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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)
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21 Apr 2016, 23:18

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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

Number of factors of (2^a)*(3^b)*(5^c) ... = (a+1)(b+1)(c+1) ...

If m, p and t are the distinct prime numbers, then the number is already represented in its prime factorization form Number of factors = (3+1)(1+1)(1+1) = 16 Out of these, one factor would be 1.

Hence different positive factors greater than 1 = 15 Correct Option: D

Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)
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22 Apr 2016, 04:42

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4

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

Let Number is (m^3)(p)(t) = (2^3)(3)(5) = 120

We can write 120 as product of two numbers in following ways 1*120 2*60 3*40 4*30 5*24 6*20 8*15 10*12

8 cases = 8*2 i.e. 16 factors (including 1)

Factors greater than 1 = 15

Answer: Option D
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)
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20 Mar 2017, 06:48

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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

To determine the total number of factors of a number, we can add 1 to the exponent of each distinct prime number 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.

Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)
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01 Mar 2018, 10:34

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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) = 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

Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)
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