# 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 $S$.

## Solutions

### Solution 1

First we'll show that $S \geq 502$, then we'll find an example $(a, b)$ that have $f(a, b)=502$.

Let $x_k$ be the remainder when $ak$ is divided by 2012, and let $y_k$ be defined similarly for $bk$. First, we know that, if $x_k > y_k >0$, then $x_{2012-k} \equiv a(2012-k) \equiv 2012-ak \equiv 2012-x_k \pmod 2012$ and $y_{2012-k} \equiv 2012-y_k \pmod 2012$. This implies that, since $2012 - x_k \neq 0$ and $2012 -y_k \neq 0$, $x_{2012-k} < y_{2012-k}$. Similarly, if $0< x_k < y_k$ then $x_{2012-k} > y_{2012-k}$, establishing a one-to-one correspondence between the number of $k$ such that $x_k < y_k$. Thus, if $n$ is the number of $k$ such that $x_k \neq y_k$ and $y_k \neq 0 \neq x_k$, then $S \geq \frac{1}{2}n$. Now I'll show that $n \geq 1004$.

If $gcd(k, 2012)=1$, then I'll show you that $x_k \neq y_k$. This is actually pretty clear; assume that's not true and set up a congruence relation: $$ak \equiv bk \pmod {2012}$$ Since $k$ is relatively prime to 2012, it is invertible mod 2012, so we must have $a \equiv b \pmod {2012}$. Since $0 < a, b <2012$, this means $a=b$, which the problem doesn't allow, thus contradiction, and $x_k \neq y_k$. Additionally, if $gcd(k, 2012)=1$, then $x_k \neq 0 \neq y_k$, then based on what we know about $n$ from the previous paragraph, $n$ is at least as large as the number of k relatively prime to 2012. Thus, $n \geq \phi(2012) = \phi(503*4) = 1004$. Thus, $S \geq 502$.

To show 502 works, consider $(a, b)=(1006, 2)$. For all even $k$ we have $x_k=0$, so it doesn't count towards $f(1006, 2)$. Additionally, if $k = 503, 503*3$ then $x_k = y_k = 1006$, so the only number that count towards $f(1006, 2)$ are the odd numbers not divisible by 503. There are 1004 such numbers. However, for all such odd k not divisible by 503 (so numbers relatively prime to 2012), we have $x_k \neq 0 \neq y_k$ and $2012-k$ is also relatively prime to 2012. Since under those conditions exactly one of $x_k > y_k$ and $x_{2012-k} > y_{2012-k}$ is true, we have at most 1/2 of the 1004 possible k actually count to $f(1006, 2)$, so $\frac{1004}{2} = 502 \geq f(1006, 2) \geq S \geq 502$, so $S=502$.

### Solution 2

Let $ak \equiv r_{a} \pmod{2012}$ and $bk \equiv r_{b} \pmod{2012}$. Notice that this means $a(2012 - k) \equiv 2012 - r_{a} \pmod{2012}$ and $b(2012 - k) \equiv 2012 - r_{b} \pmod{2012}$. Thus, for every value of $k$ where $r_{a} > r_{b}$, there is a value of $k$ where $r_{b} > r_{a}$. Therefore, we merely have to calculate $\frac{1}{2}$ times the number of values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$.

However, the answer is NOT $\frac{1}{2}(2012) = 1006$! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$.

So, let's start counting.

If $k$ is even, we have either $a \equiv 0 \pmod{1006}$ or $a - b \equiv 0 \pmod{1006}$. So, $a = 1006$ or $a = b + 1006$. We have $1005$ even values of $k$ (which is all the possible even values of $k$, since the two above requirements don't put any bounds on $k$ at all).

If $k$ is odd, if $k = 503$ or $k = 503 \cdot 3$, then $a \equiv 0 \pmod{4}$ or $a \equiv b \pmod{4}$. Otherwise, $ak \equiv 0 \pmod{2012}$ or $ak \equiv bk \pmod{2012}$, which is impossible to satisfy, given the domain $a, b < 2012$. So, we have $2$ values of $k$.

In total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$, so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} \neq r_{b}$ and $r_{a} \neq 0$. Thus, by our reasoning above, our solution is $\frac{1}{2} \cdot 1004 = \boxed{502}$.

Solution by $\textbf{\underline{Invoker}}$

### Solution 3

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

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


### Solution 4

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 $503$.

## See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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