If x and y are positive integers, what is the value of xy?

Given: x and y are positive integers

Target 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) = 30
Solution ii: x = 3 and y = 1, in which case xy = (3)(1) = 3
Since 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) = 30
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent