Let $N$ be a positive integer with $d\geq 4$ digits, none of which is zero. Suppose that erasing some digit of $N$ yields another number $M$ which happens to be a divisor of $N$.
Examples : 1375 divides 12375. 1875 divides 61875.
Question : is it true that $M$ must always end with 25 or 75, and that the two final digits of $M$ must be the same as those of $N$ (so that they too will be 25 or 75) ?
I have checked this with a computer for $d=4$ and $d=5$.
On
First, convince yourself that the deleted digit can't be the unit or tens digit.
Then let the last 2 digits form the number $a$, and let $N/M=b$. Then we have $$ab\equiv a\pmod{100}$$ which is to say, $a(b-1)$ is a multiple of 100. Here's what we know about $a$ and $b$: neither digit of $a$ is zero, $11\le a<100$, $2\le b<100$. Also, $b-1$ can't be relatively prime to 100. Let $e=\gcd(b-1,100)$. Then $e$ can't be 2, as that forces $a=50$; $e$ can't be 5, that forces $a$ to be a multiple of 20; $e$ can't be 10, that would force $a$ to be a multiple of 10. So $e$ is 4, 20, 25, or 50. Well, that narrows things down a bit, though I'm afraid I don't have the strength to try to push it through to a complete answer.
Let $n$ be the original number and $m$ the number after a digit is deleted. Suppose that the deleted digit is $d$, that $r$ is the number represented by the $k$ digits to the right of $d$, and that $\ell$ is the number represented by the digits to the left of $d$, so that $m=10^k\ell+r$, and $n=10^{k+1}\ell+10^kd+r$.
Suppose first that $n=(10-s)m$, where $1\le s\le 8$. Then
$$10^{k+1}\ell+10^kd+r=10^{k+1}\ell-10^ks\ell+(10-s)r\;,$$
so $(9-s)r=10^k(d+s\ell)$. Now $(9-s)r$ has at most $k+1$ digits, and $10^k(d+s\ell)$ has at least $k+1$ digits, so in fact $d+s\ell$ is a single digit, and $(9-s)r$ has $k+1$ digits. Thus, $\ell$ is a single digit, and we must have $k\ge 2$ (since $n$ has at least $4$ digits). If $s=4$, then $r=20\cdot10^{k-2}(d+s\ell)$ ends in $0$, which is impossible. Otherwise $25$ divides $(9-s)r$ and is relatively prime to $9-s$, so $25\mid r$, and since $n$ has no zero digit, $r$ must end in $25$ or $75$. (Of course the last two digits of $r$ are the last two digits of $m$ and $n$ as well.)
Now suppose that $\frac{n}m>10$. If $d$ is not the first digit of $n$, then $11\le\frac{n}m\le 19$. Suppose that $n=(10+s)m$, where $1\le s\le 9$. Then
$$10^{k+1}\ell+10^kd+r=10^{k+1}\ell+10^ks\ell+(10+s)r\;,$$
so $(9+s)r=10^k(d-s\ell)$. Clearly $d>s\ell$, so $\ell$ is a single digit, and $k\ge 2$. Thus, $25\mid(9+s)r$. If $s=6$, then $3r=20\cdot10^{k-2}(d-6\ell)$, so $10\mid r$, which is impossible. Otherwise, $25\mid r$, and we’re done, as before.
Finally, suppose that $d$ is the first digit of $n$, so $n=10^kd+m$. Suppose further that $n=sm$, so that $10^kd+m=sm$, and $10^kd=(s-1)m$. Now $k\ge 3$, so $125\mid 10^k$, and hence $25\mid m$ (and we’re done, as before) unless $25\mid s-1$. Clearly $s<100$, so we need only worry about the cases $s=26$, $s=51$, and $s=76$. In those cases we have $40d=m$, $40d=2m$, and $40d=3m$, respectively, and in each case $10\mid m$, which is impossible.