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M28-54

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M28-54 [#permalink]

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New post 16 Sep 2014, 01:44
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A
B
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E

Difficulty:

  75% (hard)

Question Stats:

53% (01:32) correct 47% (01:50) wrong based on 50 sessions

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Re M28-54 [#permalink]

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New post 16 Sep 2014, 01:44
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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, 11:33
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, 21:09
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, 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

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New post 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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Re: M28-54 [#permalink]

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New post 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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New post 24 Aug 2017, 10:46
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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M28-54   [#permalink] 24 Aug 2017, 10:46
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