Difference between revisions of "2013 AIME II Problems/Problem 14"

Revision as of 15:48, 6 April 2013

Problem 14

For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ .

Solution

Easy solution without strict proof

We can find that

$20\equiv 6 \pmod{7}$

$21\equiv 5 \pmod{8}$

$22\equiv 6 \pmod{8}$

$23\equiv 7 \pmod{8}$

$24\equiv 6 \pmod{9}$

$25\equiv 7 \pmod{9}$

$26\equiv 8 \pmod{9}$

Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get

$99\equiv 31 \pmod{34}$

$100\equiv 32 \pmod{34}$

So the sum is $5+3\times(6+...+31)+32\times 2=1512$ , so the answer is $\boxed{512}$ .

@@ Line 4: / Line 4: @@
 ==Solution==
-====Easy solution without strict proof====
+===Easy solution without strict proof===
 We can find that

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

Page

Toolbox

Search

Difference between revisions of "2013 AIME II Problems/Problem 14"

Revision as of 15:48, 6 April 2013

Contents

Problem 14

Solution

Easy solution without strict proof

See Also