# Divisor Count of 496

## Example of Use of Divisor Counting Function

$\map {\sigma_0} {496} = 10$

where $\sigma_0$ denotes the divisor counting function.

## Proof

$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$

where:

$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.

We have that:

$496 = 2^4 \times 31$

Thus:

 $\ds \map {\sigma_0} {496}$ $=$ $\ds \map {\sigma_0} {2^4 \times 31^1}$ $\ds$ $=$ $\ds \paren {4 + 1} \paren {1 + 1}$ $\ds$ $=$ $\ds 10$

The divisors of $496$ can be enumerated as:

$1, 2, 4, 8, 16, 31, 62, 124, 248, 496$

$\blacksquare$