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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.)
re(2): Solution | Comment 6 of 7 |
(In reply to re: Solution by Richard)

Thanks for catching that. Yes the sum should be from 0 instead of 1. I have corrected the html document. Were you able to follow the proof?

Edited on July 22, 2006, 10:56 pm
  Posted by Bractals on 2006-07-22 22:54:55

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