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(1) 2n is an integer --> 2n=integer: if n=1/2 then the answer is NO but if n=1 then the answer is YES. Not sufficient. (2) 3n is an integer --> 3n=integer: if n=1/3 then the answer is NO but if n=1 then the answer is YES. Not sufficient.

i was taking into the realm of LCM's but couldnt get a headway. i.e., 1 and 2 cannot deduce the fact that n is an integer.

combining both of them together 2n and 3n is an integer, it would mean that 6n would be an integer but reached a block there again as it still doesnt prove n is an integer.

Is there any other way to solve the problem except "3n - 2n" approach?

... a/b=2/3. Since a and b are integers, then a is a multiple of 2 ...

So for any integers a and b when a/b = k/m (a>k) (fractions simplifying) a is multiple of k, and b is multiple of b, isn't it ? Delicious solution, thank you!

Is there any other way to solve the problem except "3n - 2n" approach?

Yes, it is. Though the approach provided is the easiest one.

From (1): 2n=a, for some integer a --> n=a/2. From (2): 3n=b, for some integer b --> n=b/2.

a/2=b/3 --> a/b=2/3. Since a and b are integers, then a is a multiple of 2 --> a=2k, for some integer k --> 2n=2k --> n=k=integer. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel,

Can u explain how u got C?

a=2K, Why K should be integer K can also be 1/2.

Please correct me.

We have a/b=2/3. Since a and b are integers, then a is a multiple of 2 (or simply a must be an even number). Even number can be represented as a=2k, where k is an integer. If k=1/2, then a=2k=1=odd.

Is there any other way to solve the problem except "3n - 2n" approach?

Yes, it is. Though the approach provided is the easiest one.

From (1): 2n=a, for some integer a --> n=a/2. From (2): 3n=b, for some integer b --> n=b/2.

a/2=b/3 --> a/b=2/3. Since a and b are integers, then a is a multiple of 2 --> a=2k, for some integer k --> 2n=2k --> n=k=integer. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel,

Can u explain how u got C?

a=2K, Why K should be integer K can also be 1/2.

Please correct me.

Hi, I had the same concern, let me answer your question: We know that a/2=b/3 (if we had no info about a & b then a/2=b/3 could be anything, for example 1/3=0.333) and it gives us a/b = 2/3 (again, if no information about a & b it could be a = 1 and b = 1.5). Our key knowledge gives us that a = integer and b = integer and it means that minimum |a| = 2 and minimum |b|=3 and in general a is multiple of 2 and b is multiple of 3.

(1) 2n is an integer --> 2n=integer: if n=1/2 then the answer is NO but if n=1 then the answer is YES. Not sufficient. (2) 3n is an integer --> 3n=integer: if n=1/3 then the answer is NO but if n=1 then the answer is YES. Not sufficient.

(1) 2n is an integer --> 2n=integer: if n=1/2 then the answer is NO but if n=1 then the answer is YES. Not sufficient. (2) 3n is an integer --> 3n=integer: if n=1/3 then the answer is NO but if n=1 then the answer is YES. Not sufficient.

1) 2n is integer, 2n can be 1,2,3,4,5 which can give n as integer as well as fractions, A and D ruled out 2) 3n is integer same as above B ruled out. combining both if 2n is integer let say 2n=1 i.e. n=1/2 and 3n is integer let say 3n=1 i.e. n=1/3 then the common series will be divisible by 6 (which is LCM of 2 and 3) i.e. 0,6,12,18,......and so on including negative values as well, clearly n is integer here. Therefore Answer is C _________________

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