6. For a natural number $n$, $G(n)$ denotes the number of natural numbers $m$ for which $m+n$ divides $m n$. Find $G\left(10^{\mathrm{k}}\right)$.
Answer: $G\left(10^{k}\right)=2 k^{2}+2 k$.

Sketch of the solution. Since $m n=n(m+n)-n^{2}$, if $m+n$ divides $m n$, then $m+n$ divides $n^{2}$. Therefore, $G(n)$ is equal to the number of divisors of $n^{2}$ that are greater than $n$. For each divisor $d > n$ of $n^{2}$, we can associate a divisor $n^{2} / d < n$, which means the number of divisors of $n^{2}$ greater than $n$ is equal to the number of divisors of $n^{2}$ less than $n$. Therefore, $2G(n)+1$ is the number of divisors of $n^{2}$. The number of natural divisors of $10^{2k}$ is $(2k+1)^{2}$, and thus $G\left(10^{k}\right)=2 k^{2}+2 k$.