2. Find a positive integer $n$ less than 2006 such that $2006 n$ is a multiple of $2006+n$.
(1 mark)求一個小於 2006 的正整數 $n$, 使得 $2006 n$ 是 $2006+n$ 的倍數。
2. 1475
2. Note that $\frac{2006 n}{2006+n}=\frac{2006^{2}+2006 n}{2006+n}-\frac{2006^{2}}{2006+n}=2006-\frac{2006^{2}}{2006+n}$.
Hence, in order for $2006 n$ to be a multiple of $2006+n, 2006+n$ must be a factor of $2006^{2}$. As $1 \leq n \leq 2005$, we have $2007 \leq 2006+n \leq 4011$. As $2006^{2}=2^{2} \times 17^{2} \times 59^{2}$, the only factor of $2006+n$ between 2007 and 4011 is $59^{2}=3481$. Hence we must have $2006+n=3481$, or $n=1475$.