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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: ( You must be logged in to post comments.)
  Subject Author Date
re(3): SolutionRichard2006-07-23 04:00:33
re(2): SolutionBractals2006-07-22 22:54:55
re: SolutionRichard2006-07-22 22:09:57
A bet winning solutionFederico Kereki2006-07-22 13:27:55
SolutionSolutionBractals2006-07-22 12:44:06
Hints/Tipsre: A betBractals2006-07-22 09:39:17
Some ThoughtsA betOld Original Oskar!2006-07-22 09:28:54
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