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

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10 Jun 2008, 17:23

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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.

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.

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

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.

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.

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.

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

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

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

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

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.

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.

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

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22 Sep 2015, 13:56

Hello from the GMAT Club BumpBot!

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

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28 Sep 2016, 19:47

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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