SOLUTION
Given: Number P expressed as y+\(1000^x\), x and y are integers.
We need to determine the reminder of P when divided by 9.
So, P = [ y+\(1000^x\) ]/9 =\(\frac{y}{9}\) + \(\frac{1000^x}{9}\) and add the individual remainders to determine the final answer.
Irrespective of the value of p, if we divide 1000^x by 9, we will always get the remainder 1 –> hence, our final answer doesn’t depend on the value of p.
(1) 2\(y^2\) = 32
y^2 = 16
y= 4
From what we calculated above -> \(\frac{4}{9}\) + \(\frac{1000^x}{9}\)-> we now know the remainder of both individual fractions.
( You don't actually need to calculate the reminder!)
SUFFICIENT!
(2) 3\(x^2\)=27
x^2 = 9
x = 3
\(\frac{y}{9}\) + \(\frac{1000^3}{9}\).......we already knew that no matter what the value of x will be, the remainder would always be 1 but we know nothing about y -> so not possible to find the remainder of \(\frac{y}{9}\).
INSUFFICIENT!
Thus, the answer is option A.
P.S: In DS questions, if the question stem is not analyzed in detail, before jumping onto statements, then you would end up getting incorrect answers (trap answer in this question is option C)