olymmath

OBM 2000 Problem 2

    Let s(n)s(n) be the sum of all positive divisors of nn , so s(6)=12s(6) = 12 . We say nn is almost perfect if s(n)=2n−1s(n) = 2n - 1 . Let mod  (n,k)\mod(n, k) denote the residue of nn modulo kk (in other words, the remainder of dividing nn by kk ). Put t(n)=mod  (n,1)+mod  (n,2)+⋯+mod  (n,n)t(n) = \mod(n, 1) + \mod(n, 2) + \cdots + \mod(n, n) .

    Show that nn is almost perfect if and only if t(n)=t(n−1)t(n) = t(n-1) .

    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 br-obm/2000/p2 · Tag EVY72JWFAK · Edit this page on GitHub · Keys: j next, k previous