If x any y are positive integers,is x even? (1) x^2 + y^2=98 (2) x = y

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

If x is an even integer, then we have \(x = 2k\) for an integer \(k\).
We have \(y^2 = 98 - x^2 = 98 - 4k^2 = 4(24-k^2) + 2\) from condition 1).
A square of an integer \(y^2\) can't have a remainder \(2\) when it is divided by \(4\) for the following reasoning.
Thus, \(x\) can not be even.

If \(y\) is an odd integer, \(y = 2a + 1\), then \(y^2 = (2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2+a) + 1\) and \(y^2\) has a remainder \(1\).
If \(y\) is an even integer, \(y = 2a\), then \(y^2 = (2a)^2 = 4a^2 = 4a^2 + 0\) and \(y^2\) has a remainder \(0\).

Condition 1) yields a unique answer 'no'.

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)

Since condition 2) does not yield a unique solution obviously, it is not sufficient.

Therefore, A is the answer.

If \(x\) and \(y\) are positive integers, is \(x\) even ?


(1) \(x^{2} + y^{2} = 98\)
This combination is possible only when x & y = 7
Sufficient

(2) \(x = y\)
Not sufficient

IMO A