17. From any $n$-digit $(n>1)$ number $a$, we can obtain a $2 n$-digit number $b$ by writing two copies of $a$ one after the other. If $\frac{b}{a^{2}}$ is an integer, find the value of this integer.
(2 marks)
對於任意 $n$ 位數 $a$ （其中 $n>1$ ）, 把 $a$ 連寫兩次可得到一個 $2 n$ 位數 $b \circ$ 若 $\frac{b}{a^{2}}$ 為
整數，求此整數的值。
(2 分)
17. 7
17. Note that $b=a\left(10^{n}+1\right)$, so $\frac{b}{a^{2}}=\frac{10^{n}+1}{a}$. Let this be an integer $d$. Noting that $10^{n-1} \leq a1$, we must have $1<d<11$. Since $10^{n}+1$ is not divisible by 2,3 and 5 , the only possible value of $d$ is 7. Indeed, when $a=143$, we have $b=143143$ and $d=7$.