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

Question

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

OptimusPrepJanielle · Accepted Answer

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

GMATinsight · Answer

Method 1:

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

Method 2:

CONCEPT : If  N = a^p*b^q*c^r... 
where a, b, c... are distinct primes
Number of factors of  N = (p+1)*(q+1)*(r+1)*...

So factors of  (m^3)(p)(t)  will be  (3+1)(1+1)(1+1) = 16   but these factors include 1 as well

therefore factors of  (m^3)(p)(t)  greater than 1 = 16-1 = 15

Answer: Option D

stonecold · Answer

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

Regards
StoneCold

Konstantin1983 · Answer

I wonder that 16 is not among the answers))

KarishmaB · Answer

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

Konstantin1983 · Answer

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

kimbercs · Answer

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

KarishmaB · Answer

Check out this video: https://youtu.be/Kd-4cH4cqHw

JeffTargetTestPrep · Answer

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.

Answer: D

KrishnakumarKA1 · Answer

total number of factors = (3+1)(1+1)(1+1) = 16
except 1 number of factors = 16-1 = 15

Shiv2016 · Answer

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

BrentGMATPrepNow · Answer

----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

RELATED VIDEO
 https://www.youtube.com/watch?v=lVZ9GXd6ZBE

BrentGMATPrepNow · Answer

From the GMAT Official Guide: 
A prime number is a  positive  integer that has exactly two different positive divisors, 1 and itself. 
For example, 2, 3, 5, 7, 11, and 13 are prime numbers, but 15 is not, since 15 has four different positive divisors, 1, 3, 5, and 15. 
The number 1 is not a prime number since it has only one positive divisor. 
Every integer greater than 1 either is prime or can be uniquely expressed as a product of prime factors. For example, 14 = (2)(7), 81 = (3)(3)(3)(3), and 484 = (2)(2)(11)(11).

Kinshook · Answer

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

Total number of positive factors of (m^3)(p)(t) = 4*2*2 = 16
Number of different positive factors greater than 1 = 16- 1 = 15

IMO D

AA098 · Answer

How do I make sure that with the formula of adding +1 to the exponent, I do not double count any factors? So also that 1 is counted only once

Bunuel · Answer

That's how the formula works.

Let's say the prime factorization of the number is  p^a * q^b * r^c.. . where p, q, r are prime numbers and a, b, c are their respective exponents. The total number of factors of the given number is then  (a+1)(b+1)(c+1)... . This will include 1 and the number itself.

Why? Because each factor includes zero or more instances of each prime number. If p is a factor, it might be there 0 times, 1 time, 2 times, all the way up to  a  times. Hence, there are (a+1) possibilities for p, and similarly for q, r, and so on. Since these possibilities for each factor are independent, you multiply the possibilities to find the total.

For example, if a number is 36, its prime factors are 2 and 3. In terms of prime factorization, 36 can be written as 2^2 * 3^2. So, the total number of factors will be (2+1)*(2+1) = 9. These factors are 1, 2, 3, 4, 6, 9, 12, 18, 36.

I hope this makes it easier to understand.

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

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

Prep Toolkit

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