We define a surprising number $n$ as a positive integer such that if $n$ is written on the right of any natural number, the result is a multiple of $n$.
I concluded that if $k$ is the number of digits of $n$, then $n$ must be divisor of $10^k$. Then, $n$ may be $1, 2, 5, 10$ or a power of them. It is seen that the condition is fulfilled with any power of $10$, but with $2$ and $5$ does not occur the same. With $2 ^ 2 = 4$ no longer occurs; with $5$ it happens until $5 ^ 3 = 125$, but I can not give a clear explanation of why. I think the issue is that in $2 ^ 2$ the number of digits did not increase, however, with $5$ it is increased: $5 ^ 2 = 25, 5 ^ 3 = 125$, and when it ceases to be increased ($5 ^ 4 = 625$), then from there the condition is no longer fulfilled.
What you have noticed is the key to the problem. Note that if we have a number $x$ and we add $n$ to the right of it we get a new number:
$$\bar{x} = x\cdot 10^k + n$$
As $x$ can be any number we must have that $n \mid 10^k$, where $k$ is the number of digits of $n$. This explains why we have that $5^4$ doesn't work, as $5^4 \not \mid 10^3$.