Count the number of divisors of a number given its prime factors in the form:
This is because we can choose any number of a's from 0 to p, which gives us (p+1) choices, and so on.
http://mathforum.org/library/drmath/view/55741.html
If a number n look like this: n = a^p * b^q * c^r * d^s * e^t * f^u ... , where a,b, c and so on are its prime factors
then the number of divisors of n = (p+1)*(q+1)*(r+1)*...
This is because we can choose any number of a's from 0 to p, which gives us (p+1) choices, and so on.
http://mathforum.org/library/drmath/view/55741.html
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