The prime factorization of integer N is A x A x B x C, where A, B and C are all distinct prime integers. How many factors does N have?
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3
N=A×A×B×C
N=A^2×B^1×C^1
no.of factors=(2+1)(1+1)(1+1)
=3×2×2
=12
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n=a2∗b1∗c1; a,b,c are primes.
If n=(p1)a∗(p2)b∗(p3)c∗......where p1,p2,p3 are primes then, the total number of factors of n is (a+1)∗(b+1)∗(c+1)∗........
Do, here, a,b,c are primes
The powers are 2,1,1 respectively and you want to know the number of factors of n where,
So, according to the formula, number of factors of n is (2+1)∗(1+1)∗(1+1) which turns out to be 3∗2∗2=12
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