If N is a positive integer and N^2 has 15 positive factors, how many p

IMPORTANT RULE: If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40

----NOW ONTO THE QUESTION-------------------
Scanning the answer choices, I can see that there must be TWO ways in which N² can have 15 positive factors.

Since 15 = (3)(5), we can see that, if N² = (some prime^2)(some other prime^4), then ...
...the total number of divisors of N² = (2+1)(4+1) = (3)(5) = 15
If N² = (some prime^2)(some other prime^4), then N = (some prime^1)(some other prime^2)
If N = (some prime^1)(some other prime^2), then the number of positive divisors of N = (1+1)(2+1) = (2)(3) = 6
ELIMINATE A, D and E

Also recognize that if N² = (some prime^14), then ...
...the total number of divisors of N² = (14+1) = 15
If N² = (some prime^14), then N = (some prime^7)
If N = (some prime^7), then the number of positive divisors of N = (7+1) = 8

Answer:

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