# Difference between revisions of "2003 Pan African MO Problems/Problem 4"

## Problem

Let $\mathbb{N}_0=\{0,1,2 \cdots \}$. Does there exist a function $f: \mathbb{N}_0 \to \mathbb{N}_0$ such that: $$f^{2003}(n)=5n, \forall n \in \mathbb{N}_0$$ where we define: $f^1(n)=f(n)$ and $f^{k+1}(n)=f(f^k(n))$, $\forall k \in \mathbb{N}_0$?

## Solution (credit to akhan98)

Let $n = 5^a \cdot b$, where $a,b$ are integers and $b$ is relatively prime to 5. Notice that there are an infinite number of values that $b$ can be.

Thus, we can group the values of $b$ into groups of 2004. WLOG, let one group be $\{ b_1, b_2, b_3, \cdots b_{2004} \}$. We can have $$f(5^a \cdot b_n) = \left \{ \begin{array}{l} 5^a \cdot b_{n+1}, n < 2004 \\ 5^{a+1} \cdot b_1, n = 2004\end{array} \right.$$ as the function that satisfies the initial conditions for all $b$ in the given set. Since we can apply a similar function from the other groups of 2004, we can demonstrate that there is a function $f(x)$ that satisfies the conditions for all non-negative integers.