## Frequently Asked Questions

#### How do you work out the sum of divisors?

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 145600. Fortunately, there is a simple and elegant method at hand.

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.

Although no proof is supplied here, the usefulness of the function, σ(`n`), is its multiplicativity, which means that σ(`a`×`b`×...)=σ(`a`)×σ(`b`)×..., where `a`, `b`, ..., are relatively prime.

Returning to example, we use the fact that σ(72) = σ(2^{3}×3^{2}). As 2^{3} and 3^{2} are relatively prime, we can separately work out σ(2^{3}) = 2^{4} − 1 = 15 and σ(3^{2}) = (3^{3} − 1)/2 = 13. Therefore, σ(72) = 15×13 = 195.