To solve, divide both sides of the equation by :

As a check, if you divide an odd number by another odd number, your result should be odd. 

With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables.  However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.

Therefore, our best approach is to simply analyze each answer choice.

:  Since  is odd,  is also odd, since and odd number multiplied by an odd number yields an odd product.  Since  is also odd, multiplying it by  will again yield an odd product, so this expression is always odd.

:  Since  is odd, multiplying it by 2 will yield an even number.  Subtracting this number from  will also give an odd result, since an odd number minus an even number gives an odd number.  Therefore, this answer is also always odd.

:  Since both numbers are odd, their product will also always be odd.

:  Since  is odd, multiplying it by 2 will give an even number.  Since  is odd, subtracting it from our even number will give an odd number, since an even number minus and odd number is always odd.  Therefore, this answer will always be odd.

:   Since both numbers are odd, there sum will be even.  However, dividing an even number by another even number (2 in our case) does not always produce an even or an odd number.  For example, 5 and 7 are both odd.  Their sum, 12, is even.  Dividing by 2 gives 6, an even number.  However, 5 and 9 are also both odd.  Their sum, 14, is even, but dividing by 2 gives 7, an odd number.  Therefore, this expression isn't necessarily always odd or always even, and is therefore our answer.

To find the answer to this question, calculate the total jelly beans for each person:

Portia:  * <Theodore's count of jelly beans>, which is  or 

Harvey:  * <Portia's count of jelly beans>, which is  or 

So, the total is:

(Do not forget that you need those original  for Theodore!)

When  two odd whole numbers are multipliedd, they will equal an odd whole number. The only answer that fits the requirements of being an odd and a whole number is  since it comes from  or .

To solve, pick numbers to represent  and . Let  and . Now try each of the equations given:

. 

Only  works and is thus our answer.

Begin by considering our options.  For , it is either the case that:

 is even and  is odd, or,

 is even and  is even, or,

 is odd and  is even

Now, for , it is either the case that:

 and  are even, or,

 and  are odd

Now, for , it cannot be the case that both are odd.  This means that the only viable option for  is the case when both are even.  

Therefore,  must be even, meaning that the remainder of  must be .  

Therefore, quantity B is larger.

Recall that when you multiply by an even number, you get an even product.

Therefore, we know the following from the first statement:

 is even or  is even or both  and  are even.

For the second, we know this:

Since  is even, therefore,  can be either even or odd. (Regardless of what it is, we can get an even value for .)  

Based on all this data, we can tell nothing necessarily about . If  is even, then  is even, even if  is odd. However, if  is odd while  is even, then  will be even.

In a group of philosophers,  are followers of Durandus. Twice that number are followers of Ockham. Four times the number of followers of Ockham are followers of Aquinas. One sixth of the number of followers of Aquinas are followers of Scotus. How many total philosophers are in the group?

To start, let's calculate the total philosophers:

Ockham:  * <Number following Durandus>, or 

Aquinas:  * <Number following Ockham>, or 

Scotus:  divided by , or 

Therefore, the total number is:

When two even numbers are multiplied, they must equal an even number. Also, since both variables are said to be even whole numbers, the answer must fit the requirement that its factors are two even numbers multiplied by one another. The only answer that fits both requirements is  which can be factored into the even whole numbers  and .

There are certain patterns that can be used to predict whether the product or sum of numbers will be odd or even. The sum of two odd numbers is always even, as is the sum of two even numbers. The sum of an odd number and an even number is always odd. In multiplication the product of two odd numbers is always odd. While the product of even numbers, as well as the product of odd numbers multiplied by even numbers is always even. So for this problem we need to find scenarios where the only possibile answers are even.  can only result in even numbers no matter what positive integers are used for  and , because  must can only result in even products; the same can be said for . The rules provide that the sum of two even numbers is even, so  is the answer.