GRE Math : How to multiply even numbers

Study concepts, example questions & explanations for GRE Math

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Example Questions

Example Question #1 : How To Multiply Even Numbers

If  is even, and  is odd. Which of the following must be odd?

Possible Answers:

Correct answer:

Explanation:

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

Only  works and is thus our answer.

Example Question #2 : How To Multiply Even Numbers

 is even

 is even

Quantity A:

The remainder of 

Quantity B:

1

Possible Answers:

The two quantities are equal.

The relationship between the quantities cannot be determined.

Quantity B is larger.

Quantity A is larger.

Correct answer:

Quantity B is larger.

Explanation:

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.

Example Question #3 : How To Multiply Even Numbers

 is even

 is even

Therefore, which of the following must be true about ?

Possible Answers:

It must be odd.

It must be even.

It could be either even or odd.

Correct answer:

It could be either even or odd.

Explanation:

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.

Example Question #1 : How To Multiply Even Numbers

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?

Possible Answers:

Correct answer:

Explanation:

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:

Example Question #31 : Even / Odd Numbers

Assume  and  are both even whole numbers.

What is a possible solution for ?

Possible Answers:

Correct answer:

Explanation:

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 .

Example Question #174 : Arithmetic

Which of the following integers has an even integer value for all positive integers  and ?

Possible Answers:

Correct answer:

Explanation:

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.

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