# 2520 equals Sum of 4 Divisors in 6 Ways

## Theorem

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

 $\ds 2520$ $=$ $\ds 1260 + 630 + 504 + 126$ $\ds$ $=$ $\ds 1260 + 630 + 421 + 210$ $\ds$ $=$ $\ds 1260 + 840 + 360 + 60$ $\ds$ $=$ $\ds 1260 + 840 + 315 + 105$ $\ds$ $=$ $\ds 1260 + 840 + 280 + 140$ $\ds$ $=$ $\ds 1260 + 840 + 252 + 168$

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

## Proof

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

 $\ds 1$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}$

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

 $\ds 42$ $=$ $\ds 2^1 \times 3^1 \times 7^1$ $\ds 24$ $=$ $\ds 2^3 \times 3^1$ $\ds 18$ $=$ $\ds 2^1 \times 3^2$ $\ds 30$ $=$ $\ds 2^1 \times 3^1 \times 5^1$ $\ds 20$ $=$ $\ds 2^2 \times 5^1$ $\ds 12$ $=$ $\ds 2^2 \times 3^1$

Therefore the smallest number would be:

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

$\blacksquare$