1995 AIME Problems/Problem 13
Contents
Problem
Let be the integer closest to Find
Solution
When , . Thus there are values of for which . Expanding using the binomial theorem,
Thus, appears in the summation times, and the sum for each is then . From to , we get (either adding or using the sum of consecutive squares formula).
But this only accounts for terms, so we still have terms with . This adds to our summation, giving .
Solution 2
This is a pretty easy problem just to bash. Since the max number we can get is , we just need to test values for and . Then just do how many numbers there are times , which should be
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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