2. Find the smallest positive integer $n$ for which the last three digits of $2007 n$ (in decimal notation) are 837.
求最小的正整數 $n$, 使得 $2007 n$ (在十進制中) 的最後三位數字是 837 。
2. 691
2. Since $2007 \equiv 7(\bmod 1000)$, we need $n$ for which $7 n \equiv 837(\bmod 1000)$. Noting that $7 \times 143=1001 \equiv 1(\bmod 1000)$, we multiply both sides of $7 n \equiv 837(\bmod 1000)$ by 143 to get $7 n(143) \equiv 837(143)(\bmod 1000)$, or $n \equiv 691(\bmod 1000)$. Hence the answer is 691 .
Alternative Solution. To find the smallest $n$ for which $7 n \equiv 837(\bmod 1000)$, we look for the smallest multiple of 7 in the sequence $837,1837,2837, \ldots$ It turns out that 4837 is the first term divisible by 7 , and hence the answer is $4837 \div 7=691$.