# 2520 equals Sum of 4 Divisors in 6 Ways

Jump to navigation
Jump to search

## Theorem

The number $2520$ can be expressed as the sum of $4$ of its divisors in $6$ different ways:

\(\displaystyle 2520\) | \(=\) | \(\displaystyle 1260 + 630 + 504 + 126\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1260 + 630 + 421 + 210\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1260 + 840 + 360 + 60\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1260 + 840 + 315 + 105\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1260 + 840 + 280 + 140\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1260 + 840 + 252 + 168\) |

This is the maximum possible number of ways it is possible to express an integer as the sum of $4$ of its divisors.

## Proof

We apply 1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways:

The number $1$ can be expressed as the sum of $4$ distinct unit fractions in $6$ different ways:

\(\displaystyle 1\) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}\) |

Find the maximum powers of the primes in each equation, and choose the largest that appears:

\(\displaystyle 42\) | \(=\) | \(\displaystyle 2^1 \times 3^1 \times 7^1\) | |||||||||||

\(\displaystyle 24\) | \(=\) | \(\displaystyle 2^3 \times 3^1\) | |||||||||||

\(\displaystyle 18\) | \(=\) | \(\displaystyle 2^1 \times 3^2\) | |||||||||||

\(\displaystyle 30\) | \(=\) | \(\displaystyle 2^1 \times 3^1 \times 5^1\) | |||||||||||

\(\displaystyle 20\) | \(=\) | \(\displaystyle 2^2 \times 5^1\) | |||||||||||

\(\displaystyle 12\) | \(=\) | \(\displaystyle 2^2 \times 3^1\) |

Therefore the smallest number would be:

- $2^3 \times 3^2 \times 5^1 \times 7^1 = 2520$

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $2520$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $2520$