olymmath

Putnam 1984 A6

  • Session A

Let nn be a positive integer, and let f(n)f(n) denote the last nonzero digit in the decimal expansion of n!n! . (a)(\text a) Show that if a1,a2,…,aka_1,a_2,\ldots,a_k are distinct nonnegative integers, then f(5a1+5a2+…+5ak)f(5^{a_1}+5^{a_2}+\ldots+5^{a_k}) depends only on the sum a1+a2+…+aka_1+a_2+\ldots+a_k . (b)(\text b) Assuming part (a)(\text a) , we can define g(s)=f(5a1+5a2+…+5ak),g(s)=f(5^{a_1}+5^{a_2}+\ldots+5^{a_k}), where s=a1+a2+…+aks=a_1+a_2+\ldots+a_k . Find the least positive integer pp for which g(s)=g(s+p),for all s≥1,g(s)=g(s+p),\enspace\text{for all }s\ge1, or show that no such pp exists.

Dataset text AI-MO AoPS. Quoted with source.

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator putnam/1984/a6 · Tag M0B7A2NS7J · Edit this page on GitHub · Keys: j next, k previous