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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.

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.

You can check Gmat Club Math Book link below for quant topics.
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(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

(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 \(k\).

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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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.

You can check Gmat Club Math Book link below for quant topics.

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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Consider giving me kudos if you find my explanations helpful so i can learn how to express ideas to people more understandable.

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

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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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- Stne

Last edited by stne on 23 Jul 2013, 00:05, edited 1 time in total.

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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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 \(4 = 2^3\)

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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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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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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