Ooh interesting. Let's see...
Key facts: x is a perfect square.
In the prime factorization of a perfect square, each distinct prime number will be raised to an even power.
Example: 144 = 2^4 * 3^2 so that the root is found by halving each power i.e. 12 = 2^2 * 3
Any non-perfect square will have ATLEAST one odd exponent in its prime factorization.
So, if the powers of p,q,r and s are even, they MAY be distinct (but may not be also)
But, if any of the powers is ODD, then in order for x to be a perfect square it must be true that this one must be the same prime as another one, which also has an ODD power, so that the net power is even.
eg. If p = q = 3. Then perhaps a = 5 and b =7 so that a+b = 12. Thus , 3^5 * 3^7 reduces to 3^12, leading us to the conclusion that they are all not in fact distinct.
So if any of a,b,c and d are found to be odd, we can say with confidence that p,q,r and s cannot all be distinct.
1. 18 is a factor of ab and cd. 18 well a = b= c = d = 6 satisfies this, and p,q,r, and s may or may not be distinct.
INSUFFICIENT.
2. ab and cd do not have a factor of 4. Thus if a is even, b is odd or if d is even, c is odd. i.e. 2 out of these 4 numbers is odd. Otherwise if both were even, 4 would definitely be a factor of ab and cd. (Both cannot be odd since then x cannot be a perfect square). This is sufficient for us to conclude that p,q,r and s are NOT distinct giving us a sure answer of "No"
SUFFICIENT.
Pick B.
Nice question!!
anybody
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albatross86
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albatross86
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Good point.ilikaroy wrote:It has not been mentioned that a, b, c, d are integers? Your explanation will not hold true then?
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!
~Abhay
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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Anaira Mitch
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