happyface101 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) =
16IMPORTANT: 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
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