16. Let $n$ be a positive integer. When $n$ is divided by 902,802 and 702 , the remainders are 602,502 and 402 respectively. Find the minimum remainder when $n$ is divided by 2005 .
(2 marks)設 $n$ 為正整數。當 $n$ 除以 902 、 802 及 702 時, 其餘數分別是 602 、 502 及 402 。求 $n$ 除以 2005 時的餘數的最小值。
16. 101
16. Since $n$ leaves a remainder of 502 when divided by 802 , we may write $n=802 k+502$ for some integer $k$. Let $k=5 h+r$ for some integers $h$ and $r$ such that $0 \leq r \leq 4$. Then we have
$$
n=802(5 h+r)+502=2005(2 h)+(802 r+502) .
$$

Setting $r=0,1,2,3,4$ respectively, we see that the remainders when $n$ is divided by 2005 are $502,1304,101,903$ and 1705 respectively. Finally, it is indeed possible for the remainder to be 101. The reason is as follows. According to the question, we know that $n+300$ is divisible by both 902 and 702, and hence by their L.C.M. (say $L$ ). Since $L$ and 2005 are relatively prime, the Chinese remainder theorem asserts that there exists a positive integer $n$ which is congruent to -300 modulo $L$ and to 101 modulo 2005. It follows that the answer is 101 .