IMO Team Selection Contest I

Suppose that $5^x\cdot 10^n+5^y=5^z$, where $5^y$ has $n$ digits. Then $5^{x+n}\cdot 2^n=5^y\cdot (5^{z-y}-1)$, whence $2^n=5^{z-y}-1$.

Case $n=1$ does not work.

For case $n=2$ we get $z-y=1$. Since $5^y$ has $2$ digits, the only possibility is $y=2$ and $z=3$, whence $x=0$, which is not positive.

Case $n>2$ yields $5^{z-y}\equiv 1(\mathrm{mod}8)$, thus $z-y=2k$ for an integer $k$. Now $2^n=25^k-1=24\cdot (25^{k-1}+\cdots +1)$, this is impossible, since $3\mid 24$.