Data Sufficiency

[GMAT math practice question]

m and n are positive integers. Is m^2 + n^2 is divisible by 3?

1) m = 1234
2) n = 4321

=>

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.

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 the conditions if necessary.

Recall that the remainder when an integer is divided by 3 is the same as the remainder when the sum of all of its digits is divided by 3.

The square of an integer k has remainder 0 or 1 when it is divided by 3.
If k is a multiple of 3, then k = 3a for some integer a, and k^2 = (3a)^2 = 3(3a^2) has remainder 0 when it is divided by 3.
If k has remainder 1 when it is divided by 3, then k = 3a + 1 for some integer a, and k^2 = (3a+1)^2 = 9a^2 +6a + 1 = 3(3a^2 +2a) + 1 has remainder 1 when it is divided by 3.
If k has remainder 2 when it is divided by 3, then k = 3a + 2 for some integer a, and k^2 = (3a+2)^2 = 9a^2 +12a + 4 = 3(3a^2 +4a+1) + 1 has remainder 1 when it is divided by 3.

Since m=1234 has remainder 1 when it is divided by 3, m^2 has remainder 1 when it is divided by 3. Since n^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of n. Condition 1) is sufficient, since it yields the unique answer, 'no'.

Since n=4321 has remainder 1 when it is divided by 3, n^2 has remainder 1 when it is divided by 3. Since m^2 could have remainder 0 or 1 when it is divided by 3, m^2 + n^2 is never divisible by 3, regardless of the value of m. Condition 2) is sufficient, since it yields the unique answer, 'no'.

Therefore, D is the answer.
Answer: D

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