#math trivia for #February17: #48 is very abundant — its divisors add up to 124. How do you go from 48 = 16*3 to 124 = 31*4?

— Burt Kaliski Jr. (@modulomathy) February 17, 2012

A **perfect number** *n* one whose smaller divisors add up to *n*. The first two perfect numbers are 6 (smaller divisors 1, 2 and 3) and 28 (1, 2, 4, 7, and 14).

If the sum of the smaller divisors is greater than *n*, then *n* is **abundant**; if it’s less than *n*, then *n* is **deficient**.

If we look instead at the sum of all divisors of *n*, including *n* itself, then perfect, abundant, and deficient correspond to a sum that equals, is more than, or is less than 2*n*.

The sum of the divisors of 48 is 124, well over 2*48 = 96, hence the term “very abundant”.

The problem is looking for a simple pattern that relates the sum of the divisors of 48 to its factorization, 16*3.

What are the divisors of 48? Given the prime factorization 48 = 16*3 = 24*3, it follows that any divisor *d* of 48 must have the form

d = 2^{x}*3^{y}

where 0 ≤ *x* ≤ 4 and 0 ≤ *y* ≤ 1. This leads to the following set of 10 divisors, one for each possible combination of *x* and *y* values:

- 1, 3
- 2, 6
- 4, 12
- 8, 24
- 16, 48

The sum of the first divisor is each row is 31. The sum of each row is 4 times the first divisor in the row. So the sum overall is 31*4 = 124.

A general number *n* of the form 2* ^{a}**3

*has (*

^{b}*a*+1)*(

*b*+1) divisors, again each of the form 2

**3*

^{x}*. The sum of the first “column” of divisors in this case is 1+2+4+…+2*

^{y}*, or 2*

^{a}^{a+1}-1. The sum of each row is 1+3+9+…+3

*, or (3*

^{b}^{b+1}-1)/2 times the first divisor in the row. The sum of the divisors is thus (2

^{a+1}-1)*(3

^{b+1}-1)/2.

A similar formula can be developed based on the factorization of any number *n*. One special example: Suppose *n* has the form 2* ^{a}**

*p*where

*p*= (2

^{a+1}-1) and

*p*is prime. Then the sum of the first column is 2

^{a+1}-1, the sum of each row is

*p*+1 times the first divisor in the row, and the overall sum is (2

^{a+1}-1)*(

*p*+1) =

*p**2

^{a+1}= 2

*n*. Such an

*n*is perfect.