Another Divisibility Problem

Let $k$ be the number of digits of $y$. Then concatenating $x$ and $y$ means appending $k$ decimal places after $x$, so

$$x \# y = x\cdot 10^k + y.$$

We want this to be divisible by $x+y$:

$$x+y \mid x\cdot 10^k+y.$$

The key move is to look at the expression modulo $x+y$. Since

$$y \equiv -x \pmod{x+y},$$

we get

$$x\cdot 10^k+y \equiv x\cdot 10^k-x = x(10^k-1) \pmod{x+y}.$$

So the condition becomes:

$$x+y \mid x(10^k-1).$$

Now we want a construction that works for every $x$, without searching. Since $10^k-1$ is a number like $9, 99, 999, \dots$, it is always divisible by $9$.

So let's force

$$x+y = 9x.$$

That gives

$$y = 8x.$$

Now check it:

• Let $k$ be the number of digits of $y=8x$.
• Then $10^k-1$ is divisible by $9$.
• Therefore $x(10^k-1)$ is divisible by $9x$.
• But $9x=x+y$.

So $x+y$ divides $x(10^k-1)$, which means $x+y$ also divides $x\#y$.

The bounds are fine too:

$$x < 10^8 \implies y=8x < 8\cdot 10^8 < 10^9.$$

And $y$ is positive because $x$ is positive.

So the entire problem is just print $8x$. That's it. Problem got nuked by a repunit divisibility fact.