All-Soviet-Union Mathematical Olympiad

Solution:

(a) This one must have slipped through: $121$!

(b) Answer: $16, 36, 121, 484$. Suppose the number has more than $2$ digits. Write it as $(10m + n)10^{r} + s$, where $1 \leq m \leq 9$, $0 \leq n \leq 9$, $0 \leq s < 10^{r}$. Then we have $k(m \cdot 10^{r} + s) = (10m + n)10^{r} + s$, for some $k > 1$.

$s$ does not contain the digits $0$ or $5$, so $5$ does not divide $s$. Hence $5$ divides $k-1$, and so $k$ must be $6$, $11$, or $16$ (if $k$ was $21$ or more, then the rhs would be negative). Since $25$ does not divide $k-1$, we must have $r = 1$ and $s$ is a single digit.

We look at each possibility for $k$ in turn. $k = 6$ gives no solutions. $k = 11$ gives about two dozen multiples of $11$ from $121$ to $891$. By inspection the only squares are $121$ and $484$. $k = 16$ gives $192$, which is not a square.

In addition, there is the possibility of $2$ digit solutions, which I had overlooked. It is easiest to check each of the $2$ digit squares, thus finding the additional solutions $16$, $36$.