ilikaroy wrote:It has not been mentioned that a, b, c, d are integers? Your explanation will not hold true then?
Good point.
Actually, since x is a perfect square, and p,q,r and s are all prime, let's look at a possibility in which a,b,c and d would be nonintegers:
2^(1/2) * 2^(1/2) * 2^(1/2) * 2^(1/2) = 4
I guess you are talking about a situation like this. Well spotted, but I have a feeling the answer won't change. This is because, the only case in which a,b,c, or d could be non-integers, is if p,q,r and s are non-distinct, and the exponent would add up to an even number when you group the bases together, in order to maintain that x is a perfect square.
This rule MUST hold true - the prime factorization of a perfect square has EVEN powers for each constituent prime number.
Thus, it is only the case in which a,b,c and d are integers that you can have the POSSIBILITY that p,q,r and s are distinct primes. If they were distinct and a,b,c and d were non-integers, you would have primes that did not have even powers, so x could not be a perfect square.
Statement 1 is insufficient, because in examples in which you take the case of them being integers, you arrive at the conclusion that p,q,r and s may or may not be distinct. A particular group of values of the variables when considered are enough to judge insufficiency if they provide two possible answers to the prompt.
Statement 2 - when we considered them as integers, we found they have to be non-distinct.
But we already know, that if we consider them as non-integers, whether or not they satisfy a constraint like in statement 2 : THEY MUST BE non-distinct, otherwise x cannot be a perfect square.
You cannot have 2^1/2 * 3^1/4 & 7^2/5 or whatever, because such a product cannot be a perfect square.
Statement 2 is thus still sufficient.
I hope this is clear, and thanks for your observation - it is important to think like that!
