If n=30^4, then the sum of the positive integral divisors of n may be expressed as a^3bc, where a, b and c are prime. Compute a + b + c. My answer is 113. Is this correct?

Question

brinazarski · Answer

*correction incase there's confusion: a^3 multiplied by bc. bc is not part of the exponent.

anonymous · Answer

Integral divisors are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 36, 40, 45, 48, 50, etc.  You added all of these up to and including 30^4?

anonymous · Answer

That must be, what, 125 integral divisors?  That would be (2^x)(3^y)(5^z) divisors because each of x, y, and z can range from 0 to 4 inclusive.  So, that's 5^3 = 125 divisors?  This problem looks like cruel and unusual punishment indeed!  You didn't do this by hand, did you?

brinazarski · Answer

I did... we're not allowed to use our calculator :(

brinazarski · Answer

I got 3^4, 5^4 x 2^4, which is the same as 30^4,, and then I used some formula when you do (3^5  -1)/2 which equals 121, then the same for everything else, which gave me 2,929,531

brinazarski · Answer

From there, I just plugged in random prime numbers... 11^3 (a) multiplied by 71 multiplied by 31 = 2,929,531

brinazarski · Answer

I added them up... and got 113.

anonymous · Answer

Well, this is a most interesting prime number problem.  Truly fascinating, but I'm not sure you're going to find anyone here to give confirmation after needing to add up all those numbers.  It's too labor-intensive.  And then you have to find a, b, and c as primes to fit that equation.  I would think that there is some special prime-number mathematics that help, but I'm not aware of any.  I'm also not aware of the derivation of that formula you mentioned, so sorry, I can't help you.

brinazarski · Answer

Well, thanks for trying :) It's math team, the problems are incredibly difficult, and it's supposedly Pre-Calc, which I never learned. Of course -_-