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If m > 2, is m an integer?

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Bunuel
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If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 01:30
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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.
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B
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Re: If m > 2, is m an integer? [#permalink] New post   30 Apr 2021, 01:15
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Bunuel wrote:
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.


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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.
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Re: If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 01:41
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Bunuel wrote:
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.


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

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Re: If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 01:50
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
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Re: If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 01:53
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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
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Re: If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 07:27
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

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Re: If m > 2, is m an integer? [#permalink] New post   29 Apr 2021, 14:11
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. :)
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Re: If m > 2, is m an integer? [#permalink]
29 Apr 2021, 14:11
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