Difference between revisions of "2013 AIME II Problems/Problem 14"
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| + | ==Problem 14== | ||
For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>. | For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>. | ||
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<math>100\equiv 32 \pmod{34}</math> | <math>100\equiv 32 \pmod{34}</math> | ||
| − | So the sum is <math>5+3\times(6+...+31)+32\times 2=1512</math>, so the answer is <math>\boxed{512}</math> | + | So the sum is <math>5+3\times(6+...+31)+32\times 2=1512</math>, so the answer is <math>\boxed{512}</math>. |
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| + | ==See Also== | ||
| + | {{AIME box|year=2013|n=II|num-b=13|num-a=15}} | ||
Revision as of 15:45, 6 April 2013
Problem 14
For positive integers 
and 
, let 
be the remainder when is divided by , and for 
let 
. Find the remainder when 
is divided by 
.
Solution
Easy solution without strict proof
We can find that
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
So the sum is 
, so the answer is 
.
See Also
| 2013 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
