sqrt{ABC} = 504. Is B divisible by 2?

Okay, here's my explanation to this problem. Hope this sheds some light for you.

\(\sqrt{ABC}=504\)

Let's prime factorize 504 first.

\(504 = (2)^3 (3)^2 (7)\)

So, now we have
\(\sqrt{ABC}=(2)^3 (3)^2 (7)\)

To get ABC we square both sides. So you're right in saying that you get \(504^2\) on the right hand side, but \(504^2=((2)^3 (3)^2 (7))^2\)

So, we have \(ABC = ((2)^3 (3)^2 (7))^2\)

Statement 1: C = 168

\((AB)168 = ((2)^3 (3)^2 (7))^2\)


\((AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2\)

So, we know that \((AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)}\) This doesn't say much. So it's insufficient.

Statement 2: A is a perfect square.

So, we have \(ABC = ((2)^3 (3)^2 (7))^2\) and A is a perfect square, but then again, this has conflicting values as expressed in the post above.

Statement 1 and 2 together:

\((AB)(2)^3(3)(7) = ((2)^3 (3)^2 (7))^2\)

where A is a perfect square.

Simplifying the equation there gives us \((AB) = \frac{((2)^3 (3)^2 (7))^2}{(2)^3(3)(7)} = (2)^3 (7) (3^3)\)

If A is a perfect square, it either has to be \((2)^2 or (3)^2\) and for these respective cases, we get B to be
\((2) (7) (3^3)\) or \((2)^3 (7) (3)\). And in both these cases, B is divisible by 2. So C is the solution to the problem.