4. Let $n$ be an integer greater than 1 . If all digits of $9997 n$ are odd, find the smallest possible value of $n$.
(1 mark)
設 $n$ 爲大於 1 的整數。若 $9997 n$ 的所有數字皆是奇數，求 $n$ 的最小可能值。
( 1 分)
4. 3335
4. Note that $9997 n=(10000-3) n=10000(n-1)+(10000-3 n)$. Thus $n$ must be odd to ensure that the last digit of $9997 n$ is odd, and hence $n-1$ is even. If $n<3335$, then $0<10000-3 n<10000$ and hence the ten thousands digit of $9997 n$ is the same as the unit digit of $10000(n-1)$, which is even. Therefore $9997 n$ contains an even digit. If $n=3335$, we find that $9997 \times 3335=33339995$. It follows that the smallest possible value of $n$ is 3335 .