Difference between revisions of "1995 AIME Problems/Problem 13"
(→Solution 2) |
|||
(One intermediate revision by one other user not shown) | |||
Line 3: | Line 3: | ||
== 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*} | ||
Line 13: | Line 13: | ||
==Solution 2== | ==Solution 2== | ||
− | This is a pretty easy problem just to bash. Since the max number we can get is <math>7</math>, we just need to test n values for <math>1.5,2.5,3.5,4.5,5.5</math> and <math>6.5</math>. Then just do how many numbers there are times <math>\frac{1}{\lfloor n \rfloor}</math>, which should be <math>5+17+37+65+101+145+30 = \boxed{400}</math> | + | This is a pretty easy problem just to bash. Since the max number we can get is <math>7</math>, we just need to test <math>n</math> values for <math>1.5,2.5,3.5,4.5,5.5</math> and <math>6.5</math>. Then just do how many numbers there are times <math>\frac{1}{\lfloor n \rfloor}</math>, which should be <math>5+17+37+65+101+145+30 = \boxed{400}</math> |
== See also == | == See also == |
Latest revision as of 00:25, 12 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 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.