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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 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.

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[Reveal] Spoiler: OA

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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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New post 26 Oct 2015, 19:29
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I think the answer is A.

I could derive a equation from the given information as

{2^(2K) / K} > {2^(2M) / M}...? ( Not Sure )

Statement 1- K>M, Now substituting values according to this will keep the stem true.

Hence Sufficient.

Statement 2- Not sufficient.

Hope my approach is fine.

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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) m/k is an even integer.

Kudos for a correct solution.


In simple words, 2-height is just the number of 2s in a positive integer x.

So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.

So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.

"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"

(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.

(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.

Answer (B)
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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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Although Karishma's approach (as always) is spot on and is an optimal way to solve such questions as it directly simplifies the layer of complexity in a logical way, when I attempted this question for first time I solved by plugging numbers.

Statement 1: K>M, if K=6 and M=4, then answer is No but if K=8 and M=4, the answer is Yes. Hence, insufficient.
Statement 2: M/K is an even integer, which means that M has all the factors of K (including powers of 2, i.e. 2-height of K) AND it has at least an extra 2 (because M/K results in an even integer, i.e. M=K*(an even integer)). Therefore, 2-height of K will always be smaller than 2-height of M. Hence, sufficient.

Answer - B

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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) m/k is an even integer.

Kudos for a correct solution.


The Q basically asks us if the power of 2 is greater for k when compared to m..
1) k>m..
not suff as k could be an odd number and m an even number, which would mean the 2-height of k < the 2-height of m..
and say k=32 and x=2.. the 2-height of k >the 2-height of m
insuff

2)m/k is an even integer...
this clearly shows that m is some even multiple of k..
and thus it will always be true that the 2-height of k < the 2-height of m..
suff

B
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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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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) m/k is an even integer.

Kudos for a correct solution.


My approach was simple picking numbers
case 1) K>M
K=40, M=39: 2-height = 2^5=32 => same height
K= 65 M=40: 2-height K= 6, 2-height for M=5 => different
INSUFFICIENT

Case 2) M/K is an even integer
This means M is either 0 times of N or M is atleast twice of N, since M, N are +ve Integer 0 is invalid, which means M is atleast twice of N, which leads 2-height of M will always be at least 1 greater than 2-height of N.
Sufficient.

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In the OG 2016, the "(2)" is not "m/k" but rather "k/m".

Answer explanation p.337:
"Given that k/m is an even integer, it follows that k/m = 2n for some integer n, or k = 2mn.
This implies that the 2-height of k is at least one more than the 2-height of m; SUFFICIENT."

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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) m/k is an even integer.

Kudos for a correct solution.



Lets understand the question first

2-height of any number x=power of 2 in factorial of X

Lets check the statements:

Statement1: K>m
So can we surely say k has greater power of 2 than m?

No.
Lets see why.

It can be true for k=2^3;m=2^2
it can be false for k=(2^3)*5 and m=(2^3)*3

Notice both above examples fulfil the statement 1 but give uncertain results.
Hence Insufficient

Statement 2: m/k is even integer
since k and m are positive integer. this means m/k=2 times some integer
or m has more power of 2 than k.
hance 2-height of m will be greater than2- height of k
hence sufficient. Ans:B

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New post 27 Aug 2016, 05:00
Can someone please explain what is being asked? Preferably with algebraic equation.

Thanks

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sashlev wrote:
Can someone please explain what is being asked? Preferably with algebraic equation.

Thanks


Questions has explained the meaning of 2-height of x.

It says the 2-height of x is defined to be the greatest non negative integer n such that 2^n is a factor of x. or we have n>=0 such that 2^n is a factor of x.

Say x=2, so we know that 2^1 ihe greatest factor of x. So, its 2-height will be 1.

Similarly, if we say x =28, we have 2^2 as the greatest factor in x, so n = 2, and so on.

Now, I hope the question is clear.

Moving on to the statements :

Statement 1 : k>m. Say K =5 and m is 4. in this case 2-height of m is greater than that is k.

But say k=8 and m=4, its the other way. Hence, insufficient.

Statement 2 : m/k is an even integer.

for m/k to be an integer, we must have k<=m.

Also, since it is even,we can say m must have 2 in it.

Or I can directly say m is a multiple of k such that m is even.

=> 2 height if m will always be greater than that of k. Take any values of m and k such that the above equation i satisfied you will always get the result.

hence, B is insufficient
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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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New post 06 Jun 2017, 10:06
VeritasPrepKarishma 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) m/k is an even integer.

Kudos for a correct solution.


In simple words, 2-height is just the number of 2s in a positive integer x.

So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.

So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.

"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"

(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.

(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.

Answer (B)



Thank you very much for this . I didn't even comprehend the question so I certainly had no chance of answering correctly. I was a bit demoralized till I read your great explanation. Does the GMAT use height to represent that amount of times a prime shows up when doing prime factorization often?

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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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New post 11 Jun 2017, 06:18
for 1,
for 3>2, 2-heights are 0 and 1
4>2, 2-heights are 2 and 1
insufficient

for 2,
m/k is even integer so m will have atleast 2-height 1 more than k
sufficient

Answer B

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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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NanaA wrote:
VeritasPrepKarishma 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) m/k is an even integer.

Kudos for a correct solution.


In simple words, 2-height is just the number of 2s in a positive integer x.

So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.

So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.

"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"

(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.

(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.

Answer (B)



Thank you very much for this . I didn't even comprehend the question so I certainly had no chance of answering correctly. I was a bit demoralized till I read your great explanation. Does the GMAT use height to represent that amount of times a prime shows up when doing prime factorization often?


I am surprised that nobody has answered this question. I am a native English speaker, and someone who has spent a substantial amount of time studying GMAT quantatitive questions and I have never encountered this terminology ("height") when referring to factors of a number. For this reason, I had no idea how to answer the question.

I can only hope this obscure phraseology was used to increase the difficulty of the question.

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ilovefrankee wrote:

Thank you very much for this . I didn't even comprehend the question so I certainly had no chance of answering correctly. I was a bit demoralized till I read your great explanation. Does the GMAT use height to represent that amount of times a prime shows up when doing prime factorization often?

Quote:
I am surprised that nobody has answered this question. I am a native English speaker, and someone who has spent a substantial amount of time studying GMAT quantatitive questions and I have never encountered this terminology ("height") when referring to factors of a number. For this reason, I had no idea how to answer the question.

I can only hope this obscure phraseology was used to increase the difficulty of the question.



This is not standard terminology so its no surprise that you haven't come across it before. It is a user defined function for this question only. The question gives exactly how to calculate 2-Height for a positive integer.
There could be another question which could define 2-Height as "the number of 2s that will add up to give that integer" etc.
Yes, user defined functions can increase the difficulty level of the question though note that the concepts are still very simple. The questions only look more difficult than they actually are.
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I'm confused. In DS questions, I treat both statements provided as objectively true. Therefore, if we find values that work with one of the statements, they cannot contradict the other statement. In this questions:
(1) k > m
(2) m/k is an even integer.

I understand why (2) works, but if m/k is an even integer, and both variables are positive, then how can statement (1) ever be qualified as true? The only scenario in which this works is if m = 0, but that would cause the question to become invalid since the 2-height if m=0 cannot be calculated. Am I missing something? Bunuel or VeritasPrepKarishma, your help would be appreciated.

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New post 17 Jun 2017, 04:59
bionication wrote:
I'm confused. In DS questions, I treat both statements provided as objectively true. Therefore, if we find values that work with one of the statements, they cannot contradict the other statement. In this questions:
(1) k > m
(2) m/k is an even integer.

I understand why (2) works, but if m/k is an even integer, and both variables are positive, then how can statement (1) ever be qualified as true? The only scenario in which this works is if m = 0, but that would cause the question to become invalid since the 2-height if m=0 cannot be calculated. Am I missing something? Bunuel or VeritasPrepKarishma, your help would be appreciated.


You are right. There was a typo in the second statement. (2) reads k/m is an even integer, not m/k is an even integer.

Edited. Thank you for noticing it.
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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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New post 28 Jul 2017, 03:08
VeritasPrepKarishma 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) m/k is an even integer.

Kudos for a correct solution.


In simple words, 2-height is just the number of 2s in a positive integer x.

So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.

So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.

"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"

(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.

(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.

Answer (B)


Your solution definitely needs my time to stop and say 'Thank You!'.

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New post 10 Nov 2017, 19:01
0 is also even integer, so what if k/m is 0?

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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
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New post 17 Nov 2017, 05:43
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VeritasPrepKarishma 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) m/k is an even integer.

Kudos for a correct solution.


In simple words, 2-height is just the number of 2s in a positive integer x.

So if x = 40, its 2-height will be 3 because 40 = 8*5 = 2*2*2*5
If x = 15, its 2-height is 0 because there are no 2s in 15.
and so on.

So, an even number will have a 2-height of at least 1.
An odd number will have a 2-height of 0.

"is the 2-height of k greater than the 2-height of m?" means "Does k have more 2s than m?"

(1) k > m
k could have more 2s than m or it could have fewer 2s than m.
For example, if k = 4 and m = 3, k has two 2s while m has none.
If k = 11 and m = 8, k has no 2s while m has 3.
Not sufficient.

(2) m/k is an even integer.
When m is divided by k, you get an integer. So m has all factors of k and they get cancelled out and you are left with an integer. Also, the leftover integer is even so m has at least one 2 more than k.
So 2-height of m is certainly more than the 2-height of k. So we can answer the question that k does not have more 2s than m.

Answer (B)


Responding to a pm:
Quote:
In your explanation, you mention:

"So, an even number will have a 2-height of at least 1."

Isn't 0 an even number with a 2-height (as defined in the question) of 0? That wouldn't affect the answer to the question since we know both k and m to be positive integers, but I just wanted to be clear conceptually with respect to the statement you made about even numbers in general.


Note that the function is defined for positive integers only. Hence, whatever general statements we make regarding this function, they are made for positive integers only.
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Kudos [?]: 17871 [1], given: 235

Re: For any positive integer x, the 2-height of x is defined to be the   [#permalink] 17 Nov 2017, 05:43
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