## Task A-3.1.

Determine all pairs of natural numbers $(a, b)$ such that $1<a, b \leqslant 100$ and

$$
\frac{1}{\log _{a} 10}+\frac{1}{\log _{b} 10}
$$

is a natural number.
## Solution.

It holds that

$$
\frac{1}{\log _{a} 10}+\frac{1}{\log _{b} 10}=\log a+\log b=\log a b
$$

thus the number $\frac{1}{\log _{a} 10}+\frac{1}{\log _{b} 10}$ is a natural number if and only if $a b$ is a power of 10.

Since $1<a, b \leqslant 10^{2}$, it follows that $a b \in\left\{10,10^{2}, 10^{3}, 10^{4}\right\}$.

List the pairs $(a, b)$ such that $a b=10^{n}$ for $n=1,2,3,4$.

If $n=1$, we have two such pairs $(2,5)$ and $(5,2)$.

1 point

If $n=2$, we have seven pairs

$$
(2,50),(4,25),(5,20),(10,10),(20,5),(25,4),(50,2) . \quad 1 \text { point }
$$

If $n=3$, we have six pairs

$$
(10,100),(20,50),(25,40),(40,25),(50,20),(100,10) . \quad 1 \text { point }
$$

Finally, if $n=4$, we have only one pair (100, 100). 1 point

Note: If a student lists all solutions $(a, b)$ where $a \leqslant b$, but does not mention that there are also symmetric solutions, they should receive 5 points.

Unless it is the situation described in the previous sentence, a student who omits any solution $(a, b)$ for which $a b=10^{n}$ does not achieve the point awarded for that $n$.