5. Let $N>1$ be a positive integer, and $m$ denote the largest divisor of $N$ that is less than $N$. If $N+m$ is a power of 10, find $N$.


5. $N=75$.

Let $N=m p$. Then $p$ is the smallest prime factor of $N$.
By the problem, we know $m(p+1)=10^{k}$.
Since $10^{k}$ is not a multiple of 3, therefore, $p>2$.
Hence, $N$ and $m$ are both odd.
Thus, $m=5^{*}$.
If $s=0, N=p=10^{k}-1$ is a multiple of 9, which is a contradiction.
Then $s \geqslant 1,5 \mid N$.
Therefore, $p \leqslant 5$.
If $p=3$, then $4 \times 5^{s}=10^{k}$.
So $k=2, m=25, N=75$.
If $p=5$, we get $6110^{k}$, which is a contradiction.