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

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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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

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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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

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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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.

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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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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Responding to a pm:


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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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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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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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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