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 by Problem 4	Followed by Problem 6
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2012 USAJMO Problems/Problem 5

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Problem

Solution

Alternate, formal argument

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