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If n is an integer, then n is divisible by how many positive

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

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New post 24 Dec 2013, 03:48
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) n is the product of two different prime numbers.
(2) n and 2^3 are each divisible by the same number of positive integers.

Data Sufficiency
Question: 3
Category: Arithmetic Properties of numbers
Page: 153
Difficulty: 650


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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 24 Dec 2013, 03:49
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SOLUTION

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:

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

(1) n is the product of two different prime numbers --> n=ab, where a and b are primes, so # of factors is (1+1)(1+1)=4. Sufficient.

(2) n and 2^3 are each divisible by the same number of positive integers --> 2^3 has 4 different positive factors (1, 2, 4, and 8) so n has also 4. Sufficient.

Answer: D.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 25 Dec 2013, 12:06
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Statement 1: If n is a product of two prime number, then it must be divisible by 4 positive integers. Sufficient.

Statement 2: 2^3 is divisible by 4 positive integers and since n and 2^3 are divisible by same number of positive integers, this statement is sufficient as well.

Hence the answer is D.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 26 Dec 2013, 22:52
IMO - A
To find the # of positive integers that divide n, we need to know the primes in n
(1) 2 primes For e.g. 3 and 5 so the n is 15 and divisible by 4 positive int (1,3,5 & 15) so suff.
(2) Not sure about statement 2
For e.g. 2^3 is 8.Now, if n is also 8 then sufficient.
But if n is 16 then even though both 2^3 and 16 are divisible by 4 positive numbers.n has one more positive int 16.so insuff.

Pls correct my understanding of stmt 2.Thanks
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 27 Dec 2013, 19:36
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himapm1l wrote:
IMO - A
To find the # of positive integers that divide n, we need to know the primes in n
(1) 2 primes For e.g. 3 and 5 so the n is 15 and divisible by 4 positive int (1,3,5 & 15) so suff.
(2) Not sure about statement 2
For e.g. 2^3 is 8.Now, if n is also 8 then sufficient.
But if n is 16 then even though both 2^3 and 16 are divisible by 4 positive numbers.n has one more positive int 16.so insuff.

Pls correct my understanding of stmt 2.Thanks

It should be D mate.... Statement 2 is also sufficient. More information can be found here: math-number-theory-88376.html
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 30 Dec 2013, 03:11
If n is an integer, then n is divisible by how many positive integers?

(1) n is the product of two different prime numbers.
(2) n and 2^3 are each divisible by the same number of positive integers.

Sol:
Given n is an integer and so n is divisible by how many positive integers or how many factors does n have.

Using st 1, we have n is the product of 2 different prime numbers a and b
so n =ab....let a =2 b =3 then n =6 = 2^1 *3^1 and no. of factors of n are 4 ( 1,2,3 and 6)

Consider n is of the form n = a^2*b then n= 12 and no. of factors are 6 (1,2,3,4,6 and 12 )
St 1 is not sufficient as n is a multiple of 2 prime nos. but to what powers are the prime nos. raised, we don't know. Hence A and D are ruled out

In general, when a no. p can be represented by the form p= q^a*r^b*z^c where q,r and z are prime factors raised to the powers a, b and c respectively then the number of factors will be (a+1)*(b+1)*(c+1)


St 2 says n and 2^3 ie 8 have same factors which is 4. Hence no. of factors of n are 4

Ans B.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 02 Jun 2014, 14:56
Hi Bunuel, In statement-1, how do we know that the powers of two prime numbers is 1? I interpreted st-1 as n is a product of two different primes but their powers could be anything. So n could be ab, a^1*b, or a*b^1. Not Sufficient.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 03 Jun 2014, 08:43
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MensaNumber wrote:
Hi Bunuel, In statement-1, how do we know that the powers of two prime numbers is 1? I interpreted st-1 as n is a product of two different primes but their powers could be anything. So n could be ab, a^1*b, or a*b^1. Not Sufficient.


Well, first of all \(a^1*b=a*b^1=ab\).

As for the powers: ask yourself can we says that 12=2^2*3 is the product of two different prime numbers? No.

n is the product of two different prime numbers means n = (prime 1)*(prime 2).

Hope it's clear.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 03 Jun 2014, 08:50
Bunuel wrote:
Well, first of all \(a^1*b=a*b^1=ab\).


Haha! Off course. Thats why silly errors are killing me. :)

Bunuel wrote:
As for the powers: ask yourself can we says that 12=2^2*3 is the product of two different prime numbers? No.

n is the product of two different prime numbers means n = (prime 1)*(prime 2).

Hope it's clear.


Yes makes sense. thanks!
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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New post 28 Nov 2016, 04:39
D:

1. n = 1. p1 Xp2 so 3 . Sufficient
2. n and 2^3 implies, 2X2X2 so 3. sufficient
hence D

Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) n is the product of two different prime numbers.
(2) n and 2^3 are each divisible by the same number of positive integers.

Data Sufficiency
Question: 3
Category: Arithmetic Properties of numbers
Page: 153
Difficulty: 650


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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

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New post 09 Jun 2017, 11:26
Bunuel wrote:
SOLUTION

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.


Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: http://gmatclub.com/forum/math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:

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

(1) n is the product of two different prime numbers --> n=ab, where a and b are primes, so # of factors is (1+1)(1+1)=4. Sufficient.

(2) n and 2^3 are each divisible by the same number of positive integers --> 2^3 has 4 different positive factors (1, 2, 4, and 8) so n has also 4. Sufficient.

Answer: D.


Hi Bunuel

Here, are not we only considering positive factors. What about negative factors ?

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
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Re: If n is an integer, then n is divisible by how many positive  [#permalink]

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Re: If n is an integer, then n is divisible by how many positive   [#permalink] 08 Sep 2019, 21:38
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