Bunuel wrote:
If x and y are positive integers, what is the value of xy?
(1) \(x! = 6y!\)
(2) \(\frac{\frac{x!}{y}}{(x-2)!} = \frac{30}{y}\)
Given: x and y are positive integersTarget question: What is the value of xy? Statement 1: \(x! = 6y!\) Divide both sides of the equation by \(y!\) to get: \(\frac{x!}{y!} = 6\)
At this point we can see that there are two possible solutions
Solution i: x = 6 and y = 5, in which case
xy = (6)(5) = 30Solution ii: x = 3 and y = 1, in which case
xy = (3)(1) = 3Since we can’t answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: \(\frac{\frac{x!}{y}}{(x-2)!} = \frac{30}{y}\)Multiply both sides of the equation by \(y\) to get: \(\frac{x!}{(x-2)!}=30\)
Some nice cancellations happens here. To see how, notice that we can express the two factorials as follows: \(\frac{(x)(x-1)(x-2)(x-3).....(3)(2)(1)}{(x-2)(x-3).....(3)(2)(1)}=30\)
Simplify: \((x)(x-1)=30\)
Expand left side: \(x^2 - x = 30\)
Set the quadratic equation equal to \(0\) to get: \(x^2 - x - 30 = 0\)
Factor: \((x - 6)(x + 5) = 0\), which means either \(x = 6\) or \(x = -5\)
Since we're told x is
positive, it must be the case that \(x = 6\)
HOWEVER, since we don't know the value of \(y\), we can’t answer the
target question with certainty
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined Statement 1 tells us that EITHER \(x = 6\) and \(y = 5\) OR \(x = 3\) and \(y = 1\)
Statement 2 tells us that \(x = 6\)
So, it must be the case that x = 6 and y = 5, which means
xy = (6)(5) = 30Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent