gmatJP wrote:
I dnt understand,
if squ rt of ABC = 504 doesnt ABC becomes 504*504?
And how did you get =( 2*2*2*3*3*7)^2 for below figure...
(2*2*2*3*7)*AB =( 2*2*2*3*3*7)^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.