu.be/E7DZ1oILqE0 ~savannahsolver
Video Solution 2
https://youtu.be/AvOXHvvTlio Soo, DRMS, NM
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.