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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(3): Solution Comment 7 of 7 |
(In reply to re(2): Solution by Bractals)

It looks correct to me now.  I do intend to do a further verification (what I would term an acid test) by specializing to n=(2^3)*(3^2) and working out each step numercially. I would urge others to try the same.
  Posted by Richard on 2006-07-23 04:00:33

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