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

Home > Numbers
Cube first, or square second (Posted on 2006-07-22) Difficulty: 3 of 5
For any number n, let

A= The square of the sum of the number of divisors of each of the divisors of n.

B= The sum of the cubes of the number of divisors of each of the divisors of n

Prove A=B

For example, let's have n=6. Its divisors are 1, 2, 3, and 6. These numbers respectively have 1, 2, 2, and 4 divisors. So A=(1+2+2+4)²=81 and B=1³+2³+2³+4³= 81.

No Solution Yet Submitted by Jer    
Rating: 4.2857 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 3 of 7 |

See

www.geocities.com/bractals/squarecube.html

for a solution.


  Posted by Bractals on 2006-07-22 12:44:06
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information