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

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Joined: 10 Apr 2018
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If m is a positive integer, is m a perfect square.  [#permalink]

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Updated on: 30 Aug 2018, 11:03
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Difficulty:

45% (medium)

Question Stats:

76% (01:48) correct 24% (01:34) wrong based on 47 sessions

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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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Probus

~You Just Can't beat the person who never gives up~ Babe Ruth

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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Posts: 8005
Re: If m is a positive integer, is m a perfect square.  [#permalink]

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30 Aug 2018, 02:27
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]

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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
Please change the OA to B.
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Re: If m is a positive integer, is m a perfect square.  [#permalink]

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30 Aug 2018, 06:03
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>)

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

Regards,
fskilnik.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: If m is a positive integer, is m a perfect square.   [#permalink] 30 Aug 2018, 06:03
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