2520 equals Sum of 4 Divisors in 6 Ways

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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