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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)=81 and B=1+2+2+4= 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.)
Hints/Tips re: A bet | Comment 2 of 7 |
(In reply to A bet by Old Original Oskar!)

You would win that bet. I used that and the Unique Factorization Theorem to solve the problem. The only thing is that the proof is loaded with superscripts and subscripts. Here would be a nice place for the comments section to allow HTML tags.
  Posted by Bractals on 2006-07-22 09:39:17

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