Divisor Sum of 210

From ProofWiki
Jump to navigation Jump to search

Example of Divisor Sum of Square-Free Integer

$\map {\sigma_1} {210} = 576$

where $\sigma_1$ denotes the divisor sum function.

Proof

We have that:

$210 = 2 \times 3 \times 5 \times 7$


Hence:

\(\ds \map {\sigma_1} {210}\) \(=\) \(\ds \paren {2 + 1} \paren {3 + 1} \paren {5 + 1} \paren {7 + 1}\) Divisor Sum of Square-Free Integer
\(\ds \) \(=\) \(\ds 3 \times 4 \times 6 \times 8\)
\(\ds \) \(=\) \(\ds 3 \times 2^2 \times \paren {2 \times 3} \times 2^3\)
\(\ds \) \(=\) \(\ds 2^6 \times 3^2\)
\(\ds \) \(=\) \(\ds \paren {2^3 \times 3}^2\)
\(\ds \) \(=\) \(\ds 24^2\)
\(\ds \) \(=\) \(\ds 576\)

$\blacksquare$