9.5. To a natural number $N$, the largest divisor of $N$ less than $N$ was added, and the result was a power of ten. Find all such $N$. (N. Agakhanov)
Answer: 75.

Solution. Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=mp$, where $p$ is the smallest prime divisor of the number $N$. We have $N+m=10^{k}$, that is, $m(p+1)=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is an odd number, and therefore $m$ is also odd. Thus, since $10^{k}$ is divisible by $m$, we get $m=5^{s}$. If $m=1$, then $N=p=10^{k}-1$, which is impossible since $10^{k}-1$ is divisible by 9, and thus is not a prime. Therefore, $s \geqslant 1$, the number $N$ is divisible by 5, and hence $p \leqslant 5$. If $p=3$, we get the equation $4 \cdot 5^{s}=10^{k}$, from which $k=2, m=25$, and $N=75$. If $p=5$, then $p+1=6$, and the number $10^{k}$ is divisible by 3, which is impossible.