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# If k is a positive integer, then 20k is divisible by how man

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Joined: 20 Aug 2013
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If k is a positive integer, then 20k is divisible by how man [#permalink]

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22 Aug 2013, 04:14
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57% (01:34) correct 43% (00:30) wrong based on 476 sessions

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If k is a positive integer, then 20k is divisible by how many different positive integers?

(1) k is prime
(2) k = 7

[Reveal] Spoiler:
Divisible by a positive integer -> factor
No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)

20k = (2^2)(5^1)(k)

Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Hence answer is D, but that is not the OA. What am I missing?
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Aug 2013, 04:17, edited 1 time in total.
Edited the question.
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Re: If k is a positive integer, then 20k is divisible by [#permalink]

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22 Aug 2013, 04:21
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spjmanoli wrote:
If k is a positive integer, then 20k is divisible by how many different positive integers?
1. k is prime
2. k is 7

Divisible by a positive integer -> factor
No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)

20k = (2^2)(5^1)(k)

Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Hence answer is D, but that is not the OA. What am I missing?

What you are missing in F.S 1, is that we don't know the value of k.

Scenario I: k=2, the total no of factors for 20k = $$2^2*5*2 = 2^3*5 = (3+1)*(1+1) = 8$$

Scenario II: k=3, the total no of factors for 20k = $$2^2*5*3 = (2+1)*(1+1)*(1+1) = 12.$$
Hence, 2 different answers, thus, Insufficient.
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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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22 Aug 2013, 04:21
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spjmanoli wrote:
If k is a positive integer, then 20k is divisible by how many different positive integers?

(1) k is prime
(2) k = 7

[Reveal] Spoiler:
Divisible by a positive integer -> factor
No of factors for a number in the form (a^x)(b^y)(c^z) is given by (x+1)(y+1)(z+1)

20k = (2^2)(5^1)(k)

Stmt 1 says k is prime. so 20k = (2^2)(5^1)(k^1). Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Stmy 2 says k = 7 so again Total # of factors is (2+1)(1+1)(1+1). So sufficient.

Hence answer is D, but that is not the OA. What am I missing?

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.

Back to the original question:

If k is a positive integer, then 20k is divisible by how many different positive integers?

$$20k=2^2*5*k$$.

(1) k is prime. If $$k=2$$, then $$20k=2^3*5$$ --> the # of factors = $$(3+1)(1+1)=8$$ but if $$k=5$$, then $$20k=2^2*5^2$$ --> the # of factors = $$(2+1)(2+1)=9$$. Not sufficient.

(2) k = 7 --> $$20k=2^2*5*7$$ --> the # of factors = $$(2+1)(1+1)(1+1)=12$$. Sufficient.

Hope it's clear.
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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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27 Oct 2014, 08:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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24 Feb 2016, 11:43
Bunuel wrote:
Back to the original question:

If k is a positive integer, then 20k is divisible by how many different positive integers?

$$20k=2^2*5*k$$.

(1) k is prime. If $$k=2$$, then $$20k=2^3*5$$ --> the # of factors = $$(3+1)(1+1)=8$$ but if $$k=5$$, then $$20k=2^2*5^2$$ --> the # of factors = $$(2+1)(2+1)=9$$. Not sufficient.

(2) k = 7 --> $$20k=2^2*5*7$$ --> the # of factors = $$(2+1)(1+1)(1+1)=12$$. Sufficient.

Hope it's clear.

Thanks, the outcome is clear but this method of picking random numbers to test with makes me very uneasy. If you get lucky and the 2 numbers you pick yield different results, then all is fine. But if they yield the same result, you don't know anything. Do you pick a 3rd candidate? A 4th?
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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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14 Mar 2016, 00:09
Here Statement ! is not sufficient as the prime may be 2 or 5 or any other prime.
But statement 2 is sufficient as we Number of divisors now = 2*3*2= 12
Hence B
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Re: If k is a positive integer, then 20k is divisible by how man [#permalink]

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20 Apr 2017, 06:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If k is a positive integer, then 20k is divisible by how man   [#permalink] 20 Apr 2017, 06:58
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