Explanation

We need to know whether nm is odd. Both are integers, so each will be odd or even. On to the statements, separately first.

Statement (1) will succumb to analysis by cases. n is either even or odd. If n is even, m is odd, and the product nm will be an even times an odd number, which is always even. When two numbers are multiplied, the product contains all of the factors of the two numbers multiplied. That's why any integer multiplied by an even number must be even; the factor of 2 carries over into the product. Moving on: if n is odd, then m is even, and nm is again even. These cases are exhaustive, so nm must be even. Statement (1) is sufficient.

Statement (2) tells us that . Hence, . That means that m must be even, since the factor of 2 in that number we are calling "even" will be contained within m. Since m is even, and n is an integer, nm must be even. We have answered the question definitively, so Statement (2) is sufficient.

The correct answer is (D).

Explanation

This question involves odd and even number properties, so it should be a piece of cake, using the easily recallable rules. On to the data statements, separately first.

Statement (1) says that n + m is not even, which means that n + m is odd, because both numbers are integers. If n + m is odd, that means that one of them is even and one is odd; we can recall this by the rule or by examining a number of cases and convincing ourselves that it will be true in all cases. And with one even and one odd, when they are multiplied, the result will be even. Again, that's true by rule and can be seen by considering cases. In all cases, the answer to the question is "no," so Statement (1) gives sufficient information to answer the question.

Statement (2) indicates that n is even, since an integer and it's not odd. If n is even, nm will be even. So Statement (2) definitively answers the question and is sufficient.

The correct answer is (D).