Difference between revisions of "1995 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
− | When <math>\left(k - \frac {1}{2}\right)^4 \leq n < \left(k + \frac {1}{2}\right)^4</math>, | + | When <math>\left(k - \frac {1}{2}\right)^4 \leq n < \left(k + \frac {1}{2}\right)^4</math>, <math>f(n) = k</math>. Thus there are <math>\left \lfloor \left(k + \frac {1}{2}\right)^4 - \left(k - \frac {1}{2}\right)^4 \right\rfloor</math> values of <math>n</math> for which <math>f(n) = k</math>. Expanding using the [[binomial theorem]], |
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 11:18, 6 December 2019
Contents
[hide]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 n 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.