## Frequently Asked Questions

#### How many divisors does a number have?

Suppose you wish to find the number of divisors of 48. Starting with 1 we can work through the set of natural numbers and test divisibility in each case, noting that divisors can be listed in factor pairs.

48 = 1×48 = 2×24 = 3×16 = 4×12 = 6×8

Hence we can see that 48 has exactly ten divisors. It should also be clear that, using this method, we only ever need to work from 1 up to the square root of the number.

Although this method is quick and easy with small numbers, it is tedious and impractical for larger numbers. Fortunately there is a quick and accurate method using the divisor, or Tau, function.

Let d(`n`) be the number of divisors for the natural number, `n`.

We begin by writing the number as a product of prime factors: `n` = `p`^{a}`q`^{b}`r`^{c}...

then the number of divisors, d(`n`) = (`a`+1)(`b`+1)(`c`+1)...

To prove this, we first consider numbers of the form, `n` = `p`^{a}. The divisors are 1, `p`, `p`^{2}, ..., `p`^{a}; that is, d(`p`^{a})=`a`+1.

Now consider `n` = `p`^{a}`q`^{b}. The divisors would be:

1 | p | p^{2} | ... | p^{a} |

q | pq | p^{2}q | ... | p^{a}q |

q^{2} | pq^{2} | p^{2}q^{2} | ... | p^{a}q^{2} |

... | ... | ... | ... | ... |

q^{b} | pq^{b} | p^{2}q^{b} | ... | p^{a}q^{b} |

Hence we prove that the function, d(`n`), is multiplicative, and in this particular case, d(`p`^{a}`q`^{b})=(`a`+1)(`b`+1). It should be clear how this can be extended for any natural number which is written as a product of prime factors.

The number of divisor function can be quickly demonstrated with the example we considered earlier: 48 = 2^{4}×3^{1}, therefore d(48)=5×2=10.