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For any positive integer x, the 2-height of x is defined to be the

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Joined: 10 Dec 2011
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Concentration: Finance, Economics
GMAT Date: 09-28-2012
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For any positive integer x, the 2-height of x is defined to be the  [#permalink]

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New post 25 Nov 2017, 07:57
Good question. My solution
x,m,k>=1.
If X/2^n = 1, n is greatest. (Let's ignore x and do this in terms of m & k only)
n>=0.
Now, the understanding is that the higher the number of 2s in k or m, the greater the n. And, n is least for odd. n=0 for odd k or m.

(1) k>m.
Well, if K & m are even and 2s in k > 2s in m, 2 height (n) of k > m i denote this as n(k) > n(m)
However, if k=odd, and m=even, n(m)>n(k) because n(k) = 0, but n(m)>=1
Insufficient

(2) k/m = even
This implies, k is even. m can be odd or even. And when m=even, k = 2(*>=1) * m
Thus, k always has higher 2s than m. Therefore, n(k) > n(m)
Sufficient.

Answer B
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For any positive integer x, the 2-height of x is defined to be the  [#permalink]

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New post 27 Aug 2018, 16:04
Bunuel wrote:
For any positive integer x, the 2-height of x is defined to be the greatest nonnegative integer n such that 2^n is a factor of x. If k and m are positive integers, is the 2-height of k greater than the 2-height of m?

(1) k > m
(2) k/m is an even integer.

Kudos for a correct solution.


chetan2u wrote:
Cheryn wrote:
0 is also even integer, so what if k/m is 0?



Hi, always read the entire q thoroughly so as not to miss out on minor details.
It is given m and k are positive integers so they cannot be 0


Bunuel chetan2u

I think Cheryn means that the ratio k/m can be equal to 0. If that's the case, then B would be insufficient. I don't any condition given in the question stem that does not allow m to be equal to n. Please let me know if I'm missing something. Thanks!

For example:
Case 1: No
k = m = 3
2-height of k greater = 2-height of m greater = 0

Case 2: Yes
k = 6, m = 3
(2-height of k greater = 1) > (2-height of m greater = 0)

For stmt 1+2, from stmt 1 we have k>m so case 1 from stmt 2 is not valid. Hence the answer is C.
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Re: For any positive integer x, the 2-height of x is defined to be the  [#permalink]

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New post 27 Aug 2018, 18:29
1
dabaobao wrote:
Bunuel wrote:
For any positive integer x, the 2-height of x is defined to be the greatest nonnegative integer n such that 2^n is a factor of x. If k and m are positive integers, is the 2-height of k greater than the 2-height of m?

(1) k > m
(2) k/m is an even integer.

Kudos for a correct solution.


chetan2u wrote:
Cheryn wrote:
0 is also even integer, so what if k/m is 0?



Hi, always read the entire q thoroughly so as not to miss out on minor details.
It is given m and k are positive integers so they cannot be 0


Bunuel chetan2u

I think Cheryn means that the ratio k/m can be equal to 0. If that's the case, then B would be insufficient. I don't any condition given in the question stem that does not allow m to be equal to n. Please let me know if I'm missing something. Thanks!

For example:
Case 1: No
k = m = 3
2-height of k greater = 2-height of m greater = 0

Case 2: Yes
k = 6, m = 3
(2-height of k greater = 1) > (2-height of m greater = 0)

For stmt 1+2, from stmt 1 we have k>m so case 1 from stmt 2 is not valid. Hence the answer is C.



Hi the 2-height of k or m will never be 0.
You are mistaking them as quantity of 2 in k and m. But 2-heoght is 2^n in them ...
So if k=m=3 ,but then k/m=3/3=1, whi h is NOT an even number so does not satisfy statement II.
Also, 2-height in k is 2^0=1 and similarly 1 for m
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: For any positive integer x, the 2-height of x is defined to be the &nbs [#permalink] 27 Aug 2018, 18:29

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