Divisor Count of 15
Jump to navigation
Jump to search
Example of Use of Divisor Count Function
- $\map {\sigma_0} {15} = 4$
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:
- $15 = 3 \times 5$
Thus:
\(\ds \map {\sigma_0} {15}\) | \(=\) | \(\ds \map {\sigma_0} {3^1 \times 5^1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 + 1} \paren {1 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4\) |
The divisors of $15$ can be enumerated as:
- $1, 3, 5, 15$
$\blacksquare$