# If x is an integer, which of the following must be an even integer?

If \$x\$ is an integer, which of the following must be an even integer?

1. \$x^2-x-1\$
2. \$x^2-4x+6\$
3. \$x^2-5x+5\$
4. \$x^2+3x+8\$
5. \$x^2+2x+10\$

So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 9 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of Counting Methods can be kind of tricky, but never fear, PrepScholar has got your back!

## Survey the Question

Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on your paper.

The question asks us which answer choices must be an even integer, so it likely tests our Integers math skill. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.

## What Do We Know?

Let’s carefully read through the question and make a list of the things that we know.

1. We know that \$x\$ is an integer
2. We need to figure out which answer choice with \$x\$ in it must be even

## Develop a Plan

We want to know which answer choice MUST be even. Looking at the answer choices, we have addition, multiplication, and subtraction all throughout. So let’s quickly review our rules for addition, subtraction, and multiplication when it comes to odd and even integers.

## Concept Refresher – Even-Odd Arithmetic Rules

Even numbers are integers with a units digit of \$0, 2, 4, 6, \or \;8\$, whereas odd numbers are integers with a units digit of \$1, 3, 5, 7, \or \;9\$. The rules for adding even and odd numbers are the same as the rules for subtracting them, so we’ll write them both together:

\$\$\Even ± \Even = \Even\$\$
\$\$\Even ± \Odd = \Odd\$\$
\$\$\;\;\,\Odd ± \Odd = \Even\$\$

So to get an odd number when adding or subtracting two numbers together, we must have exactly one of the numbers be odd. Now let’s review the rules for multiplying even and odd numbers

\$\$\Even · \Even = \Even\$\$
\$\$\;\,\Even · \Odd = \Even\$\$
\$\$\;\;\Odd · \Odd = \Odd\$\$

So to get an odd number when multiplying two numbers together, we must have both numbers be odd.

Now let’s get back to the question at hand.

We know that \$x\$ is an integer, but we don’t know if it is even or odd. Let’s test each answer choice twice, first with \$x\$ being even and then with \$x\$ being odd. We’ll choose as the correct answer the answer choice that gives us an even result for both \$x\$ being an odd integer AND \$x\$ being an even integer. We can eliminate any answer choice whenever we get an odd result.

## Solve the Question

Let’s test each answer choice, being sure to use PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction) to do the Exponents first, then the multiplication, and finally the addition and subtraction.

### Test A: \$x^2-x-1\$

First let’s test when \$x\$ is odd:

\$\$x^2-x-1 = \odd^2 – \odd -\odd\$\$

Since \$\odd^2 = \odd·\odd = \odd\$, we next get:

\$\$x^2-x-1 = \odd – \odd -\odd\$\$

Next, \$\odd-\odd=\even\$:

\$\$x^2-x-1 = \even -\odd\$\$

And finally, we know that \$\even + \odd = \odd\$. Since we were able to get an odd integer as a result, we know that A is incorrect. Let’s test the other answer choices, but we’ll move through them a bit quicker since we’re more comfortable with our even-odd arithmetic rules now.

### Test B: \$x^2-4x+6\$

First let’s test when \$x\$ is odd:

 \$x^2-4x+6\$ \$=\$ \$\odd^2 – \even·\odd +\even\$ \$x^2-4x+6\$ \$=\$ \$(\odd – \even) +\even\$ \$x^2-4x+6\$ \$=\$ \$(\odd) +\even\$ \$x^2-4x+6\$ \$=\$ \$\odd\$

So we can rule out B since we got an odd result. Let’s try C.

### Test C: \$x^2-5x+5\$

First let’s test when \$x\$ is odd:

 \$x^2-5x+5\$ \$=\$ \$\odd^2 – \odd·\odd +\odd\$ \$x^2-5x+5\$ \$=\$ \$(\odd – \odd) +\odd\$ \$x^2-5x+5\$ \$=\$ \$(\even) +\odd\$ \$x^2-5x+5\$ \$=\$ \$\odd\$

So we can rule out C since we got an odd result. Let’s try D.

### Test D: \$x^2+3x+8\$

First let’s test when \$x\$ is odd:

 \$x^2+3x+8\$ \$=\$ \$\odd^2 + \odd·\odd +\even\$ \$x^2+3x+8\$ \$=\$ \$(\odd + \odd) +\even\$ \$x^2+3x+8\$ \$=\$ \$(\even) +\even\$ \$x^2+3x+8\$ \$=\$ \$\even\$

Hmmm…we got an even result. We still need to check \$x=\even\$ in D before choosing it as the right answer:

 \$x^2+3x+8\$ \$=\$ \$\even^2 + \odd·\even+\even\$ \$x^2+3x+8\$ \$=\$ \$(\even+ \even) +\even\$ \$x^2+3x+8\$ \$=\$ \$(\even) +\even\$ \$x^2+3x+8\$ \$=\$ \$\even\$

Ah! So D will ALWAYS be even. The correct answer is D, \$x^2+3x+8\$.

## What Did We Learn

Well, that certainly could have taken us a LOT of time if we were moving slowly. Looks like it’s important to practice these odd-even arithmetic rules a bit so that we can move quickly through these types of questions. Also, let’s try to make the rules more concise to speed up this odd-even arithmetic.

When multiplying odd and even integers, if we have at least one even integer then the product is even. When adding and subtracting odd and even integers, we need to have an odd number of integers that are odd to get an odd result.

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