If k is an integer greater than 1, is k wqual to 2^r for : GMAT Data Sufficiency (DS)
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# If k is an integer greater than 1, is k wqual to 2^r for

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

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19 Sep 2005, 18:37
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This topic is locked. Open discussion of the question: if-k-is-an-integer-greater-than-1-is-k-equal-to-2-r-for-88539-20.html#p1631245

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

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19 Sep 2005, 19:06
Not sure what does "some positive integer means"??
if it mean for ANY integer r > 0, ans will be C.

if it mean for ALL integers r > 0, answer will be E.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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19 Sep 2005, 19:30
"Some positive integer" means "any positive integer".

OA is B. Unfortunately GMATPrep doesn't provide any OEs.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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19 Sep 2005, 20:20
coffeeloverfreak wrote:
"Some positive integer" means "any positive integer".

OA is B. Unfortunately GMATPrep doesn't provide any OEs.

ii)

if k = 2, not divisible by x, x > 1 and x is odd

for some integer r = 1, k = 2 ^ r

ans is YES

now if k = 6

k <> 2^r

ans is NO

ii) therefore, is INSUFFICIENT.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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19 Sep 2005, 20:27
But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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19 Sep 2005, 20:31
coffeeloverfreak wrote:
But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.

You right. This is really tricky if one doesnt know these number system properties (especially if she is not good in number picking like me).
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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20 Sep 2005, 01:37
1
KUDOS
Under the unique factorisation theorem, any positive integer > 1, like k can be expressed as the product of prime powers.

So

k = 2^a * 3^b * 5^c * 7^d * .... *

where a,b,c,d >=0 integers.

We are asked if k = 2^r
this will be when r=a, b=c=d=0

(1) says that a>=6
but it does not give any information about b,c,d, ...
So not sufficient.
(examples a=6, b=c=d=0, k=64, and
a=6, b=1, c=d=0, k=192)

(2) says that b=c=d=0
but k>1, so k>=2, so a>=1
So sufficient

B

This abstract approach is not the quickest way,
but if you don't know this conceptual material,
you could struggle with a real question.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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20 Sep 2005, 05:59
I think the easiest way to see that A is not sufficient is by considering that 2^6*3 can not be represented as 2^n.

And B is suffiecient for the reason explained by coffee.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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05 Oct 2005, 16:15
B it is...

baiscally you want to know if K is multiple of 2 ONLY....

which means numbers like 6, 10, 12 are out cause they are multiples of other primes such as 3, 5 etc...

as long as K is a multiple of 2...we can express it as 2^r...where r is some number (positive)
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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05 Oct 2005, 18:24
i think it is E.

II) states that it is not divisible by any odd number > 1, which implies k is even. however, what we need to say if K is a power of 2. For example, 10 is even , but is not a power of 2. I am assuming here that the notation, 2^ r , denotes, 2 raised to r and not 2 multiplied by r.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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05 Oct 2005, 18:26
sorry , i understood my mistake. B it is.
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Re: If k is an integer greater than 1, is k wqual to 2^r for [#permalink]

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05 Oct 2005, 19:04
We're given k > 1, r > 0. We're asked if k = 2^r

From statement 1, we're told k is a multiple of 2^6. This could be 3(2^6), 4(2^6). Since there are so many values of k, we can't tell.

From statement 2, we're told k is not divisbile by any odd integer greater than 1. This means k is even and a power of 2. Any other even number that has a prime factor other than 2 will be divisible by any odd integer > 1. For e.g., if k=18, then 18 = 3*3*2. It's divisible by 3. However, if k = 16, then 2^4 and this is equal to 2^r if r is 4.

The answer B is correct if we're asked if k can be represented in the form 2^r where r is a positive value.

However, we're asked here if k EQUALS 2^r and we do not have information for r.
Re: If k is an integer greater than 1, is k wqual to 2^r for   [#permalink] 05 Oct 2005, 19:04
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