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tricky question!

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tricky question!

by san2009 » Thu Jun 03, 2010 1:30 am
For any positive integer x, the 2-height of x is defined to be the greatest non-negative 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


The way that I rephrased this question is: is K>M? Because if K is greater than M then, 2^n will be larger for K, and thus the 2-height of K will be greater than the 2-height of M.

OA is B
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Source: — Data Sufficiency |
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by jube » Thu Jun 03, 2010 1:55 am
St. 1: Let k = 10, 2-height is 1
Let m=6, 2-height is 1

Let k= 12, 2-height is 2
Let k =10, 2-height is 1

- insufficient

St. 2: k/m= 2a
or k = 2am i.e. k will have one more 2 in it than m, hence it's 2-height will always be 1 more.

Hence, B
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by Testluv » Thu Jun 03, 2010 2:09 am
The way that I rephrased this question is: is K>M? Because if K is greater than M then, 2^n will be larger for K, and thus the 2-height of K will be greater than the 2-height of M.
cool question!

Unfortunately, your rephrase is a bit off. If the question were rephraseable as "is k>m?", then (1) is clearly sufficient since it tells us that k>m.

The question is just asking us whether there are more 2s in k than in m. Let's proceed from there.

(1) k>m

Knowing that k is larger than m doesn't tell us whether there are more 2s in k than in m. Certainly, if k is 8 and m is 4, then there are more 2s in k than in m. But m could be 8, and k could be 30 (only one 2), or 31 (no 2s).

(2) k/m is an even integer

An odd number can't be divided by an even one. And an odd number divided by another odd number results in either a non-integer or else another odd number. Thus, k is even. Regardless of m's even/odd status, k must have more 2s in it.
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