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

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Bunuel
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For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   26 Oct 2015, 09:58
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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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KarishmaB
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For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   Updated on: 01 Sep 2023, 04:01
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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.


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)

Check this video for numbers and their factors: https://youtu.be/DxIH8rjhpKY

Originally posted by KarishmaB on 27 Oct 2015, 02:16.
Last edited by KarishmaB on 01 Sep 2023, 04:01, edited 1 time in total.
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   23 Dec 2015, 22:15
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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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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   23 Dec 2015, 23:01
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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] New post   28 Jul 2016, 08:08
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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

Kudos if you like my solution
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   27 Aug 2016, 05:13
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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] New post   06 Jun 2017, 10:06
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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)



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] New post   16 Jun 2017, 06:42
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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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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   17 Jun 2017, 01:04
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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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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   10 Nov 2017, 19:01
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0 is also even integer, so what if k/m is 0?
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   10 Nov 2017, 19:32
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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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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] 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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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   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] New post   27 Aug 2018, 18:29
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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] New post   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] New post   24 Feb 2019, 05:15
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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] New post   13 Jun 2021, 03:15
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Video solution from Quant Reasoning starts at 20:00
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   09 Jul 2022, 00:11
KarishmaB

Even if "greatest nonnegative integer n" is replaced by "n is a positive integer", statement 1 is still insufficient.

K=2^2*7 > M=2^1*7 (2-h of k is 2 and 2-h of m is 1)
K=2*9 > M=2^2*3 (2-h of k is 1 and 2-h of m is 2)

So statement 1 is still insuff.
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   13 Jul 2022, 06:00
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Sneha2021 wrote:
KarishmaB

Even if "greatest nonnegative integer n" is replaced by "n is a positive integer", statement 1 is still insufficient.

K=2^2*7 > M=2^1*7 (2-h of k is 2 and 2-h of m is 1)
K=2*9 > M=2^2*3 (2-h of k is 1 and 2-h of m is 2)

So statement 1 is still insuff.


It doesn't make sense to define 2-height as a positive integer. Many positive integers x will be odd (no 2 in them). This means n must be 0. If n is a positive integer, 2-height will not be defined for them. Then you cannot define the 2-height of "any integer x." Then you will need to give that "the 2-height is defined for multiples of 2 such that ..." etc.
Hence, in the question, it doesn't make sense to replace non-negative with positive. And once you understand what 2-height is, it is easiest to see that an odd number (n =0) can be greater than an even number (n >= 1).
Of course, we know that multiples of 2 only can be greater than multiples of 4 e.g. 6 > 4
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink] New post   28 Feb 2024, 21:34
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Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
28 Feb 2024, 21:34
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