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# If p is an integer, then p is divisible by how many positive

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If p is an integer, then p is divisible by how many positive [#permalink]

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08 Feb 2009, 13:03
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If p is an integer, then p is divisible by how many positive integers?

(1) p=2^x, where x is a prime number.

(2) p=x^2, where x is a prime number.
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08 Feb 2009, 13:41
This one is kind of similar like the other one you posted I think.
1. This one should be easy to figure out, as x gets larger there will be more divisible integers so we don't know wtf the exact number is going to be
2. This one we just have to factor X^2 -> X,X -> oh wait that's it, since x is a prime number. Answer is B.

Factoring is the right terminology right?

Last edited by vladmoney on 06 Mar 2009, 13:08, edited 1 time in total.
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08 Feb 2009, 15:09
IMO B. The approach is slightly different.

The number of possible factors of a number is the product of (powers of each distinct prime factor + 1)

For ex, Possible factors of 24 can be found as follows -

$$24 = 2^3 * 3$$ ==> powers of 2 and 3 are 3, 1 ==> Possible factors of 24 are $$(3+1) * (1 + 1) = 4 * 2 = 8.$$

Now coming back to our question, it is asking what is the number of possible factors of p.

From stmt 1, it is given p = $$2^x$$. Possible factors will be x + 1. depending of x, the value keeps changing. Hence insufficient.

From Stmt 2, it is given that $$p = x^2$$ where x is a prime. So the possible factors of P will be 2 + 1 = 3. Sufficent.
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08 Feb 2009, 15:15
lumone wrote:
If p is an integer, then p is divisible by how many positive integers?

(1) p=2^x, where x is a prime number.

(2) p=x^2, where x is a prime number.

B.

p= x^2 = 1.x.x ( x is prime number)

P always divisible by 3 positive integers.. (1,x,x^2)
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Re: DS- 12   [#permalink] 08 Feb 2009, 15:15
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