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Is n an integer?

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Is n an integer?

(1) 2n is an integer.
(2) 3n is an integer.

Please explain your answeres
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Apr 2012, 04:57, edited 1 time in total.
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Is n an integer?

(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)+(2) Subtract (1) from (2): 3n-2n=integer-integer --> n=integer (since integer-integer=integer). Sufficient.

Answer: C.

Hope it's clear.
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Re: Is n an integer? [#permalink]

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New post 08 Apr 2012, 05:19
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+1 C

Bunuel like the shortcut for getting "n" while solving both. I was trying to use a fraction to break the question stem
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New post 08 Apr 2012, 07:25
thanks Bunuel!!!

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.

ur method is short and easy. thanks!!

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Re: Is n an integer? [#permalink]

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New post 28 May 2013, 09:35
Is there any other way to solve the problem except "3n - 2n" approach?

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VadimKlimenko wrote:
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.
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Re: Is n an integer? [#permalink]

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New post 28 May 2013, 22:17
Bunuel wrote:
VadimKlimenko wrote:
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!

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VadimKlimenko wrote:
Bunuel wrote:
VadimKlimenko wrote:
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 ?


Yes, that's correct.
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rajasami4u wrote:
Bunuel wrote:
VadimKlimenko wrote:
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.

Hope it's clear.
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Re: Is n an integer? [#permalink]

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rajasami4u wrote:
Bunuel wrote:
VadimKlimenko wrote:
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.

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Is n an integer? [#permalink]

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New post 02 Jun 2013, 10:38
IMO the answer is C.

For statement 1, n=1/2 results in 2n = 1; not suff
For statement 2, n=1/3 results in 3n =1; not suff

Considering both the statements, 3n-2n=n; therefore n is an integer because the difference of two integer numbers is always integer. Suff.

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Re: Is n an integer? [#permalink]

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New post 11 Aug 2017, 23:12
Bunuel wrote:
Is n an integer?

(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)+(2) Subtract (1) from (2): 3n-2n=integer-integer --> n=integer (since integer-integer=integer). Sufficient.

Answer: C.

Hope it's clear.



What if the IInd statement says its 5N then also the Answer will be C ??

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New post 12 Aug 2017, 01:31
ashwin1590 wrote:
Bunuel wrote:
Is n an integer?

(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)+(2) Subtract (1) from (2): 3n-2n=integer-integer --> n=integer (since integer-integer=integer). Sufficient.

Answer: C.

Hope it's clear.



What if the IInd statement says its 5N then also the Answer will be C ??


Yes.

2n = integer and 5n = integer gives 5n - 2n = 3n = integer. 2n = integer and 3n = integer gives 3n - 2n = n = integer.

Or: 2n = integer, means 2*2n = 4n = 2*integer = integer. So, if 5n = integer, then 5n - 4n = n = integer.

Hope it's clear.
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Is n an integer? [#permalink]

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New post 12 Aug 2017, 12:02
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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Is n an integer?   [#permalink] 12 Aug 2017, 12:02
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