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M28-54

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M28-54  [#permalink]

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New post 16 Sep 2014, 00:44
1
5
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

48% (01:25) correct 52% (01:58) wrong based on 143 sessions

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Re M28-54  [#permalink]

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New post 16 Sep 2014, 00:44
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.


Answer: A
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Re M28-54  [#permalink]

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New post 27 Nov 2016, 10:33
I think this is a high-quality question and I agree with explanation.
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Re: M28-54  [#permalink]

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New post 18 Dec 2016, 20:09
45 sec..just took various numbers and got it
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Re: M28-54  [#permalink]

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New post 17 Apr 2017, 06:24
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
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Re: M28-54  [#permalink]

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New post 17 Apr 2017, 06:25
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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Re: M28-54  [#permalink]

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New post 24 Aug 2017, 09:33
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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Re: M28-54  [#permalink]

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New post 24 Aug 2017, 09:46
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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]

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New post 20 Dec 2018, 07:48
Bunuel wrote:
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.



I started by listing values

S1) 6, 18, 30, 42, 54, 68, 80, 92, 104, 116, 128, 140, 152, 164, 178, 190, 202, ...
Now, the obvious observation is that the answer to the question is always NO. The less obvious observation is that the prime factorization of every number here has an odd power of 2, so the answer will continue to be NO
SUFFICIENT
S2) 2, 16 Already we have one No and one Yes answer. INSUFFICIENT

ANSWER A
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Re: M28-54 &nbs [#permalink] 20 Dec 2018, 07:48
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