Last visit was: 03 Aug 2024, 18:56 It is currently 03 Aug 2024, 18:56
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94778
Own Kudos [?]: 646406 [345]
Given Kudos: 86853
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67093 [223]
Given Kudos: 436
Location: Pune, India
Manager
Joined: 13 Feb 2011
Posts: 63
Own Kudos [?]: 183 [14]
Given Kudos: 3386
General Discussion
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11502
Own Kudos [?]: 34856 [5]
Given Kudos: 329
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
3
Kudos
2
Bookmarks
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
Intern
Joined: 27 Jul 2016
Posts: 7
Own Kudos [?]: 15 [5]
Given Kudos: 2
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
3
Kudos
1
Bookmarks
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
Board of Directors
Joined: 18 Jul 2015
Status:Emory Goizueta Alum
Posts: 3595
Own Kudos [?]: 5477 [4]
Given Kudos: 346
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
4
Kudos
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
Intern
Joined: 18 May 2017
Posts: 9
Own Kudos [?]: 14 [2]
Given Kudos: 116
WE:Corporate Finance (Health Care)
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
2
Kudos
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.

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?
Intern
Joined: 15 Jan 2017
Status:Studying for GMAT (April '19)
Posts: 10
Own Kudos [?]: 7 [1]
Given Kudos: 44
Location: United Kingdom
Concentration: Finance, General Management
GPA: 3.5
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
1
Kudos
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.

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.
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67093 [2]
Given Kudos: 436
Location: Pune, India
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
1
Kudos
1
Bookmarks
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.
Intern
Joined: 11 Sep 2017
Posts: 26
Own Kudos [?]: 51 [2]
Given Kudos: 49
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
2
Kudos
0 is also even integer, so what if k/m is 0?
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11502
Own Kudos [?]: 34856 [5]
Given Kudos: 329
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
5
Kudos
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
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67093 [1]
Given Kudos: 436
Location: Pune, India
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
1
Kudos
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.

Responding to a pm:
Quote:

"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.
Director
Joined: 24 Oct 2016
Posts: 581
Own Kudos [?]: 1373 [0]
Given Kudos: 143
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
GMAT 3: 690 Q48 V37
GMAT 4: 710 Q49 V38 (Online)
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
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.
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11502
Own Kudos [?]: 34856 [1]
Given Kudos: 329
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
1
Kudos
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
Manager
Joined: 27 Nov 2015
Posts: 87
Own Kudos [?]: 39 [0]
Given Kudos: 325
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
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?
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67093 [0]
Given Kudos: 436
Location: Pune, India
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
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.
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 1754 [2]
Given Kudos: 66
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
2
Kudos
Video solution from Quant Reasoning starts at 20:00
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
Senior Manager
Joined: 20 Dec 2020
Posts: 286
Own Kudos [?]: 30 [0]
Given Kudos: 499
Location: India
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
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.
Tutor
Joined: 16 Oct 2010
Posts: 15181
Own Kudos [?]: 67093 [1]
Given Kudos: 436
Location: Pune, India
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
1
Kudos
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
Non-Human User
Joined: 09 Sep 2013
Posts: 34228
Own Kudos [?]: 858 [0]
Given Kudos: 0
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: For any positive integer x, the 2-height of x is defined to be the [#permalink]
Moderator:
Math Expert
94778 posts