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

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

10 Jun 2008, 18:23

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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)m/k is an even integer.

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Re: DS- 2-height [#permalink]

10 Jun 2008, 19:02

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goalsnr wrote:

Not possible

the stem says that m and k are positive integers.

m/k is an even integer, and k > m , so the only solution would be m = 0, but that can not be true because m must be greater than 0

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Re: DS- 2-height [#permalink]

10 Jun 2008, 19:15

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Answer is B

Statement 1 : K >M consider k=5 and m=4 and answer is no because greatest n for 5 is 0 and greatest n for 4 is 2
but if k=8 and m=4 then answer is yes because greatest integer for k is 3 and greatest integer for m is 2
INSUFF.

Statement 2 : m/k is even integer is possible only if m > k and also both m and k are even or m is even and k is odd. In both cases greatest integer n such that x/2^n is integer is higher for m .
Therefore answer is always no.
for example m=6 and k=3 answer NO
m=8 and k=2 answer NO
SUFF.

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Re: DS- 2-height [#permalink]

10 Jun 2008, 19:55

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goalsnr wrote:

Good one. Took a little more than 2 min to be sure that 1) was not possible.
B for me.

Let the function be TH(x) = n where 2-power-n is a factor of x.

(1) k > m
consider k =12 , m=4 ( TH(k) = TH( m) = 2 in both cases ).
So condition is false.
consider k=8, m=4 ( TH(8)= 3 ; TH(4)= 2 ) . So condition is true.
So condition is insufficient.

(2)if m/k is even there are 2 possibilities.
m and k are odd
OR
m and k are even
But the problem is such that m and k must be even if they must be divisible by 2-power-n.
So with m and k even and m/k being an integer, lets try m=16; k=8
We get TH(x) to be 4 and 3 for m and k.

Try m=20,n=10; Again TH(20) = 2 and TH(10)= 1

So B is sufficient.

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Re: DS- 2-height [#permalink]

11 Jun 2008, 00:41

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.

Guys i m sorry if it sounds stupid but i m not clear with the question

can somebody please explain
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Re: DS- 2-height [#permalink]

11 Jun 2008, 06:35

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vdhawan1 wrote:

Take examples, assume m = 10 and k = 12.
The 2-height of 10: 10 = 2 x 5
The greatest non negative n such as 2^n is a factor of 10: n = 1 because 2^1 is a factor of 10
The 2-height of 12: 12 = 2^2 x 3
The greatest non negative n such as 2^n is a factor of 12: n = 2 because 2^2 is a factor of 12

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Re: DS- 2-height [#permalink]

11 Jun 2008, 06:46

businessm wrote:

vdhawan1 wrote:

Hey thanks man

i understood .....
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Re: DS- 2-height [#permalink]

11 Jun 2008, 06:52

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goalsnr wrote:

B

1: is insuff. b/c k could be 8 or 10 and m could be 4.

2: assuming you meant k/m b/c otherwise it would clearly contradict the first statment which IS NOT POSSIBLE ON THE GMAT!.

we cannot have 12/4 for example.

Thus inorder to have an even integer. the numerator must have a larger number of 2^n's.

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Re: DS- 2-height [#permalink]

14 Jun 2008, 08:03

Good job guys.
The OA is B

Re: DS- 2-height [#permalink] 14 Jun 2008, 08:03

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

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

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