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

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

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New post 22 Aug 2018, 01:19
1
4
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

84% (01:03) correct 16% (01:36) wrong based on 49 sessions

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

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New post Updated on: 22 Aug 2018, 01:27
1
From statement 1: P = 2^x. X can be 2,3,5,7 etc. Then P will be 4,8,32,128. Therefore the number of positive integers is not same. Hence 1 is insufficient.

From statement 2: P = x^2. X is a prime. Then X can take values 2,3,5,7 etc..
Then P = 4,9,25,49 and so on. In all such cases P has only 3 positive integers.
Example 4 has 1,2 and 4 as factors.
9 has 1,3 and 9 has factors.
Hence B is sufficient.

B is the answer.

Posted from my mobile device

Originally posted by Afc0892 on 22 Aug 2018, 01:26.
Last edited by Afc0892 on 22 Aug 2018, 01:27, edited 1 time in total.
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If p is an integer, then p is divisible by how many positive integers?  [#permalink]

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New post 22 Aug 2018, 01:26
Bunuel 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.



we are looking for the number of factors of p.

Statement 1: \(p = 2^x\) , where x is prime. check it.

\(p = 2^2 or 2^3 or 2^5\)

when \(2^2\), number of factors will be 2 + 1 = 3

when \(2^3\) , number of factors will be 3 + 1 = 4.

So, Not sufficient.

Statement 2:\(p = x^2\), x is a prime.

\(2^2 or 3^2 or 5^2\).

when \(2^2\) , factors = 2 + 1 = 3

when \(3^2\) factors = 2 + 1 = 3.

Sufficient.

The best answer is B.
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Re: If p is an integer, then p is divisible by how many positive integers?  [#permalink]

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New post 22 Aug 2018, 01:34
Bunuel 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.


St1:- p = 2^x, where x is a prime number
Possible values of p: \(2^2,2^3,2^5,2^7\)
No of factors of p: 3,4,6,8 etc.
Insufficient

St2:- p = x^2, where x is a prime number
Here x can't be factorized further and power of x is a constant.
Hence, the no of factors is constant i.e, 3.
Sufficient.

Ans. (B)
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If p is an integer, then p is divisible by how many positive integers?  [#permalink]

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New post 22 Aug 2018, 10:53
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For those who know the "Unique Factors Trick" statement 2 should be really quick without any number picking. In the Unique Factors Trick you break a number down to its primes and express them as exponents (so like 12 = 2^2 * 3^1), then add 1 to each exponent and multiply the exponents together (so with 12 = 2^2 * 3^1 you'd add 1 to each exponent and have (2+1)(1+1) = 3 * 2 = 6 factors (and those factors are 1, 2, 3, 4, 6, and 12).

Well here even if you don't know x, you do know that it's prime, so you've already taken the first step of the Unique Factors Trick: you have its prime base, x, taken to an exponent. Just add one to that exponent and you're set. x^2 --> 2 + 1 factors = 3 factors.

For this one, too, it's not a bad idea to know the "rule" of statement 1 in reverse: if an integer has exactly three unique factors, it's the square of a prime number.
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If p is an integer, then p is divisible by how many positive integers? &nbs [#permalink] 22 Aug 2018, 10:53
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