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HMMT February

Number theory Difficulty 5.2 AIME, harder Find the answer

For positive integers nn and kk, let ℧(n,k)\mho(n, k) be the number of distinct prime divisors of nn that are at least kk. Find the closest integer to ∑n=1∞∑k=1∞℧(n,k)3n+k−7\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{\mho(n, k)}{3^{n+k-7}}

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

A prime pp is counted in ℧(n,k)\mho(n, k) if p∣np \mid n and k≤pk \leq p. Thus, for a given prime pp, the total contribution from pp in the sum is 37∑m=1∞∑k=1p13pm+k=37∑i≥p+113i=37−p23^{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∈{2,3,5,7,…}p \in\{2,3,5,7, \ldots\} we get ∑n=1∞∑k=1∞℧(n,k)3n+k−7=352+342+322+302+ε=167+ε\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 ε<∑i=11∞37−i2=1108≪12\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.

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