When multiplying odd/even numbers, we know that odd * odd = odd, and odd * even = even. We also know that odd + odd = even. We will proceed to evaluate each answer choice, knowing that m, n and p are odd.

(m + 1) * n

m + 1 becomes even. This gives us even * odd = even.

m + n + p + 1

Odd + odd + odd + odd = even + odd + odd = even + even = even.

(m - 2 )* n * p

m - 2 is stil odd. This gives us odd * odd * odd = odd * odd = odd.

m * (n + p)

Odd + odd is even, so here we have odd * even = even.

m * p * (n -1)

n - 1 becomes even so we have odd * odd * even = odd * even = even.

The correct answer is therefore m * p * (n -1).

For this problem, we must recognize under what arithmetic conditions an even or odd number is produced. We do not know what the values for x and y are, but we do know for example that any two numbers added can be either even or odd, and and any number multiplied by two must be even (e.g. 2 * 2 = 4; 3 * 2 = 6), or an odd number multiplied by another odd is always even. 

In this situation, we have to ensure that an arithmetic operation (subtraction) must result only in an even number, by further ensuring that the variables themselves are fixed to either odd or even values. 

In this situation, we know that due to the "*2 (= 2 * 2)" principle, the 4x component always being an even value. The only situation in which the operation "4x – y²" will be even is if the "y²" term is also even, since an even minus an odd would be odd. The only situation in which the "y²" term is even is if the y itself is even, since an odd number squared always results in an odd value (e.g. 3² = 9; 7² = 25). To ensure that the y itself is even, we must also double it to "2y"(²).

We cannot add 1 to a random variable y as it may still result in an odd y-value. Similarly, since we already stated that a squared value may still be odd, we cannot be sure that squaring the y will also result in an even number.

Doing this by counting the odd numbers will take way too long for the GRE.  And if you look at the answer choices, you see five consecutive odd numbers, so one little mistake in counting will give you the wrong answer!  Instead, we should use the formula for finding odd numbers: the n^th odd number is 2n – 1.  (The n^th even number is 2n.)  So the 65^th odd number is 2 * 65 – 1 = 129.

Even numbers come in the form 2x, and odd numbers come in the form (2x + 1), where x is an integer.  If this is confusing for you, simply plug in numbers such as 1, 2, 3, and 4 to find that:

Any odd number + any even number = odd number

Any odd number + any odd number = even number

Any even number x any number = even number

Any odd number x any odd number = odd number

If there are  students taking Greek, then there are  or  students taking Latin. However, the question asks how many total students there are in the school, so you must add these two values together to get:

 or  total students.

An odd whole number minus an odd whole number will result in an even whole number. Since , the subtraction will result in a negative even whole number. The only answer that fits these requirements is .

If you have trouble with problems like this, try stacking the numbers, and subtracting one place at a time.  First, seven minus one is six, so that's your ones digit.  Next, three minus three is zero, so that's your tens digit.  Seven minus two is five, so that's your hundreds digit, and finally, one minus zero is one so that's you're thousands digit.  

To solve, simply subtract.  If you have trouble subtracting, try splitting the numbers up and adding your results:

To perform this operation, simply subtract.  If you are having trouble, you can split the problem into two easier to solve portions, by splitting the numbers up:

To solve, simply subtract.  If you have trouble subtracting, you can split the problem up and add your results back together: