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Solution to puzzle 116: Factorial divisors

Skip restatement of puzzle.Show that, for each n greater than or equal to 3,  n! can be represented as the sum of n distinct divisors of itself.  (For example, 3! = 1 + 2 + 3.)


We will use mathematical induction on n to prove the slightly stronger result that, for each n greater than or equal to 3,  n! can be represented as the sum of n distinct divisors of itself, with 1 as one of the divisors..

We have the base case, for n = 3:  3! = 1 + 2 + 3, with 1 as one of the divisors.

For the induction step, we assume the inductive hypothesis, that we have a representation for n = k, with 1 as one of the divisors.
That is, k! = 1 + d2 + d3 + ... + dk, with 1 < d2 < ... < dk.

Then (k + 1)! = (k + 1) + (k + 1)d2 + ... + (k + 1)dk
 = 1 + k + (k + 1)d2 + ... + (k + 1)dk

That is, the sum of k + 1 distinct divisors of (k + 1)!, with 1 as one of the divisors.

It follows by induction that, for each n greater than or equal to 3,  n! can be represented as the sum of n distinct divisors of itself.


A generalization (3 star)

Show that any given positive integer less than or equal to n! can be represented as the sum of at most n distinct divisors of n!.

Hint  -  Solution

Source: Mathematical Miniatures, by Svetoslav Savchev and Titu Andreescu. See Chapter 15.

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