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#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 = 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^b has (a+1)*(b+1) divisors, again each of the form 2^x*3^y.  The sum of the first “column” of divisors in this case is 1+2+4+…+2^a, or 2^a+1-1.  The sum of each row is 1+3+9+…+3^b, or (3^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 = 2n.  Such an n is perfect.