Divisor Count of 17,796,126,877,482,329,126,045

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Example of Use of Divisor Count Function

$\map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045} = 48$

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:

$17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045 = 5 \times 17^2 \times 47 \times 53 \times 4 \, 944 \, 062 \, 119 \, 125 \, 691$


Thus:

\(\ds \) \(\) \(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 045}\)
\(\ds \) \(=\) \(\ds \map {\sigma_0} {5^1 \times 17^2 \times 47^1 \times 53^1 \times 4 \, 944 \, 062 \, 119 \, 125 \, 691^1}\)
\(\ds \) \(=\) \(\ds \paren {1 + 1} \times \paren {2 + 1} \times \paren {1 + 1} \times \paren {1 + 1} \times \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 3 \times 2^4\)
\(\ds \) \(=\) \(\ds 48\)

$\blacksquare$