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If x and y are positive integers, is 2x a multiple of y?
(1) 2x + 2 is a multiple of y
(2) y is a multiple of x
(1) 2x + 2 is a multiple of y
(2) y is a multiple of x
15 Jul 2014, 23:29
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kashishh
Use some logic and some number plugging:
Question: Is 2x a multiple of y?
(1) 2x+2 is a multiple of y
(2x + 2) is a multiple of y.
2x and (2x+2) are two consecutive even numbers and hence share only 2 factors: 1 and 2
So 2x will also be a multiple of y only if y is 1 or 2. Not sufficient.
(2) y is a multiple of x
Here are the multiples: x....y....2x
If y is x or 2x, 2x will be a multiple of y, else it will not be. Not sufficient.
Using both together:
If y = 2 and x = 2, all conditions are met and 2x IS a multiple of y.
If y = 6 and x = 2, all conditions are met but 2x IS NOT a multiple of y.
Not sufficient.
Answer (E)
13 Aug 2011, 16:10
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\(x, y > 0\), integers
"Is \(\frac{2x}{y}\) an integer?"
"Is \(2x\) a multiple of (divisible by) \(y\)?"
"Is \(y\) a factor of \(2x\)?"
Statement 1) This statements says that \(\frac{2x + 2}{y} = \frac{2(x+1)}{y}\) is an integer. In other words, \(2(x+1)\) is a multiple of \(y\).
In order for both \(2x\) and \(2(x+1)\) to be divisible by \(y\), \(y\) must be either \(1\) (factor of \(2\) and both \(x\) and \((x+1)\)) or \(2\) (factor of \(2\)). Thus if \(y\) equals \(1\) or \(2\), then \(\frac{2x}{y}\) is an integer. If \(y\) has any other value under this constraint, \(\frac{2x}{y}\) is not an integer.
You can also break this statement into two expressions: \(\frac{2x + 2}{y} = \frac{2x}{y} + \frac{2}{y}\). If \(\frac{2x}{y} + \frac{2}{y}=integer\), then in order for \(\frac{2x}{y}\) to be an integer, \(\frac{2}{y}\) must also be an integer. Thus if \(y\) equals \(1\) or \(2\), then \(\frac{2x}{y}\) is an integer. If \(y\) has any other value, \(\frac{2x}{y}\) is not an integer.
Insufficient.
Statement 2) This statement says that \(\frac{y}{x}\) is an integer. So \(y = x*k\), where \(k\) is some integer.
In order for \(\frac{2x}{y} = \frac{2*x}{k*x}\) to be an integer, \(\frac{2}{k}\) must be an integer, ie \(k\) must equal either \(1\) or \(2\). Thus if \(y = 2x\) or \(y = x\), \(2x\) will be a multiple of \(y\). With any other value of \(k\), \(2x\) will not be a multiple of \(y\).
Insufficient.
Combined)
From statement 1, \(\frac{2x}{y}\) is an integer when \(y={1,2}\), and not an integer when \(y\) has another value.
From statement 2: \(\frac{2x}{y}\) is an integer when \(y=2x\) or \(y=x\), and not an integer when \(y\) is a different multiple of \(x\) than \(2\).
So under the constraints of both statements, \(\frac{2x}{y}\) is an integer when \((x,y)\) = \((1,1)\), \((2,2)\), or \((1,2)\).
It only takes one example that follows the two constraints but is not one of these 3 possibilities to prove the combined statements insufficient. A couple examples are \((x,y) = (1,4)\), or \((2,6)\).
"Is \(\frac{2x}{y}\) an integer?"
"Is \(2x\) a multiple of (divisible by) \(y\)?"
"Is \(y\) a factor of \(2x\)?"
Statement 1) This statements says that \(\frac{2x + 2}{y} = \frac{2(x+1)}{y}\) is an integer. In other words, \(2(x+1)\) is a multiple of \(y\).
In order for both \(2x\) and \(2(x+1)\) to be divisible by \(y\), \(y\) must be either \(1\) (factor of \(2\) and both \(x\) and \((x+1)\)) or \(2\) (factor of \(2\)). Thus if \(y\) equals \(1\) or \(2\), then \(\frac{2x}{y}\) is an integer. If \(y\) has any other value under this constraint, \(\frac{2x}{y}\) is not an integer.
You can also break this statement into two expressions: \(\frac{2x + 2}{y} = \frac{2x}{y} + \frac{2}{y}\). If \(\frac{2x}{y} + \frac{2}{y}=integer\), then in order for \(\frac{2x}{y}\) to be an integer, \(\frac{2}{y}\) must also be an integer. Thus if \(y\) equals \(1\) or \(2\), then \(\frac{2x}{y}\) is an integer. If \(y\) has any other value, \(\frac{2x}{y}\) is not an integer.
Insufficient.
Statement 2) This statement says that \(\frac{y}{x}\) is an integer. So \(y = x*k\), where \(k\) is some integer.
In order for \(\frac{2x}{y} = \frac{2*x}{k*x}\) to be an integer, \(\frac{2}{k}\) must be an integer, ie \(k\) must equal either \(1\) or \(2\). Thus if \(y = 2x\) or \(y = x\), \(2x\) will be a multiple of \(y\). With any other value of \(k\), \(2x\) will not be a multiple of \(y\).
Insufficient.
Combined)
From statement 1, \(\frac{2x}{y}\) is an integer when \(y={1,2}\), and not an integer when \(y\) has another value.
From statement 2: \(\frac{2x}{y}\) is an integer when \(y=2x\) or \(y=x\), and not an integer when \(y\) is a different multiple of \(x\) than \(2\).
So under the constraints of both statements, \(\frac{2x}{y}\) is an integer when \((x,y)\) = \((1,1)\), \((2,2)\), or \((1,2)\).
It only takes one example that follows the two constraints but is not one of these 3 possibilities to prove the combined statements insufficient. A couple examples are \((x,y) = (1,4)\), or \((2,6)\).
