Rewrite  by using factors.  Simplify until the answer cannot be reduced any further.

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:

We know that when you multiply any integer by an even number, the result is even. Thus, if  is odd but when you multiply this by  the number  is even, we know that  must be even. You cannot say anything about the sign value of any of the numbers. Likewise, it is impossible for either  or  to be even.

Three consecutive integers will include at least one, and possibly two, even integers. Since the addition of even a single even integer to a chain of integer products will make the final product even, we know the product must be even.

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.

If Steve has 10 chores, Jeff has 7 chores (3 less than Steve).  Tom has twice as many as Steve, so Tom has 14 chores. 

Create an equation for the situation described and solve:

Create an equation to decribe the above situation and solve.

Total presents given out is represented by .

Presents given to Sally is represented by .

Presents given to Billy is represented by .

Thus our equation becomes,

Always start by gathering your facts. You know that if  is odd, then neither of those numbers are even. (If either were, then the product would be even.)  Now, the rules for subtracting are just like adding. If the difference of two numbers is even, then they must either both be odd or both be even. Thus if  is even and we know that  is odd, then we know that  must be odd. Thus if  is odd, it is also true that  is even, for both  and  are even.