12. Let $\mathbb{N}$ be the set of all positive integers. A function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfies $f(m+$ $n)=f(f(m)+n)$ for all $m, n \in \mathbb{N}$, and $f(6)=2$. Also, no two of the values $f(6), f(9), f(12)$ and $f(15)$ coincide. How many three-digit positive integers $n$ satisfy $f(n)=f(2005)$ ?
12. Ans: 225
Since $f(6+n)=f(f(6)+n)=f(2+n)$ for all $n$, the function $f$ is periodic with period 4 starting from 3 onwards. Now $f(6), f(5)=f(9), f(4)=f(12)$ and $f(3)=f(15)$ are four distinct values. Thus in every group of 4 consecutive positive integers $\geq 3$, there is exactly one that is mapped by $f$ to $f(2005)$. Since the collection of three-digit positive integers can be divided into exactly 225 groups of 4 consecutive integers each, there are 225 three-digit positive integers $n$ that satisfies $f(n)=f(2005)$.