#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 2n.
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 = 2x*3y
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 2a*3b has (a+1)*(b+1) divisors, again each of the form 2x*3y. The sum of the first “column” of divisors in this case is 1+2+4+…+2a, or 2a+1-1. The sum of each row is 1+3+9+…+3b, or (3b+1-1)/2 times the first divisor in the row. The sum of the divisors is thus (2a+1-1)*(3b+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 2a*p where p = (2a+1-1) and p is prime. Then the sum of the first column is 2a+1-1, the sum of each row is p+1 times the first divisor in the row, and the overall sum is (2a+1-1)*(p+1) = p*2a+1 = 2n. Such an n is perfect.