# Difference between revisions of "2012 USAJMO Problems/Problem 5"

## Problem

For distinct positive integers $a$, $b < 2012$, define $f(a,b)$ to be the number of integers $k$ with $1 \le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012. Let $S$ be the minimum value of $f(a,b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012. Determine $S$.

## Solution

The key insight in this problemo is noticing that when ak is higher then bk, a(2012-k) is lower than b(2012-k), except at 2(mod 4) residues*. Also, they must be equal quite a lot. 2012=2^2*503. We should have multiples of 503. After trying all three pairs and getting 503 as our answer, we win. But look at the 2(mod 4) idea. What if we just took 2 and plugged it in. with 1006. We get 502.

```--Va2010 11:12, 28 April 2012 (EDT)va2010
```

## Alternate, formal argument

Say that the problem is a race track with 2012 spots. To intersect the most, we should get next to each other a lot so the negation is high. As 2012=2^2*503, we intersect at a lot of multiples of 503.

 2012 USAJMO (Problems • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions