Divisor Count of 40,315

Example of Use of Divisor Count Function

$\map {\sigma_0} {40 \, 315} = 8$

where $\sigma_0$ denotes the divisor count function.

Proof

From Divisor Count Function from Prime Decomposition:

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

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:

$40 \, 315 = 5 \times 11 \times 733$

Thus:

\(\ds \map {\sigma_0} {40 \, 315}\)

\(=\)

\(\ds \map {\sigma_0} {5^1 \times 11^1 \times 733^1}\)

\(\ds \)

\(=\)

\(\ds \paren {1 + 1} \paren {1 + 1} \paren {1 + 1}\)

\(\ds \)

\(=\)

\(\ds 8\)

The divisors of $40 \, 315$ can be enumerated as:

$1, 5, 11, 55, 733, 3665, 8063, 40 \, 315$

$\blacksquare$