Question

The prime factorization of integer N is AX
AxBxCx D x D, where A, B, C and D are
all distinct prime integers. How many
factors does N have?​

Answer 1

Answer:

4 maybe.............

Answer 2

Answer:

Essentially, you want to know the number of factors of n where,

n=a2∗b1∗c1; a,b,c are primes.

There's a rule which says:

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)∗....... .

so, here, a,b,c are primes, and their powers are 2,1,1 respectively.

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 .