A positive integer n is called an abundant number if (n) > 2n. Determine all abundant numbers
less than 50.
start by "crossing numbers off the list". for example:
σ(p) = p+1 < 2p (if p is an odd prime).
σ(pq) = pq + p + q + 1 < pq + p + p + 1 = pq + 2p + 1 ≤ pq + pq = 2pq (when p,q are distinct odd primes with p > q)
(to see the last inequality, note that if q is an odd prime: q - 2 ≥ 1, so p(q - 2) ≥ 1, so pq ≥ 2p + 1).
σ(p2) = p2 + p + 1 < p2 + 2p < p2 + p2 = 2p2 (for any odd prime p).
can you think of some more ways to eliminate numbers without actually computing σ(n) for every single number?