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If k is an integer greater than 1, is k equal to 2^r for

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If k is an integer greater than 1, is k equal to 2^r for some positive integer r?

(1) k is divisible by 2^6.
(2) k is not divisible by any odd integer greater than 1.
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Aug 2012, 00:39, edited 1 time in total.
Edited the question and added the OA.

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Re: Power of 2. [#permalink]

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s1 is insufficient because k is at least 64 if it is divisible by 2^6. so, k cannot equal to 2, 2^2 (4), 2^3 (8), 2^4 (16), and 2^5 (32). However, it can equal to 2^6, 2^7, 2^8 ....

s2 is sufficient because if k is larger than 1 and cannot divisible by any odd number, then 2 and any power of 2 are the only factors of k. Therefore, k can equal to any power of 2.

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If k is an integer greater than 1, is k equal to 2^r for some positive integer r?

Given: \(k=integer>1\), question is \(k=2^r\).

Basically we are asked to determine whether \(k\) has only 2 as prime factor in its prime factorization.

(1) k is divisible by 2^6 --> \(2^6*p=k\), if \(p\) is a power of 2 then the answer is YES and if \(p\) is the integer other than 2 in any power (eg 3, 5, 12...) then the answer is NO.

(2) k is not divisible by any odd integers greater then 1. Hence \(k\) has only power of 2 in its prime factorization. Sufficient.

Answer: B.

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Re: Power of 2. [#permalink]

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New post 29 Dec 2009, 14:54
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Re: Power of 2. [#permalink]

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New post 13 Sep 2010, 09:25
(1) k is divisible by 2^6 - NOT SUFFICIENT - I Agree

(2) k is not divisible by any odd integer greater than 1 - SUFFICIENT - I disagree
This means k is divisible by 2,4,6,8,10,12,14 and so on.
so k can have 2,3,5,7 in its prime factors.

If k can only have 2's or it can have 2's and 3/5/7

So, this is NOT SUFFICIENT.

Please reply if I am missing something here.

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Re: Power of 2. [#permalink]

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nizam wrote:
(1) k is divisible by 2^6 - NOT SUFFICIENT - I Agree

(2) k is not divisible by any odd integer greater than 1 - SUFFICIENT - I disagree
This means k is divisible by 2,4,6,8,10,12,14 and so on.
so k can have 2,3,5,7 in its prime factors.

If k can only have 2's or it can have 2's and 3/5/7

So, this is NOT SUFFICIENT.

Please reply if I am something here.


OA for this question is B: please read the solution above.

Also red part in your reasoning is not correct.

Statement (2): \(k\) is not divisible by any odd integers greater then 1, so how it can be divisible by 6=2*3 or 10=2*5 or 12=4*3 or 14=2*7?

For example if it's divisible by 6 then it would mean that it's divisible by every factor of 6 too, so by 3 as well, but we are told that \(k\) is not divisible by any odd integers greater then 1 so it can not be divisible by 6. The above statement means that \(k\) is not divisible by any number which has an odd factor greater than 1, so 2 is only prime factor of \(k\).

Hope it's clear.
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Re: Power of 2. [#permalink]

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New post 15 Sep 2010, 21:43
Hi Bunuel,


Suppose if \(k=64\) and \(r=2\). In such case \(k=2^r\) does not hold true.

What I was thinking that we must have to get value of r in order to prove this..

Please, comment.

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Re: Power of 2. [#permalink]

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samark wrote:
Hi Bunuel,


Suppose if \(k=64\) and \(r=2\). In such case \(k=2^r\) does not hold true.

What I was thinking that we must have to get value of r in order to prove this..

Please, comment.

Cheers!


I'm not sure I understand your question. If k=64 why it's necessary r to equal to 2?

From (2) we have that the only prime factor of \(k\) is 2 which means \(k=2^r\), for some integer \(r\geq{1}\) --> \(k=64=2^6\), so in this case \(r=6\).
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Re: Power of 2. [#permalink]

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New post 16 Sep 2010, 06:46
ok, I got confused in question. Silly mistake. :|
Thanks!
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Re: Power of 2. [#permalink]

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New post 16 Sep 2010, 09:34
I got B too. The question asks for "some" r. I basically test the next number after 6 from 2^7.

(1) If k is divisible by 2^6, k will also be divisible by an int*2^6. If 2 is that int, then k = 2^7 and r = 7. >> SUFF

(2) If k is not divisible by any odd int > 1, it means that it could be divisible by 2. Then follow same logic as (1). >> SUFF

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Re: Power of 2. [#permalink]

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New post 25 Nov 2010, 23:36
Bunuel wrote:
Given: \(k=integer>1\), question is \(k=2^r\).

Basically we are asked to determine whether \(k\) has only 2 as prime factor in its prime factorization.

(1) \(2^6*p=k\), if \(p\) is a power of 2 then the answer is YES and if \(p\) is the integer other than 2 in any power (eg 3, 5, 12...) then the answer is NO.

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i made some careless mistake as i did not perform some word translation: k = 2^6 * p (p = 3,5,...).
The equalation k = 2^6 * p will help us to see the trap of (1)
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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arjunrampal wrote:
If k is an integer greater than 1, is k equal to 2^r for some positive integer r?

(1) k is divisible by 2^6.
(2) k is not divisible by any odd integer greater than 1.


Question: Its a YES or NO question.

Either k=2^r or not, we need to get a definite answer thats all, r>0

(1) k is divisible by 2^6.

2^6 * x =k

x could be anything, 1,2,3,4 etc =>Not sufficient

(2) k is not divisible by any odd integer greater than 1.

well okay, so 3,5,7 etc are all out of the loop, anything that involves odd no's, then that leaves us only even no's to work with.

and even no's are all multiple of 2, so yeah, for some positive r? k=2^r is possible.

Sufficient.

Ans: B
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 22 Jul 2013, 23:47
Question does require some clarification for me.
Lets just focus on B
k is not divisible by any odd integer greater than 1

I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to \(2^r\) for some positive integer r?
If K=2 and r=2,then is \(2 = 2^2\) answer is No \(2 \neq 4\) { there is no restriction on the value of k other than it has no odd factors greater than 1 and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is \(4 = 2^3\) answer is No \(4 \neq 8\) { there is no restriction on the value of k other than it has no odd factors greater than 1 and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is \(4 = 2^2\) Answer is Yes \(4 = 4\) { there is no restriction on the value of k other than it has no odd factors greater than 1 and r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,\(2^r\)changes .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2

If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.
Please do correct, if there is anything wrong with the reasoning above, not entirely confident in challenging solution of GMAT prep.
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 22 Jul 2013, 23:59
stne wrote:
Question does require some clarification for me.
Lets just focus on B
k is not divisible by any odd integer greater than 1

I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to \(2^r\) for some positive integer r?
If K=2 and r=2,then is \(2 = 2^2\) answer is No \(2 \neq 4\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is \(4 = 2^3\) answer is No \(4 \neq 8\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is \(4 = 2^2\) Answer is Yes \(4 = 4\) { there is no restriction on the value of k other than it has no odd factors r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,\(2^r\)changes .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2

If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.


I think you misunderstood the question.

The question basically asks: can k be written as some power of 2. (2) implies that k is some power of 2, thus it's sufficient.
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 23 Jul 2013, 00:09
Bunuel wrote:
stne wrote:
Question does require some clarification for me.
Lets just focus on B
k is not divisible by any odd integer greater than 1

I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to \(2^r\) for some positive integer r?
If K=2 and r=2,then is \(2 = 2^2\) answer is No \(2 \neq 4\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is \(4 = 2^3\) answer is No \(4 \neq 8\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is \(4 = 2^2\) Answer is Yes \(4 = 4\) { there is no restriction on the value of k other than it has no odd factors r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,\(2^r\)changes .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2

If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.


I think you misunderstood the question.

The question basically asks: can k be written as some power of 2. (2) implies that k is some power of 2, thus it's sufficient.


Bunuel question says is \(K = 2^r\)
if K = 4 and r=3 then how is \(4 = 2^3\)

The stress is on " equal to "
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 23 Jul 2013, 00:17
stne wrote:
Bunuel wrote:
stne wrote:
Question does require some clarification for me.
Lets just focus on B
k is not divisible by any odd integer greater than 1

I can be sure of only one thing here, that is K is even, or it can be raised to the power of 2.
What is the definite value of K cannot be obtained by statement 2, under any circumstances, I think all of us agree.
e.g, K could be 2, 4, 8,16,32 etc
Now question is
Is k equal to \(2^r\) for some positive integer r?
If K=2 and r=2,then is \(2 = 2^2\) answer is No \(2 \neq 4\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=2 and r=2 are both valid}
If K=4 and r=3,then is \(4 = 2^3\) answer is No \(4 \neq 8\) { there is no restriction on the value of k other than it has no odd factors and r has no restriction other than it is an integer, so k=4 and r=3 are both valid}
If K=4 and r=2,then is \(4 = 2^2\) Answer is Yes \(4 = 4\) { there is no restriction on the value of k other than it has no odd factors r has no restriction other than it is an integer, so k =4 and r=2 are both valid}
So we can see depending upon r ,\(2^r\)changes .
so we can get both a Yes and a No depending upon K and r, can we not?
We do not have a definite K and a definite r, from 2

If question could have stated does K have only 2 as prime factors then of course solution could be more justified. I think some of us do agree that solution is debatable.


I think you misunderstood the question.

The question basically asks: can k be written as some power of 2. (2) implies that k is some power of 2, thus it's sufficient.


Bunuel question says is \(K = 2^r\)
if K = 4 and r=3 then how is \(4 = 2^3\)

The stress is on " equal to "


Again, you misinterpret the question.

The question asks whether we can find some positive integer r so that to represent k in form of 2^r. Or to put it simply, the question asks whether the prime factorization of k contain only 2's.
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 23 Jul 2013, 01:04
Again It could have been better if we could not assume "equal to " to mean the same as " if K can be raised to some power of 2"
if question was can k be written as \(2^r\) - of course solution makes sense
if question was whether the prime factorization of k contain only 2's- of course solution makes sense.

but question is, is \(K = 2^r\) or is k "equal to" \(2^r\)

Why are we changing the " equal to " in the original question to " can be written as" to suit the OA?
Is \(k=2^r\) and can k " be written as \(2^r\) are two very different questions I think.

The language " equal to " is a bit ambiguous here I think.
To make the answer b of course we have to assume "Is K = \(2^r\)" to mean the same as " can K be written as \(2^r\)"

Anyway I get what you mean, I was just hoping that some one could agree that the term " equal to " is a bit ambiguous.
Because if question was "is K = \(2^r\)"

and if K = 2 and r= 3 , \(2 \neq 8\) , \(2 \neq 2^3\) \(k \neq 2^r\)
if K=4 and r=2 then, \(4 = 4\) , \(2 = 2^2\) \(k= 2^r\)

Without the OA high chance of someone misinterpreting this, I think.
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Re: If k is an integer greater than 1, is k equal to 2^r for [#permalink]

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New post 23 Jul 2013, 01:06
stne wrote:
Again It could have been better if we could not assume "equal to " to mean the same as " if K can be raised to some power of 2 "
if question was can k be written as \(2^r\) - of course solution makes sense
if question was whether the prime factorization of k contain only 2's- of course solution makes sense.
but question is, is \(K = 2^r\) or is k "equal to" \(2^r\)

Why are we changing the " equal to " in the original question to " can be written as" to suit the OA?
Is \(k=2^r\) and can k " be written as \(2^r\) are two very different questions I think.

The language " equal to " is a bit ambiguous here I think.
To make the answer b of course we have to assume "Is K = \(2^r\)" to mean the same as " can K be written as \(2^r\)"

Anyway I get what you mean, I was just hoping that some one could agree that the term " equal to " is a bit ambiguous.
Because if question was "is K = \(2^r\)"

and if K = 2 and r= 3 , \(2 \neq 8\) , \(2 \neq 2^3\) \(k \neq 2^r\)
if K=4 and r=2 then, \(4 = 4\) , \(2 = 2^2\) \(k= 2^r\)

Without the OA high chance of someone misinterpreting this, I think.


It's a GMAT Prep question. The wording is fine.
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If k is an integer greater than 1, is k wqual to 2^r for some positive [#permalink]

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New post 16 Jan 2016, 12:03
If k is an integer greater than 1, is k equal to 2^r for some positive integer r?

(1) k is divisible by 2^6

(2) k is not divisible by any odd integer greater than 1
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Last edited by ENGRTOMBA2018 on 16 Jan 2016, 12:09, edited 1 time in total.
Corrected the typo in the question.

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If k is an integer greater than 1, is k wqual to 2^r for some positive   [#permalink] 16 Jan 2016, 12:03

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