HMMT-Feb 2020 alg, Problem 4

For positive integers $n$ and $k$, let $\mho(n, k)$ be the number of distinct prime divisors of $n$ that are at least $k$. For example, $\mho(90,3)=2$, since the only prime factors of 90 that are at least 3 are 3 and 5 . Find the closest integer to \[ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n, k)}{3^{n+k-7}} \]

Solution: Solution: A prime $p$ is counted in $\mho(n, k)$ if $p \mid n$ and $k \leq p$. Thus, for a given prime $p$, the total contribution from $p$ in the sum is \[ 3^{7} \sum_{m=1}^{\infty} \sum_{k=1}^{p} \frac{1}{3^{p m+k}}=3^{7} \sum_{i \geq p+1} \frac{1}{3^{i}}=\frac{3^{7-p}}{2} \] Therefore, if we consider $p \in\{2,3,5,7, \ldots\}$ we get \[ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n, k)}{3^{n+k-7}}=\frac{3^{5}}{2}+\frac{3^{4}}{2}+\frac{3^{2}}{2}+\frac{3^{0}}{2}+\varepsilon=167+\varepsilon \] where $\varepsilon<\sum_{i=11}^{\infty} \frac{3^{7-i}}{2}=\frac{1}{108} \ll \frac{1}{2}$. The closest integer to the sum is 167 . $\fbox{167}$.

Answer: 167