Max@Math Revolution wrote: If n is the product of the squares of 4 different prime numbers, how many factors does n have?

A. 8
B. 16
C. 27
D. 64
E. 81

-----------ASIDE------------------
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

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
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n is the product of the squares of 4 different prime numbers
So, let's say n = (p²)(q²)(r²)(s²), where p, q, r and s are 4 different prime numbers
By the above rule, the number of positive divisors of n = (2+1)(2+1)(2+1)(2+1) =(3)(3)(3)(3) = 81

Answer: E

Cheers,
Brent

Max@Math Revolution wrote: ↑

Fri Jun 15, 2018 12:54 am

[GMAT math practice question]

If n is the product of the squares of 4 different prime numbers, how many factors does n have?

A. 8
B. 16
C. 27
D. 64
E. 81

We can use 2, 3, 5, and 7:

So, we have 2^2 x 3^2 x 5^2 x 7^2

To determine the total number of factors, we add 1 to each exponent attached to each base and multiply those values together.

(2 + 1) x (2 + 1) x (2 + 1) x (2 + 1) = 81

Answer: E