All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 SUM = PR0DUCT (Posted on 2017-10-18)
List all the integers N, below 10000, such that the sum of their proper divisors (i.e. N excluded) equals their product.

Counterexample:
take 12: s(1,2,3,4,6)=16; p(1,2,3,4,6)=144;
so 12 is not on the list.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Analytic Solution | Comment 2 of 3 |
Assume that N has two divisors greater than 2, call them f and g with f<g.

Then f+g+3<3*g<=f*g implies the product of the divisors will always be greater than the sum.  Adding more large divisors will only increase the difference.

Therefore N must have at most one divisor f greater than 2.  Then the possible sets of divisors of N are {1}, {1,f}, {1,2}, {1,2,f}.

The first set implies that N is a prime number.

The second set and third sets fail the sum=product requirement.

The fourth set implies 3+f = 2*f, which makes f=3 and N=6.

Then the solution set for this puzzle consists of 6 and all prime numbers.

Edited on October 20, 2017, 11:49 am
 Posted by Brian Smith on 2017-10-18 12:02:57

Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information