How many different positive integer solutions does the equation $x y=10^{n}$ have?
$x$ and $y$ are each divisors of $10^{n}=2^{n} \cdot 5^{n}$. It is sufficient to find how many divisors $10^{n}$ has. These are, on the one hand, the terms of the sequences

$$
\begin{aligned}
& 1,2,2^{2}, \ldots 2^{n-1}, 2^{n} \\
& 1,5,5^{2}, \quad 5^{n-1}, 5^{n}
\end{aligned}
$$

and, on the other hand, they arise from the product of any two terms of these two sequences, so they are the terms of the product

$$
\left(1+2+2^{2}+\ldots+2^{n-1}+2^{n}\right)\left(1+5+5^{2}+\ldots+5^{n-1}+5^{n}\right)
$$

The number of these is: $(n+1)(n+1)=(n+1)^{2}$.

Tibor Hoffmann (Szent-István g. VI. o. Bp. XIV.)

Solved by: Baka S., Deák A., Freud G, Juhász Kató, Lipsitz I., Prack Éva.

The exchange of values of $x$ and $y$ was not considered a separate solution: Haraszthy A., Hódi E., Kornis Edit, Pallós K., Trellay J.

Some of the solutions received correctly state the number of positive integer roots but do not prove it. Naturally, these were not acceptable.