Cubes over the terminating denominators

Let SS be the set of positive integers kk for which 1k\tfrac1k has a terminating decimal expansion (e.g. 2S2 \in S since 12=0.5\tfrac12 = 0.5, but 7S7 \notin S). Evaluate kS1k3\displaystyle\sum_{k \in S} \frac{1}{k^3}.

Show hints (2)+
  1. 1k\tfrac1k terminates     \iff k=2a5bk=2^a5^b. Factor the sum over aa and bb separately.
  2. (8a)(125b)=87125124=250217\big(\sum 8^{-a}\big)\big(\sum 125^{-b}\big)=\tfrac87\cdot\tfrac{125}{124}=\tfrac{250}{217}.

Answer

Reveal answer →

1.1521 (± 0.005)

Want the full step-by-step worked solution? It's part of Premium - along with a worked solution for every question in the bank.

Asked at: Jane Street, SIG

Related questions