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A caracterization of almost-perfect numbers

Source: Brazil Math Olympiad - 2000

March 3, 2006
floor functionnumber theory proposednumber theory

Problem Statement

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).
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