# Functions Problem 2

### Solution

Let's represent each number $$13^m$$ as $$13^m = q_m2013 + r_m$$.

There are 2049 natural numbers $$\le 2048$$. So there are 2049 numbers of the form $$13^m$$. However, there are only 2013 remainders when you divide numbers by 2013. So two of the $$13^m$$ must have the same remainder mod 2013.

Let's choose a and b as the names of two numbers such that $$13^a$$ and $$13^b$$ have the same remainder mod 2013. Let's call this shared remainder R. That is, we have $$13^a = q_a2013 + R$$ and $$13^b = q_b2013 + R$$.

But then $$13^a - 13^b = q_a2013 - q_b2013 = (q_a - q_b)2013$$. Since $$q_a$$ and $$q_b$$ are integers, $$q_a -q_b$$ is an integer. So this means $$13^a - 13^b$$ is divisible by 2013, which is what we needed to show.