If x and y are both integers greater than 1, is x a multiple of y?
(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y
OA A
Source: GMAT Prep
If x and y are both integers greater than 1, is x a multiple
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Target question: Is x a multiple of y?BTGmoderatorDC wrote:If x and y are both integers greater than 1, is x a multiple of y?
(1) 3y^2 + 7y = x
(2) x^2 -x is a multiple of y
OA A
Source: GMAT Prep
Asking whether x is a multiple of y is the same as asking whether x = (y)(some integer)
For example, 12 is a multiple of 3 because 12 = (3)(4)
So, let's rephrase the question as...
REPHRASED target question: Does x = (y)(some integer)?
Statement 1: 3y² + 7y = x
Factor to get x = y(3y + 7)
If y is an integer, then (3y + 7) must be an integer
In other words: x = y(some integer)
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: x² - x is a multiple of y
There are several values of x and y that satisfy this condition. Here are two:
Case a: x = 4 and y = 2 (this satisfies statement 2 because x² - x = 12, and 12 is a multiple of 2). In this case, x IS a multiple of y
Case b: x = 5 and y = 2 (this satisfies statement 2 because x² - x = 20, and 20 is a multiple of 2). In this case, x is NOT a multiple of y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
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Brent
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(1) \(3y^2 + 7y = x\Rightarrow y (3y + 7) = x\)
So \(x\) must be a multiple, \((3y+7)\) times, of \(y\). Sufficient \(\color{green}\checkmark\)
(2) \(x^2 -x\) is a multiple of \(y\)
\(x^2 - x = yk\) where \(k\) is an integer.
\(x (x-1) = yk\)
From this, we do not know whether \(x - 1\) or \(x\) is equal to \(k\). So, insufficient \(\color{red}\chi\)
Therefore, the correct answer is __A__
So \(x\) must be a multiple, \((3y+7)\) times, of \(y\). Sufficient \(\color{green}\checkmark\)
(2) \(x^2 -x\) is a multiple of \(y\)
\(x^2 - x = yk\) where \(k\) is an integer.
\(x (x-1) = yk\)
From this, we do not know whether \(x - 1\) or \(x\) is equal to \(k\). So, insufficient \(\color{red}\chi\)
Therefore, the correct answer is __A__