If m 2, is m an integer?

Question

If m > 2, is √m an integer?

(1) m = 2n, where n is a positive integer.
(2) m − 1 is the product of two consecutive odd integers.

GMATinsight · Answer

Question: Is √m an integer?

Statement 1 : m = 2n, where n is a positive integer.

@n = 8, m = 16, √m is an Integer
@n = 3, m = √6, √m is NOT an Integer

NOT SUFFICIENT

Statement 2 : m − 1 is the product of two consecutive odd integers. 
m-1 could be 1*3 or 3*5 or 5*7 or 7*9 ... etc
m could be 4 or 16 or 36 or 64 ... etc
i.e. √m will always be an Integer

Algebraic Explanation of statement-2 
 m-1 = (2k-1)*(2k+1) = 4k^2-1 
i.e.  m = 4k^2  i.e. a  Perfect square

SUFFICIENT

Answer: Option B

sumitkrocks · Answer

If m > 2, is √m an integer?

Statement :1 m = 2n, where n is a positive integer.
m = 2, 4 , 6 , 8 ..
Not sufficient

Statement :2 m − 1 is the product of two consecutive odd integers.

m-1=(2n-1)(2n+1) where n is an integer 
m=4n^2

so,  \sqrt{m}  will be equal to 2n i.e an integer 
Hence, sufficient

B is the answer

Aks111 · Answer

If m > 2, is √m an integer?
or is  m = n^2  where  n >=2

Statement (1) m = 2n, where n is a positive integer.
if  m = 2* (2*2*2) , then  m = 4^2  and answer to D.S is Yes
if  m = 2 * (3) , then  √m  is not an integer and answer to D.S is No
Hence, insufficient.

Statement (2) m − 1 is the product of two consecutive odd integers.
Let the two consecutive odd integers be 2n-1 and 2n+1, where n is +ve because  m>2 
 m - 1 = (2n-1) * (2n+1)
=> m - 1 = 4n^2 - 1
=> m = 4n^2 
=> √m is an integer always.
Hence sufficient.
 Answer: B

asishron29181 · Answer

If m > 2, is √m an integer?

(1) m = 2n, where n is a positive integer.

Let n=3 ,m=6 ,√m=not an integer 
Let n=8,m=16, √m=4(integer)
Not sufficient.

(2) m − 1 is the product of two consecutive odd integers.
Let us take few example to check this out. 
m-1= 1*3
m=4
√m = integer 
m-1=3*5
m=16
√m=4
Sufficient 
IMO B

Posted from my mobile device

TarunKumar1234 · Answer

If m > 2, is √m an integer?

Stat1: m = 2n, where n is a positive integer.
at n= 1, √m= Not integer; n=8, √m= Integer. Not sufficient

Stat2: m − 1 is the product of two consecutive odd integers.
m-1= 3, m= 4; m-1= 15, m= 16; m-35=1, m= 36. So, √m= integer. Sufficient

So, I think B.

Bunuel · Answer

If m>2, is √m an integer?

(1) m=2n, where n is a positive integer. If n = 2, then m = 4 and the answer is YES but if n = 3, then m = 6, and the answer is NO. Not sufficient.

(2) m−1 is the product of two consecutive odd integers. Two consecutive odd integers can be written as  2x - 1  and  2x + 1 , for some integers x. Thus  m - 1 = (2x - 1)(2x + 1) =4x^2 - 1  -->  m = 4x^2  -->  \sqrt{m}=2x=integer . Sufficient.

Answer: B.

Hope it's clear.

If m > 2, is m an integer?

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