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

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

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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.

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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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27 Aug 2018, 15:56
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.

Bunuel karishma

I 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)
2-height of k = 2-height of m = 0

Case 2) No
k = 6, m = 3
k/m = 2 (even int)
(2-height of k = 1) > (2-height 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 2-height of x is defined to be the  [#permalink]

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

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22 Feb 2019, 23:50
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.

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 2-height of x is defined to be the  [#permalink]

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24 Feb 2019, 05:15
nausherwan 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.

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 2-height.
The 2-height = n where n is the greatest non negative integer such that 2^n is a factor of x.

So if k = 10, the 2-height of k is 1 (because 2^1 is divisible by 10)
If k = 24, the 2-height of k is 3 (because 2^3 = 8 is a factor of 24)
If k = 80, the 2-height of k is 4 (because 2^4 = 16 is a factor of 80)
If k = 81, the 2-height of k is 0 (because 2^0 = 1 is a factor of 81)

The more 2s a number has, the more is its 2-height.

(2) k/m is an even integer.
If k/m is an even integer, k has more 2s as factors than does m. So 2-height of k would be more than 2-height of m.
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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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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 2-height of x is defined to be the   [#permalink] 14 Apr 2019, 19:17

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