If a natural number p has 8 factors, then which of the

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK OT THE ORIGINAL QUESTION:

We are told that \(p\) has 8 factors: 8=2*4=2*2*2.

If \(p\) has only one prime in its prime factorization, say \(a\), then \(p=a^7\) --> number of factors of \(p\) is \((7+1)=8\) --> \(p^3=(a^7)^3=a^{21}\) in this case would have \((21+1)=22\) factors --> the difference is \(22-8=14\);

If \(p\) has two primes in its prime factorization, say \(a\) and \(b\), then: \(p=a^3*b\) --> number of factors of \(p\) is \((3+1)(1+1)=8\) --> \(p^3=a^9*b^3\) in this case would have \((9+1)(3+1)=40\) factors --> the difference is \(40-8=32\);

If \(p\) has three primes in its prime factorization, say \(a\), \(b\) and \(c\), then: \(p=a*b*c\) --> number of factors of \(p\) is \((1+1)(1+1)(1+1)=8\) --> \(p^3=a^3*b^3*c^3\) in this case would have \((3+1)(3+1)(3+1)=64\) factors --> the difference is \(64-8=56\).

p cannot have more than 3 factors, since the least number of factors a number with four primes can have is 16>8 (for example if \(p=abcd\), then number of factors of \(p\) is \((1+1)(1+1)(1+1)(1+1)=16\)).

Answer: B.