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If m is a positive integer, is m a perfect square.

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Probus
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If m is a positive integer, is m a perfect square. [#permalink] New post   Updated on: 30 Aug 2018, 11:03
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If m is a positive integer, is m a perfect square.


(1) \(2\)\((m-1)^2\)\(-\)\((m-2)^2\)\(+\)\(2\)= \(m^2\)

(2) \((\sqrt{m}-1)(\sqrt{m}+1)\) = \(48\)
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B

Originally posted by Probus on 30 Aug 2018, 00:45.
Last edited by Probus on 30 Aug 2018, 11:03, edited 2 times in total.
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Re: If m is a positive integer, is m a perfect square. [#permalink] New post   30 Aug 2018, 02:27
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If m is a positive integer, is m a perfect square.


(1) \(2\)\((m-1)^2\)\(-\)\((m-2)^2\)\(+\)\(2\)= \(m^2\)
Let m =2...
\(2(2-1)^2-(2-2)^2+2=2^2........2*1^2-0^2+2=2^2.........2+2=4\).. but m is not a perfect square
Let m =4
\(2*3^2-2^2+2=4^2........2*9-4+2=16......16=16\)
Rather for all values of m, it will satisfy as if you simplify the equation, all terms in m get cancelled out..
\(2(m-1)^2-(m-2)^2+2=m^2.......2(m^2-2m+1)-(m^2-4m+4)+2=m^2.......2m^2-4m+2-m^2+4m-4+2=m^2.......0=0\)
Insufficient

(2) \((\sqrt{m}-1)(\sqrt{m}+1)\) = \(48[/m\\
]\\
[m](\sqrt{m})^2-1^2=48.........m-1=48.......m=49=7^2\)
Yes
Sufficient

B

Pl check the OA provided
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Re: If m is a positive integer, is m a perfect square. [#permalink] New post   30 Aug 2018, 03:19
Probus wrote:
If m is a positive integer, is m a perfect square.


(1) \(2\)\((m-1)^2\)\(-\)\((m-2)^2\)\(+\)\(2\)= \(m^2\)

(2) \((\sqrt{m}-1)(\sqrt{m}+1)\) = \(48\)


From statement 1: 2(m - 1)^2 – (m - 2)^2 +2 = m^2
=> 2(m^2 – 2m +1) – (m^2 -4m +4) + 2 = m^2
=> 2m^2 -4m +2 –m^2 +4m -4 +2 = m^2
=> m^2 = m^2
So it doesn’t give any information about whether m is a perfect square – INSUFFICIENT
From Statement 2 :
(√m-1) * (√m+1) = 48
=> (√m)^2-1^2 = 48
=> m = 49 =7^2
Hence m is perfect square – Sufficient
Hence answer is B.
Please change the OA to B.
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Re: If m is a positive integer, is m a perfect square. [#permalink] New post   30 Aug 2018, 06:03
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Probus wrote:
If m is a positive integer, is m a perfect square.

(1) \(2\)\((m-1)^2\)\(-\)\((m-2)^2\)\(+\)\(2\)= \(m^2\)

(2) \((\sqrt{m}-1)(\sqrt{m}+1)\) = \(48\)

\(m \geqslant 1\,\,\,\operatorname{int}\)

\(m\,\,\,\mathop { = \,}\limits^? \,\,{\left( {\operatorname{int} } \right)^2}\)

(1) Expanding the LHS we get m^2, hence all values of m (integers or not, by the way) satisfy this statement. INSUFFICIENT
(Take m = 1 and m = 2 for one possible BIFURCATION.)

(2) \(LHS\,\, = \,\,{\left( {\sqrt m } \right)^2} - {\left( 1 \right)^2} = m - 1\) hence m-1 = 48 and m = 49. SUFFICIENT (<YES>)

Answer: B

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: If m is a positive integer, is m a perfect square. [#permalink]
30 Aug 2018, 06:03
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