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If m, p, and t are distinct positive prime numbers

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If m, p, and t are distinct positive prime numbers

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If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t) has how many different positive factors greater than 1?

a. 8
b. 9
c. 12
d. 15
e. 27

The OA is the option D.

How can I know the number of factor without knowing the value of m, p and t?

Experts, may you give me some help here? Please.

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GMAT/MBA Expert

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M7MBA wrote:
If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t) has how many different positive factors greater than 1?

a. 8
b. 9
c. 12
d. 15
e. 27
----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!!----------------------------

(m^3)(p)(t) = (m^3)(p^1)(t^1)
So, the number of positive divisors of (m^3)(p)(t) = (3+1)(1+1)(1+1) = (4)(2)(2) = 16

IMPORTANT: We have included 1 as one of the 16 factors in our solution above, but the question asks us to find the number of positive factors greater than 1.
So, the answer to the question = 16 - 1 = 15

Answer: D

Cheers,
Brent

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Brent Hanneson – Creator of GMATPrepNow.com
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GMAT/MBA Expert

Post
M7MBA wrote:
If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t) has how many different positive factors greater than 1?

a. 8
b. 9
c. 12
d. 15
e. 27
To determine the total number of factors of a number, we can add 1 to the exponent of each distinct prime factor and multiply together the resulting numbers.

Thus, (m^3)(p)(t) = (m^3)(p^1)(t^1) has (3 + 1)(1 + 1)(1 + 1) = 4 x 2 x 2 = 16 total factors. Since 1 is one of those 16 factors, there are actually 15 different positive factors greater than 1.

Answer:D

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Scott Woodbury-Stewart Founder and CEO

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