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# M28-54

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Math Expert
Joined: 02 Sep 2009
Posts: 46167

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16 Sep 2014, 01:44
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Difficulty:

85% (hard)

Question Stats:

49% (01:26) correct 51% (01:53) wrong based on 66 sessions

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Is $$x$$ the square of an integer?

(1) When $$x$$ is divided by 12 the remainder is 6.

(2) When $$x$$ is divided by 14 the remainder is 2.

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:44
Official Solution:

The question basically asks whether $$x$$ is a perfect square (a perfect square, is an integer that is the square of an integer. For example $$16=4^2$$, is a perfect square).

Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

(1) When $$x$$ is divided by 12 the remainder is 6. Given that $$x=12q+6=6(2q+1)=2*3*(2q+1)$$. Now, since $$2q+1$$ is an odd number then the power of 2 in $$x$$ will be odd (1), thus $$x$$ cannot be a perfect square. Sufficient.

(2) When $$x$$ is divided by 14 the remainder is 2. Given that $$x=14p+2$$. So, $$x$$ could be 2, 16, 30, ... Thus, $$x$$ may or may not be a perfect square. Not sufficient.

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Joined: 08 Aug 2016
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27 Nov 2016, 11:33
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 22 Nov 2014
Posts: 29

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18 Dec 2016, 21:09
45 sec..just took various numbers and got it
Intern
Joined: 24 Dec 2016
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17 Apr 2017, 07:24
Can someone please explain to me how we can conclude from this point
"Given that x=12q+6=6(2q+1)=2∗3∗(2q+1)x=12q+6=6(2q+1)=2∗3∗(2q+1). Now, since 2q+1 is anodd number"
How do we we know this number to be an odd number
Math Expert
Joined: 02 Sep 2009
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17 Apr 2017, 07:25
kmfaraaz@gmail.com wrote:
Can someone please explain to me how we can conclude from this point
"Given that x=12q+6=6(2q+1)=2∗3∗(2q+1)x=12q+6=6(2q+1)=2∗3∗(2q+1). Now, since 2q+1 is anodd number"
How do we we know this number to be an odd number

Notice that 2q = 2*integer = even, thus 2q + 1 = even + odd= odd.
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Joined: 27 Dec 2016
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Concentration: Marketing, Social Entrepreneurship
GPA: 3.65
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24 Aug 2017, 10:33
Bunuel , can we also define the statement 2 using similar method from statement 1?

- $$\frac{x}{14}$$ = $$Q+2$$ --> $$14Q+2 = X$$--> $$2*(7Q+1) = X$$. Since $$(7Q+1)$$ can be even or odd, thus INSUFFICIENT.
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Math Expert
Joined: 02 Sep 2009
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24 Aug 2017, 10:46
1
septwibowo wrote:
Bunuel , can we also define the statement 2 using similar method from statement 1?

- $$\frac{x}{14}$$ = $$Q+2$$ --> $$14Q+2 = X$$--> $$2*(7Q+1) = X$$. Since $$(7Q+1)$$ can be even or odd, thus INSUFFICIENT.

$$x=14p+2=2(7p+1)$$. If 7p + 1 is odd, x will not be a perfect square if it's even it could be, but you should still test some values to find whether it actually gives a prefect square (if p = 1, it is).
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Re: M28-54   [#permalink] 24 Aug 2017, 10:46
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# M28-54

Moderators: chetan2u, Bunuel

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