Divisor Sum of 570
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Example of Divisor Sum of Square-Free Integer
- $\map {\sigma_1} {570} = 1440$
where $\sigma_1$ denotes the divisor sum function.
Proof
We have that:
- $570 = 2 \times 3 \times 5 \times 19$
Hence:
\(\ds \map {\sigma_1} {570}\) | \(=\) | \(\ds \paren {2 + 1} \paren {3 + 1} \paren {5 + 1} \paren {19 + 1}\) | Divisor Sum of Square-Free Integer | |||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 4 \times 6 \times 20\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 2^2 \times \paren {2 \times 3} \times \paren {2^2 \times 5}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^5 \times 3^2 \times 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1440\) |
$\blacksquare$