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If n=2^5*3^4*5^3*7^2*11, how many factors of...

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by Brent@GMATPrepNow » Thu Nov 02, 2017 9:46 am
AAPL wrote:If
$$n=2^5*3^4*5^3*7^2*11$$
how many factors of n are there?

A. 180
B. 360
C. 540
D. 720
E. 810
----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
-----------ONTO THE QUESTION--------------------

Given: n = (2^5)(4^4)(5^3)(7^2)(11^1)
So, the number of positive divisors of n = (5+1)(4+1)(3+1) (2+1) (1+1)
=(6)(5)(4)(3)(2)
= 720
= D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Fri Nov 01, 2019 6:53 pm
AAPL wrote:If
$$n=2^5*3^4*5^3*7^2*11$$
how many factors of n are there?

A. 180
B. 360
C. 540
D. 720
E. 810

The OA is D.

I don't have clear this PS question. Please, can any expert assist me with it? Thanks.
The procedure for determining the total number of factors of a number is to first prime factorize the number, then add 1 to each prime factor's exponent, and then multiply those values. Thus, the number of factors of n is:

(5 + 1)(4 + 1)(3 + 1)(2 + 1)(1 + 1)

6 x 5 x 4 x 3 x 2 = 720

Answer: D

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