"Let x and y be integers such that...", "If y is a positive integer, what is...?" If you've taken a practice test or a real ACT before, these types of questions may look familiar to you. You've likely come across several questions on the ACT that mention the word "integer." And if you don't know what that word means, they will be difficult problems for you to solve.

Questions involving integers are common, so it's important to have a solid grasp of what integers are as you continue in your ACT math study. But what are "integers" and how do they fit into the larger ACT math picture?

**This article will be your guide to basic integers for the ACT, what they are, how they change, and how you'll see them used on the test.** For the more advanced integer concepts--including absolute values, exponents, roots, and more--look to our advanced guide to ACT integers.

## What is an Integer?

An integer is a whole number. This means an integer is any number that is NOT expressed with a decimal or a fraction.

Integers include all negative whole numbers, all positive whole numbers, and zero.

**Examples of Integers:**

-32, -2, 0, 17, 2,035

**NOT integers:**

π, $2/3$, 0.478

*Think of an integer as an object that cannot be divided into pieces. For example, you can't have half an egg in a basket.*

## Positive and Negative Integers

A number line is used to demonstrate how numbers relate to each other and to zero. All numbers to the right of zero are positive numbers. All numbers to the left of zero are negative numbers.

**Positive numbers get larger the farther they are from zero.**

154 is larger than 12 because 154 is farther along the number line in a positive direction (to the right).

**Negative numbers get smaller the farther away they are from zero.**

-154 is SMALLER than -12 because -154 is a farther along the number line in a negative direction (to the left).

**And a positive number is always larger than any negative number.**

1 is larger than -10,109

Because we don't have a reference for 0, we cannot say for sure whether A is positive or negative, which eliminates answers F, G, and K.

We do know that any number to the left of another number will be less, so **the answer must be H**, A is less than B.

*The very opposite of a number line.*

## Typical Integer Questions on the ACT

Most ACT math integer questions are a combination of word problem and equation problem. The question will usually present you with an equation and tell you that you must use "integers" in place of a variable. You must know that an integer means a whole number (and that integers also include negative numbers and zero) to solve these problems.

When x≠0, there are two possible integer values for x such that y=x(1+x). What is a possible value for y?

(A) −30

(B) −1

(C) 0

(D) 15

(E) 20

(We'll walk through how to solve this problem in the next section.)

Sometimes you’ll have to answer more abstract questions about how integers relate to one another when you add, subtract, multiply and divide them. You don't need to find a *numerical* answer for these types of questions, but you must instead identify whether certain equations will be even or odd, positive or negative.

For these types of questions, you can either guess and check how integers change in relation to one another by plugging in your own numbers and solving, or you can memorize the rules for how integers interact. How you do it is completely up to you and depends on how you learn and/or like to solve math problems.

For example, in the charts below, you'll see that: a positive number * a positive number = a positive number, each and every time. If you forget this rule (or simply don't want to learn it in the first place), you can always try it by saying 2 * 3 = 6.

Because you can always find these results by plugging in your own numbers, these rules are categorized as “good to know,” but not “necessary to know.”

negative * negative = positive | -2 * -3 = 6 |

positive * positive = positive | 2 * 3 = 6 |

negative * positive = negative | -2 * 3 = -6 |

Another way to think of this is, “When multiplying numbers, the result is always positive unless you’re multiplying a positive number and a negative number.”

odd * odd = odd | 3 * 5 = 15 |

even * even = even | 2 * 4 = 8 |

odd * even = even | 3 * 4 = 12 |

Another way to think of this is, “When multiplying numbers, the result is always even until multiplying an odd number and an odd number.”

odd +/- odd = even | 5 + 7 = 12 |

even +/- even = even | 10 - 6 = 4 |

odd +/- even = odd | 5 + 6 = 11 |

Another way to think of this is, “When adding or subtracting numbers, the result is always even unless adding or subtracting an odd number and an even number.”

With these understandings in mind, let us look again at the above ACT math problem.

Choice A is incorrect, because *b* is an even integer. And we know that an even number * an odd number = an even number.

Choice B is incorrect because *a* is an odd integer. And we know that an odd number + an odd number = an even number.

Choice C is incorrect because *a* is an odd integer and *b* is an even integer. An even number + an odd number = an odd number. And an odd number * an even number (in this case 2) = an even number.

Choice D is correct. Twice *b* will be even, because an even number * an even number = an even number. And the final result will be odd because an odd number (*a*) + an even number (2*b*) = an odd number.

Choice E is incorrect. Twice an odd number (*a*) will be an even number, because an even number * an odd number = an even number. And an even number + an even number = an even number.

So **your final answer is D**, a + 2b.

You can see how you could also solve this by double-checking these rules by using your own numbers. If you assign an odd number to *a* and an even number to *b*, you can test out each option in about the same amount of time it would take you to go through your rules like this.

So for this question, you could have said *a* was 5 and *b* was 6. Then option D would have looked like this:

5 + 2(6) = 17

Again, because you can figure out these kinds of questions using real numbers, these rules are classified as "good to know," not "necessary to know."

*If you follow the right steps, solving an integer problem is often much easier than it appears.*

## Steps to Solving an ACT Math Integer Problem

**#1: Identify if the problem is, in fact, an integer problem.**

If you must use integers to solve a problem, the ACT will explicitly use the word "integer" in the question so that you don't waste your time and effort looking for decimal or fraction solutions. For example, questions may begin with: "x is a positive integer such that...", "For all negative integers...", or "How many integers give the solution to...?"

For any problem that *doesn’t* specify that the variables (or the solution) are “integers," your answer or the variables can be in decimals or fractions.

So let's look again at the problem from earlier:

When x ≠ 0, there are two possible integer values for x such that y = x(1+x). What is a possible value for y?

(A) −30

(B) −1

(C) 0

(D) 15

(E) 20

We are told that x ≠ 0, so we know that our y cannot be 0. Why not? Because the only integer values that can give you y = 0 are x = 0 and x = −1 because 0(1+0) = 0 and (−1)(1+(−1)) = 0.

BUT we were told that x ≠ 0. So y can not equal 0 either, as the question told us that there were TWO integer values for x, neither of which is 0.

This means we can cross off C from the answer choices.

We can also cross off A and B. Why? Because there is no possible way to have x(1+x) equal a negative. Even when x is negative, we would distribute the problem to look like:

y = (1x) + (x * x)

We know that a negative * a positive = a negative, so 1x would be negative if x were negative.

BUT a positive * a positive = a positive. And a negative * a negative = a positive. So x * x would be positive, whether x was positive *or* negative. And adding the original negative value for x will not be a large enough number to take away from the positive square and make the final answer a negative.

For example, we already saw that:

x =−1 makes our y zero.

x =−2 gives us −2(1+−2) = y = 2.

x =−3 gives us −3(1+−3) = y = 6, etc.

So we are left with answer choices D and E.

Now how could we get 15 with x(1+x)? We know x must not be very large to get y = 15, so let's test a few small numbers for x.

If x = 2, then x(1+x) = 2(1+2) = 6. This means x = 2 is too small.

If x = 3, then x(1+x) = 3(1+3) = 12. So x = 3 is too small.

If x = 4, then x(1+x) = 4(1+4) = 20. This means there is *no* positive integer value that could give us 15.

But we did manage to get y = 20, so answer choice E is looking pretty good!

Now we can tell that if we kept going higher with x, the y value would keep getting larger (x = 5 would give us y = 30, etc.). This means we probably need a negative integer to give us our second value for x.

So let's try to get y = 20 with a negative value for x this time.

We already saw above that x = −2 gave us y = 2, and x = −3 gave us y = 6. So let's try some more negative values for x.

If x = −4, then x(1+x) = −4(1+−4) = 12

If x = −5, then x(1+x) = −5(1+−5) = 20

We were able to get y = 20 with both x = 4 and x = −5

**So our final answer is E**, y = 20

**#2: If the problem asks you to identify equations that are always true, test out multiple different kinds of integers.**

If the question asks you to identify whether certain equations or inequalities are true for ALL integers, the equation must work equally with 10 as with 0 and -5. **A good rule of thumb is to try -1, 0, and 1 with variable questions like these.** These numbers often have special properties that make or break conditions. I'll explain what that means with a practice example.

If *x* is an integer, which of the following equations MUST be true?

I. $x^3 ≥ (-x)^3$

II. ${x^3}/x ≥ {x^2}/x$

III. $x(x + 1) ≤ -x + x^3$

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) I, II, and III

For questions like these, we should test out our sample numbers, as it can get confusing to use our rules of integer behaviors with complex problems such as these.

So for option I, let use our test numbers of -1, 0, and 1.

$−1^3= (−1)(−1)(−1) = −1$

$(−−1)^3 = $1^3 = (1)(1)(1) = 1$

-1 is NOT greater than +1.

This automatically eliminates option I. And by eliminating option I, we can eliminate answer choices A, D, and E right away.

Now let's look at choice II with our same test numbers.

${(-1)^3}/{-1} = {(-1)(-1)(-1)}/{-1} = {-1}/{-1} = 1$

${(-1)^2}/{-1} = {(-1)(-1)}/{-1} = 1/{-1} = -1$

1 > -1

This means that option II works so far when we use a negative number. So let's try it with our positive number, 1.

${1^3}/1 = {(1)(1)(1)}/1 = 1/1 = 1$

${1^2}/1 = {(1)(1)}/1 = 1/1 = 1$

1 = 1.

So option II still works.

Lastly, we should test if the equation still works with 0.

${0^3}/0 = 0$

$0^2/0 = 0$

Option II works for all answer choices, **so our final answer is B**, II only. Because we know that option I doesn't work, we have eliminated all other answer choices.

But if you want to make *absolutely* sure you didn't make a mistake somewhere, you can test out option III as well.

−1(−1+1) = 0

$−(−1)+(−1)^3 = 1+(−1)(−1)(−1) = 1+−1 = 0$

0 = 0

The two are equal, which means that option III works so far. Now let's try it with 1.

1(1+1) = 2

$−1+1^3 = −1+(1)(1)(1) = −1+1 = 0$

2 > 0

When we used a positive number, the equation was incorrect. This means that answer choice C is eliminated and our choice of B has been confirmed to be the only correct answer.

**#3: If the problem asks you to find the answer to long calculations, use your rules that you learned above or test it out with smaller numbers.**

*a, b, c, d, e, f* are odd integers such that *a* > *b* > *c* > *d* > *e* > *f*. Which statement(s) must be true?

I. *abcdef* is odd

II. *a* + *b* + *c* + *d* + *e* + *f* is odd

III. *a(b + c + d + e + f)* is odd

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) I, II, and III

Now you can approach this problem in one of two ways: by using your number rules or by using your own numbers.

First, let's use our number rules to test option I.

We know that each letter represents an odd integer and that the product of an odd number and another odd number is an odd number. Because an odd * an odd will always be odd, we know that option I is true. This means we can also eliminate answer choices B and C.

Now let's look at option II.

We know that an odd number + an odd number = an even number. We also know that an even number + an even number = an even number.

So if we split *a* + *b* + *c* + *d* + *e* + *f* into pairs of numbers, we'll have:

(*a* + *b)* + (*c* + *d)* + (*e* + *f)*

We know that each pair of numbers will have an even sum, so we're left with:

an even number + an even number + an even number, which will give us an even final result. So option II is incorrect. This means we can eliminate answer choice E.

Finally, let's look at option III.

As we saw before, when we have six odd numbers (in other words, an even number of odd numbers), the sum will be even. Now, our parenthesis holds five (an odd number) of odd numbers, and an even number + an odd number = an odd number. So we know the number in the parenthesis will be odd.

We also know that an odd number (*a*) * an odd number (the sum of *b, c, d, e, f*) = an odd number. So option III is correct.

This means that **our final answer is D**, I and III only.

The other way you could solve this problem would be to test out these rules with small numbers and extrapolate to find the larger answer. In other words, use small numbers in place of the variables.

So for option I, if you didn't know an odd * an odd = an odd, you could replace *a* and *b* with the numbers 5 and 3. 5*3=15, so you know that an odd * an odd = an odd number, no matter how many times you multiply it. So option I is correct.

For option II, again test it out with smaller numbers. 7+5=12, and 7+5+3=15. So you know that adding odd numbers an *even* number of times gets you an even answer and adding an *odd* number of times gets you an odd answer. There are six odd numbers, so the final answer must be even. Option II is incorrect.

Taking what you learned by testing option II, you know that adding odd numbers an *even* number of times gets you an odd answer. And, taking what you learned from testing option I, you know that an odd number * an odd number = an odd number. This means your final answer must be odd, so option III is correct.

This means **the final answer is D**, I and III only.

*Whoo! There are many ways to solve integer problems and whichever way works for you is perfect.*

## The Take-Aways

In order to solve both the basic and advanced ACT integer questions, you must first understand what an integer *is*. Only then can you build up your integer knowledge to the more advanced concepts. But simply knowing that an integer is a whole number (and that 0 and negative numbers are also integers), will allow you to solve some of the more basic questions about how to plug integers into equations and how integers relate to one another.

For the more advanced integer concepts, including absolute values, exponents, etc., be sure to check out our advanced guide to ACT integers.

## What's Next?

Now that you’ve learned about what integers are, you may want to check out the advanced guide to ACT integers where we will go through absolute values, prime numbers, and exponents (among other concepts). Make sure that you also have a solid understanding of all the ACT math concepts on the test as well as all the ACT formulas you'll need to know.

**Running out of time on ACT math?** Check out our article on how to buy yourself those extra precious seconds and minutes and complete your ACT math problems before time’s up.

** Feeling overwhelmed? **Start by figuring out your ideal score

**Check out our article on how to get a perfect score written by a 36 ACT-scorer.**

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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.

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