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

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.

Bunuel - you rock man!
Kudos!

(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.

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 \(r=6\).

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

If k is an integer greater than 1, is k equal to 2^r for some positive integer r? (D)
(1) k is divisible by 2^6
(2) k is not divisible by any odd integer greater than 1



Now here we have the word "some".
Stmnt 1 is true for some values of r = 7,8,etc

So why is it insufficient ?

I know it gives you different answers but the question also says "some". Not getting this discrepancy.

Dear shahrukh0603,
I'm happy to respond.

My friend, this is a very specialized use of the word "some," a use quite different from the use in ordinary language, but a use typical of mathematics.

When a problem is specifying a precise mathematical condition, it uses the term "some value" to mean "a unique and in existence but currently unknown value." That is the sense of the word in this problem. You see, not in GMAT math, but in higher mathematics, many many proofs are "existence proofs." For example, if you have taken calculus, the Mean Value Theorem is an existence theorem: it merely guarantees that a particular value, the mean value, exists----it doesn't give us any concrete ideas about how to find that value. With the Mean Value Theorem, it's relatively simple to do the calculation to solve, but with many problem in levels of math far beyond calculus, all we have is the proof of existence: we have no way to know how to find the value whose existence is guaranteed.

This prompt question uses the word "some" in the sense of mathematical existence, not in the ordinary colloquial sense. In other words, it is saying,

If k is an integer greater than 1, does there exist a particular positive integer r, such that k = 2^r?

The problem with Statement #1 is that it gives a maybe answer to this precise question.

If k = 2^8, then r = 8, the unique value exists. Answer = Yes.

If k = 5(2^8), then there is no possible positive integer value of r such that k = 2^r (the value of r would be 8 + ln(5)/ln(2) = 10.32192809..., which is not an integer---you do not need to know how to find that expression or value for the GMAT). Answer = No.

Since both a yes and a no answer are possible, this statement is insufficient.

Does all this make sense?
Mike

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