P, q, and r are prime #'s and a, b, and c are positve intgers, how many divisors

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Hello everyone.

I think i got this right but i want to make sure...
If p, q and r are prime numbers and a, b, and c are positive integers, how many possible divisors does p^a*p^b*r^c have?

I said...
There are a+1 divisors: 1, p, p^2...,p^a
A divisor is a product of anyone of the a+1 numbers listed above times anyone of the b+1 numbers 1, q, q^2...q^b. We also have r+1...1, r, r^2...,r^c so by the muliplication rule, there are (a+1)(b+1)(r+1) divisors in all.
 
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Yes, that looks good, except for one part. The answer should be (a+1)(b+1)(c+1). You wrote (r+1) instead of (c+1), but I think this was just a typo.
 
ahh yes! thanks for picking that up, it was a typo.
Thanks for the responce.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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