Divisor Count of 17,796,126,877,482,329,126,051

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Example of Use of Divisor Count Function

$\map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051} = 48$

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:

$17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051 = 7^5 \times 29 \times 351 \, 121 \times 103 \, 987 \, 345 \, 177$


Thus:

\(\ds \) \(\) \(\ds \map {\sigma_0} {17 \, 796 \, 126 \, 877 \, 482 \, 329 \, 126 \, 051}\)
\(\ds \) \(=\) \(\ds \map {\sigma_0} {7^5 \times 29^1 \times 351 \, 121^1 \times 103 \, 987 \, 345 \, 177^1}\)
\(\ds \) \(=\) \(\ds \paren {5 + 1} \times \paren {1 + 1} \times \paren {1 + 1} \times \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 6 \times 2^3\)
\(\ds \) \(=\) \(\ds 48\)

$\blacksquare$