If n=3*4*p where p is a prime number greater than 3,how many different

Question

If n=3*4*p where p is a prime number greater than 3,how many different positive non-prime divisors does n have, excluding 1 and n?

(A) Six
(B) Seven
(C) Eight
(D) Nine
(E) Ten

Source->Prep4Gmat

Kurtosis · Answer

n = 3 * 2^2 * p

Number of divisors = 2 * 3 * 2 = 12
The 12 divisors includes 1, n, 3, 2 and p

Number of different non prime divisors excluding 1 and n = 12 - 5 = 7

Answer: B

Bunuel · Answer

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neverplayd · Answer

Hello, I'm new to this forum so sorry if this question is in the wrong place. I was working on a question in the Prep4GMAT number properties app and ran into this question (and solution):

If n = 3 x 4 x p, where p is a prime number greater than 3, how many different positive non-prime divisors does n have, excluding 1 and n?

a) Six
b) Seven
c) Eight
d) Nine
e) Ten

Here is the solution:
 In this question, we are given n in a form^ that is almost its prime factorization, which is n = 2^2 x 3 x p. To determine the number of divisors sought, we essentially have to count different combinations of the factors. We can break down the possibilities by the number of prime factors used to form each divisor. We can't have any divisors of just one prime factor, because the question stipulates that we count "non-prime" divisors. In the case that our divisor has two factors, it could be (2) (2), or (2) (3), or (2) (p), or (3)(p) for a total of 4 possibilities. In the case that our divisor has three prime factors, it could be (2) (2) (3), or (2) (3) (p), or (2) (2) (p), for a total of 3 possibilities. The case that our divisor has four prime factors is out, because that equals n, an we have been told to exclude n. Therefore, we have 7 possibilities of the type described.

The solution provided by the app involves actually spelling out each combination which is fine with a set of factors that small but I assumed there would be a way to derive the answer from a permutation/combination formula: like 4!/2! or something similar?

abhimahna · Answer

An Easy way to solve this question :

Remember the method to find the number of factors of any integer?

If No, here it is :

If  N = X^a * Y^b  where X and Y are prime,

then number of factors of N are given by the formula (a+1)(b+1) { Notice I have simply increased the powers of X and Y by one and then multiplied them. Also Make sure X and Y are DISTINCT primes)

So, in your question, n = 3 x 4 x p, where p is prime.

I can  n=3^1*2^2*p^1

or number of factors = (1+1) * (2+1) * (1+1) = 12.

But Notice we need to find Non prime and Other than 1 and n, so we have three prime numbers (2,3,p) , 1 and N.

Excluding these, we have 12-5 = 7 such factors. Hence Answer is B

Cosmas · Answer

How did you determine the power of 'p'. Why did you assume p^1?

abhimahna · Answer

Notice we are given that p is prime.

Since a prime no. cannot be factorized further, it will always have a power of one. Take any prime number, you will be able to understand what I am saying

Posted from my mobile device

stonecold · Answer

Great Question.
Here is what i did in this one=>
n=2^2*3*p where p is a prime greater than 3
number of divisors => 3*2*2=> 12 including => 1,2,3,p,n 
hence excluding the above divisors => 12-5 => Seven 
Hence B

bumpbot · Answer

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