Divisor Sum of 6825
Jump to navigation
Jump to search
Example of Divisor Sum of Integer
- $\map {\sigma_1} {6825} = 13 \, 888$
where $\sigma_1$ denotes the divisor sum function.
Proof
We have that:
- $6825 = 3 \times 5^2 \times 7 \times 13$
Hence:
\(\ds \map {\sigma_1} {6825}\) | \(=\) | \(\ds \paren {3 + 1} \times \dfrac {5^3 - 1} {5 - 1} \times \paren {7 + 1} \times \paren {13 + 1}\) | Divisor Sum of Integer | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \times \dfrac {124} 4 \times 8 \times 14\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \times 31 \times 8 \times 14\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^2 \times 31 \times 2^3 \times \paren {2 \times 7}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^6 \times 7 \times 31\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 13 \, 888\) |
$\blacksquare$