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For any positive integer x, the 2height of x is defined to be the
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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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Re: For any positive integer x, the 2height of x is defined to be the
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27 Aug 2018, 15:56
Bunuel wrote: For any positive integer x, the 2height 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 2height of k greater than the 2height of m?
(1) k > m (2) k/m is an even integer.
Kudos for a correct solution. Bunuel karishmaI think the OA should be C and not B. I don't think B is sufficient when k=m. I don't see any condition mentioned in the question stem saying that k=m is not possible. Please let me know if I'm missing something. Thanks! For example: Case 1) Yes k = m = 3 k/m = 0 (even int) 2height of k = 2height of m = 0 Case 2) No k = 6, m = 3 k/m = 2 (even int) (2height of k = 1) > (2height of m = 0) Stmt 1+2: Since stmt 1 says k>m, case 1 in stmt 1 is not possible. Hence, C is the correct answer.
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For any positive integer x, the 2height of x is defined to be the
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27 Aug 2018, 16:04
Bunuel wrote: For any positive integer x, the 2height 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 2height of k greater than the 2height 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 chetan2uI 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 2height of k greater = 2height of m greater = 0 Case 2: Yes k = 6, m = 3 (2height of k greater = 1) > (2height 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 2height of x is defined to be the
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27 Aug 2018, 18:29
dabaobao wrote: Bunuel wrote: For any positive integer x, the 2height 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 2height of k greater than the 2height 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 chetan2uI 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 2height of k greater = 2height of m greater = 0 Case 2: Yes k = 6, m = 3 (2height of k greater = 1) > (2height 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 2height of k or m will never be 0. You are mistaking them as quantity of 2 in k and m. But 2heoght 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, 2height in k is 2^0=1 and similarly 1 for m
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Re: For any positive integer x, the 2height of x is defined to be the
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22 Feb 2019, 23:50
For any positive integer x, the 2height 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 2height of k greater than the 2height of m?
(1) k > m (2) k/m is an even integer.
I am not sure where I am going wrong...
if i take k=12 and m=3 then k/m=4 which is even. In that case, h(k)=12/2^2=3 and h(m)=3/2^0=3 and therefore h(k) is not greater than h(m), therefore it is insufficient.
can someone explain why this is incorrect please?



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Re: For any positive integer x, the 2height of x is defined to be the
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24 Feb 2019, 05:15
nausherwan wrote: For any positive integer x, the 2height 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 2height of k greater than the 2height of m?
(1) k > m (2) k/m is an even integer.
I am not sure where I am going wrong...
if i take k=12 and m=3 then k/m=4 which is even. In that case, h(k)=12/2^2=3 and h(m)=3/2^0=3 and therefore h(k) is not greater than h(m), therefore it is insufficient.
can someone explain why this is incorrect please? I think you missed the concept of 2height. The 2height = n where n is the greatest non negative integer such that 2^n is a factor of x. So if k = 10, the 2height of k is 1 (because 2^1 is divisible by 10) If k = 24, the 2height of k is 3 (because 2^3 = 8 is a factor of 24) If k = 80, the 2height of k is 4 (because 2^4 = 16 is a factor of 80) If k = 81, the 2height of k is 0 (because 2^0 = 1 is a factor of 81) The more 2s a number has, the more is its 2height. (2) k/m is an even integer. If k/m is an even integer, k has more 2s as factors than does m. So 2height of k would be more than 2height of m.
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Re: For any positive integer x, the 2height of x is defined to be the
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14 Apr 2019, 19:17
Hi, here are my two cents for this question Decoding Question: 2 Height of Integer is number of two's in that in number. We are to determine if 2 Height of K > 2 Height of M So we are asked to determine if K contains more 2's than M First we are told that K> M K can be even and M can be odd eg (6, 5) here answer to our question is Yes K can be even and M can be even eg (6, 4) here answer to our question is No Second Statement tells us that\(\frac{K}{M}\) is an even integer We know that\(\frac {E}{E}\) = Even or Odd or Not an Integer or \(\frac {E}{O}\)= Even or Not an Integer But Since we are told its an even integer We can infer that If M is even then K is even then we have \(\frac {K}{M}\)= Even or K= M * Even 2nd height of K will have one more two than 2 Height of M or If M is odd then K is even then we have \(\frac {K}{M}\)= Even or K= M * Even K will have at least one 2 and M will have none ( Since M is Odd) in which case we can say 2nd height of K> 2nd height of M hence, using statement(II) we can with certainty answer our question Hope this helps
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Re: For any positive integer x, the 2height of x is defined to be the
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