If m and n are prime numbers, is m^2 + n^2 an even number?

[Max@Math Revolution]

[GMAT math practice question]

If m and n are prime numbers, is m^2 + n^2 an even number?

1) m > 10
2) n > 20

[jpcameron17]

In order for m^2 + n^2 to be even, then either: a) m and n are both even, or b) m and n are both odd.

1) m > 10
All prime numbers (with the exception of 2) are odd, therefore m is odd. However, since we don't have enough information to know whether n is even or odd, we cannot determine whether m^2 + n^2 is even. It is therefore not sufficient.

2) n > 20
All prime numbers (with the exception of 2) are odd, therefore n is odd. However, since we don't have enough information to know whether m is even or odd, we cannot determine whether m^2 + n^2 is even. It is therefore not sufficient.

1) + 2) Together
This tells us that m and n are both odd, therefore the sum of their squares is even.

The answer is C).

[Max@Math Revolution]

=>

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.

In order for m^2 + n^2 to be even, either both m and n must be even or both m and n must be odd. Since m and n are primes, we must have both m and n odd numbers as they are both greater than 10. Thus, C is the answer.

Since this question is a statistics question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
If m = 11 and n = 3, then m^2+n^2 = 11^2+3^2 = 121 + 9 = 130 is an even number, and the answer is "yes".
If m = 11 and n = 2, then m^2+n^2 = 11^2+2^2 = 121 + 4 = 125 is an odd number, and the answer is "no".
Thus, condition 1) is not sufficient since it does not yield a unique solution.

Condition 2)
If m = 3 and n = 23, then m^2+n^2 = 3^2+23^2 = 9 + 529 = 538 is an even number, and the answer is "yes".
If m = 2 and n = 23, then m^2+n^2 = 2^2+23^2 = 4 + 529 = 533 is an odd number, and the answer is "no".
Thus, condition 2) is not sufficient since it does not yield a unique solution.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

[deloitte247]

If m and n are prime numbers
$$Is\ m^2+n^2\ an\ even\ number$$
m and n are prime number with just 2 factors which are 1 and itself ( m or n)
m = odd number = 2n +1
n = odd number = 2n+1
$$odd^2+odd^2=\left(odd\cdot odd\right)+\left(odd\cdot odd\right)$$
$$odd+odd=even$$
Irrespective of the value of m and n
$$m^2+n^2=even\ number$$ $$m^2+n^2=even\ number$$
Even if m or n is a negative integer.
Hence statement 1 ans 2 alone are SUFFICIENT.

$$answer\ is\ Option\ D$$