2014 AMC 12B Problems/Problem 23
Contents
[hide]Problem
The number is prime. Let . What is the remainder when is divided by
Solution
Note that . We have for Therefore This is simply an alternating series of triangular numbers that goes like this: After finding the first few sums of the series, it becomes apparent that and Obviously, falls in the second category, so our desired value is
Sidenote
Another way to finish, using the fact that :
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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