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If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even

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If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post Updated on: 03 Jul 2017, 19:29
00:00
A
B
C
D
E

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Question Stats:

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If x and y are positive integers and \(y(2^x)=24\), x=?

1) x≥2
2) y is even

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Originally posted by MathRevolution on 03 Jul 2017, 16:59.
Last edited by chetan2u on 03 Jul 2017, 19:29, edited 1 time in total.
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post 03 Jul 2017, 19:26
MathRevolution wrote:
If x and y are positive integers and y(2x)=24, x=?

1) x≥2
2) y is even



Hi,

The Q must be ..
\(y*2^x=24......\)
Now \(24=3*2^3=6*2^2=12*2^1\)

Now let's see the statements..
1) x≥2
So x can be 2 or 3
Insufficient

2) y is even
Means it is 6*2^2 or 12*2^1
So x can be 1 or 2
Insufficient

Combined..
Ans is 2
C
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post 03 Jul 2017, 19:37
y* 2^x = 24 or y = 24/2^x.. Both x and y have to be positive integers

If x=1, then y = 24/2 = 12
If x=2, then y = 24/4 = 6
If x=3, then y = 24/8 = 3
x cannot take any other value because then y will not be an integer.

Statement 1. x>=2.. two values of x are still possible: 2 and 3. So Insufficient

Statement 2. y is even.. two values of x are possible: 1 and 2. So Insufficient.

Combining the statements: x can only be 2. Sufficient. Hence C answer
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post 04 Jul 2017, 03:18
MathRevolution wrote:
If x and y are positive integers and \(y(2^x)=24\), x=?

1) x≥2
2) y is even


I had the answer E.

Statement 1: If x is 2, 2x2=4, therefore 4x6=24
but also, if x is 6, 2x6=12, therefore 12x2=24, so insufficient

Statement 2: As above, y can be even, but x still not be exclusively determined, hence insufficient.

Both together - same examples, x could be 2 or 6, with y being even (in those cases 4 and 12), therefore both are insufficient = E??
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post 04 Jul 2017, 06:29
aroussel wrote:
MathRevolution wrote:
If x and y are positive integers and \(y(2^x)=24\), x=?

1) x≥2
2) y is even


I had the answer E.

Statement 1: If x is 2, 2x2=4, therefore 4x6=24
but also, if x is 6, 2x6=12, therefore 12x2=24, so insufficient

Statement 2: As above, y can be even, but x still not be exclusively determined, hence insufficient.

Both together - same examples, x could be 2 or 6, with y being even (in those cases 4 and 12), therefore both are insufficient = E??


24 is comprised of 2 x 2 x 2 x 3 = 24. So, the max power of x is 3.

(1) if x is 2, then 2^x = 4, then y must be 6. If x=3, then 2^x = 8, then y must be 3
(2) y might be 24, if x=0
y might be 12, if x =1 => not sufficient

(1)+(2)

a) if x=3, then y=3 - from st.2, y is even, not odd
b) if x = 2, then y=6 - the only values of x and y, hence C
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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New post 05 Jul 2017, 00:01
==> In the original condition, there are 2 variables (a,b), and in order to match the number of variables to the number of equations there must be 2 equations as well. Since there is 1 for con 1) and 1 for con 2) C is most likely to be the answer.
By solving con 1) and con 2), you get \(24=6(2^2).\)

The answer is C.
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Re: If x and y are positive integers and y(2x)=24, x=? 1) x≥2 2) y is even  [#permalink]

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