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jallenmorris
good rule, mind telling us why we do it in addition to what we should do ?

Suppose p, q and r are different primes, and p^3 * q^4 * r^2 is the prime factorization of n. How many positive divisors should n have?

Well, any number that can be written as p^a * q^b * r^c will be a divisor of n as long as:

a = 0, 1, 2 or 3 (four choices)
b = 0, 1, 2, 3 or 4 (five choices)
c = 0, 1 or 2 (three choices)

How many different sets of exponents a, b and c could we choose? This is a straight counting problem: we multiply the choices we have for each. Thus, n will have 4*5*3 = 60 divisors.

This all works because of unique factorization into primes, which guarantees that each of the divisors we get above will be different, and that n has no other divisors besides the 60 we just found.

Short version: we add one to each power because in a divisor, the power on any prime can be zero. We multiply because essentially it's just like any other counting problem.
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If positive integer n is the product of 4 different prime numbers, including 1 and n, how many factors does n have?

N =a*b*c*d, where a,b,c,d are different prime numbers

Therefore, no. of factors = (1+1)(1+1)(1+1)(1+1)= 16

Hence C
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