To H.X.К a natural number $N$ was added its largest divisor, less than $N$, and the result was a power of ten. Find all such $N$.

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Let $m$ be the greatest divisor of the number $N$, less than $N$. Then $n=m p$, where $p$ is the smallest prime divisor of the number $N$. We have

$m(p+1)=N+m=10^{k}$. The number on the right side is not divisible by 3, so $p>2$. From this, it follows that $N$ is odd, and therefore $m$ is odd. Since $10^{k}$ is divisible by $m, m=5^{s}$.

If $m=1$, then $N=p=10^{k}-1$, which is impossible since $10^{k}-1$ is divisible by 9, and thus is not prime. Therefore, $s \geq 1$, the number $N$ is divisible by 5, and so $p \leq 5$.

If $p=3$, then $4 \cdot 5^{s}=10^{k}$, 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.

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