Imagine you wish to work out the sum of divisors of the number 72. It would not take long to list the divisors, and then find their sum: 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72 = 195. However, this method would become both tedious and difficult for large numbers like 6483658785. There is a simple and elegant method. Let σ( n ) be the sum of divisors of the natural number, n . For any prime, p : σ( p ) = p + 1, as the only divisors would be 1 and p . Consider p a : σ( p a ) = 1 + p + p 2 + ... + p a (1). Multiplying by p : p σ( p a ) = p + p 2 + p 3 + ... + p a + 1 (2). Subtracting (1) from (2): p σ( p a )−σ( p a ) = ( p −1)σ( p a ) = p a +1 − 1. Hence σ( p a ) = ( p a +1 − 1)/( p − 1). For example, σ(3 4 )=(3 5 −1)/(3−1) = 242/2 = 121, and checking: 1 + 3 + 9 + 27 + 81 = 121. Al...