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:
\(\ds 2520\) | \(=\) | \(\ds 1260 + 630 + 504 + 126\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1260 + 630 + 421 + 210\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1260 + 840 + 360 + 60\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1260 + 840 + 315 + 105\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1260 + 840 + 280 + 140\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1260 + 840 + 252 + 168\) |
This is the maximum possible number of ways it is possible to express an integer as the sum of $4$ of its divisors.
Proof
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:
\(\ds 1\) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}\) |
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:
- $2^3 \times 3^2 \times 5^1 \times 7^1 = 2520$
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2520$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2520$