Question

Write one digit on the both sides of 57 to make the number divisible by 72. How many solutions does this problem have?

Answer 1

im assuming that the other person who answered got it wrong and he/she/they got sooo

Answer 2

I'm assuming writing "one digit on the both sides of 57" means you write the same digit to either side, like in 1571?

Given a number $n$ with $k$ digits, prepending and appending the same digit $d$, $1\le d\le9$ (omit 0 because it doesn't change the starting number), is the same as multiplying $n$ by 10 and adding $(10^{k+1}+1)d$. We have $n=57$, so we're looking for $d$ such that

$570+(10^3+1)d=1001d+570\equiv0\pmod{72}$

8 and 9 are coprime, so we can use the Chinese remainder theorem.

$72=8\cdot9\implies\begin{cases}1001d+570\equiv d+2\equiv0\pmod8\\1001d+570\equiv2d+3\equiv0\pmod9\end{cases}$

which simplifies to

$\begin{cases}d\equiv6\pmod8\\d\equiv3\pmod9\end{cases}$

Now we apply the CRT. We want some number $d$ such that, taken modulo 8, returns a remainder of 6, but taken modulo 9, returns a remainder of 3. So we could try

$d=6\cdot9+3\times8=54+24=78$

Modulo 8, the second term vanishes, and $54\equiv6\pmod8$. However, modulo 9, the first term vanishes but $24\equiv6\equiv2\cdot3\pmod9$, whereas we only want 3. So we multiply the second term by the inverse of 2 modulo 9.

To find the inverse, notice that $10\equiv2\times5\equiv1\pmod9$, so $2^{-1}\equiv5\pmod9$, and so we multiply the second term by 5. Now,

$d=6\cdot9+3\times8\times5=174$

and

$174\equiv72\cdot2+30\equiv30\pmod{72}$

so we find that $d=72n+30$. In particular, the smallest positive solution is 30, which is larger than 9 so there are no single-digit choices of $d$ that makes the new number divisible by 72.