Call a positive integer n "favorable" if there is a set of n distinct positive integers whose reciprocals' sum adds to 1.

How many unfavorable numbers are there?

(In reply to Minimalist Answer / No Spoiler by Old Original Oskar!)

1 is favorable because 1/1 = 1

2 is unfavorable because 1 - 1/1 is 0 which is the reciprocal of no positive integer and 1 - 1/2 = 1/2 leading to duplicate positive integers and 1 - the reciprocal of any other positive integer is greater than 1/2 and therefore no such sum can be made.

3 is favorable because 1 = 1/2 + 1/3 + 1/6

4 is favorable because 1 = 1/2 + 1/4 + 1/6 + 1/12

By inductive proof we can show that any two consecutive favorable numbers are followed by a third ad infinitum. Namely the nth number is favorable because 1 = 1/2 + 1/2(expansion of the (n-1)st sum) assuring no repeats because 1/1 is not an element of the (n-1)st expansion.

Posted by Eric on 2006-11-02 12:14:47