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Intern  Joined: 05 Aug 2015
Posts: 40
If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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Question Stats: 75% (01:14) correct 25% (01:53) wrong based on 405 sessions

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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

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

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

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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

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GRE 1: Q169 V154 Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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Okay what does "positive " prime numbers signify here ?
Can primes be negative too ?
NEVER EVER EVER.

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

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I wonder that 16 is not among the answers))
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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Konstantin1983 wrote:
I wonder that 16 is not among the answers))

You need factors greater than 1. So the number of factors is 16 - 1 = 15.
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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VeritasPrepKarishma wrote:
Konstantin1983 wrote:
I wonder that 16 is not among the answers))

You need factors greater than 1. So the number of factors is 16 - 1 = 15.

Yes i understand this. But GMAT likes to use traps so one can forget to exclude 1 and choose 16. But this answer is not present
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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I don't get this one. Is it a formula that you add +1 to each exponent to get # of factors?
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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kimbercs wrote:
I don't get this one. Is it a formula that you add +1 to each exponent to get # of factors?

Yes, the formula has been discussed here:
https://www.veritasprep.com/blog/2010/1 ... ly-number/
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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2
2
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.

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

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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

total number of factors = (3+1)(1+1)(1+1) = 16
except 1 number of factors = 16-1 = 15
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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m^3*p*t

Total no. of factors= (3+1)(1+1)(1+1)= 4*2*2=16

But 1 is excluded. Therefore the answer is 16-1=15
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Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)  [#permalink]

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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

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

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_________________ Re: If m, p, and t are distinct positive prime numbers, then (m^3)(p)(t)   [#permalink] 04 Mar 2019, 08:05
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