Integer questions are some of the most common on the SAT, so understanding what integers are and how they operate will be crucial for solving many SAT math questions. Knowing your integers can make the difference between a score you’re proud of and one that needs improvement.

In our basic guide to integers on the SAT (which you should review before you continue with this one), we covered what integers are and how they are manipulated to get even or odd, positive or negative results. In this guide, we will cover the more advanced integer concepts you’ll need to know for the SAT.

This will be your complete guide to advanced SAT integers, including consecutive numbers, primes, absolute values, remainders, exponents, and roots—what they mean, as well as how to handle the more difficult integer questions the SAT can throw at you.

## Typical Integer Questions on the SAT

Because integer questions cover so many different kinds of topics, there is no “typical” integer question. We have, however, provided you with several real SAT math examples to show you some of the many different kinds of integer questions the SAT may throw at you.

Over all, you will be able to tell that a question requires knowledge and understanding of integers when:

#1: The question specifically mentions integers (or consecutive integers).

Now this may be a word problem or even a geometry problem, but you will know that your answer must be in whole numbers (integers) when the question asks for one or more integers.

If \$j\$, \$k\$, and \$n\$ are consecutive integers such that \$0<j<k<n\$ and the units (ones) digit of the product \$jn\$ is 9, what is the units digit of \$k\$?

A. 0
B. 1
C. 2
D. 3
E. 4

(We will go through the process of solving this question later in the guide)

#2: The question deals with prime numbers.

A prime number is a specific kind of integer, which we will discuss in a minute. For now, know that any mention of prime numbers means it is an integer question.

What is the product of the smallest prime number that is greater than 50 and the greatest prime number that is less than 50?

(We will go through the process of solving this question later in the guide)

#3: The question involves an absolute value equation (with integers)

Anything that is an absolute value will be bracketed with absolute value signs which look like this:
| |

For example: \$|-210|\$ or \$|x + 2|\$

\$|10 - k| = 3\$

\$|k - 5| = 8\$

What is a value for k that fulfills both equations above?

(We will go through how to solve this problem in the section on absolute values below)

Note: there are several different kinds of absolute value problems.

About half of the absolute value questions you come across will involve the use of inequalities (represented by \$>\$ or \$<\$). If you are unfamiliar with inequalities, check out our guide to inequalities.

The other types of absolute value problems on the SAT will either involve a number line or a written equation. The absolute value questions involving number lines almost always use fraction or decimal values. For information on fractions and decimals, look to our guide to SAT fractions.

We will be covering only written absolute value equations (with integers) in this guide.

#4: The question uses perfect squares or asks you to reduce a root value

A root question will always involve the root sign: \$√\$

\$√81\$, \$^3√8\$

You may be asked to reduce a root, or to find the square root of a perfect square (a number that is the square of an integer). You may also need to multiply two or more roots together.

We will go through these definitions as well as how all of these processes are done in the section on roots.

(Note: A root question with perfect squares may involve fractions. For more information on this concept, look to our guide on fractions and ratios.)

#5: The question involves multiplying or dividing bases and exponents

Exponents will always be a number that is positioned higher than the main (base) number:

\$2^7\$, \$(x^2)^4\$

You may be asked to find the values of exponents or find the new expression once you have multiplied or divided terms with exponents.

We will go through all of these questions and topics throughout this guide in the order of greatest prevalence on the SAT.

We promise that integers are a whole lot less mysterious than...whatever these things are.

## Exponents

Exponent questions will appear on every single SAT, and you will likely see an exponent question at least twice per test. An exponent indicates how many times a number (called a “base”) must be multiplied by itself.

So \$4^2\$ is the same thing as saying \$4 * 4\$. And \$4^5\$ is the same thing as saying \$4 * 4 * 4 * 4 * 4\$. Here, 4 is the base and 2 and 5 are the exponents.

A number (base) to a negative exponent is the same thing as saying 1 divided by the base to the positive exponent.

For example, \$2^{-3}\$ becomes \$1/2^3\$ => \$1/8\$

If \$x^{-1}h=1\$, what does \$h\$ equal in terms of \$x\$?

A. \$-x\$
B. \$1/x\$
C. \$1/{x^2}\$
D. \$x\$
E. \$x^2\$

Because \$x^{-1}\$ is a base taken to a negative exponent, we know we must re-write this as 1 divided by the base to the positive exponent.

\$x^{-1}\$ => \$1/{x^1}\$

Now we have:

\$1/{x^1} * h\$

Which is the same thing as saying:

\${1h}/x^1\$ => \$h/x\$

And we know that this equation is set equal to 1. So:

\$h/x = 1\$

If you are familiar with fractions, then you will know that any number over itself equals 1. Therefore, \$h\$ and \$x\$ must be equal.

So our final answer is D, \$h = x\$

But negative exponents are just the first step to understanding the many different types of SAT exponents. You will also need to know several other ways in which exponents behave with one another.

Below are the main exponent rules that will be helpful for you to know for the SAT.

### Exponent Formulas:

Multiplying Numbers with Exponents:

\$x^a * x^b = x^[a + b]\$

(Note: the bases must be the same for this rule to apply)

Why is this true? Think about it using real numbers.

If you have \$2^4 * 2^6\$, you have:

\$(2 * 2 * 2 * 2) * (2 * 2 * 2 * 2 * 2 * 2)\$

If you count them, this give you 2 multiplied by itself 10 times, or \$2^10\$. So \$2^4 * 2^6\$ => \$2^[4 + 6]\$ => \$2^10\$.

If \$7^n*7^3=7^12\$, what is the value of \$n\$?

A. 2
B. 4
C. 9
D. 15
E. 36

We know that multiplying numbers with the same base and exponents means that we must add those exponents. So our equation would look like:

\$7^n * 7^3 = 7^12\$

\$n + 3 = 12\$

\$n = 9\$

So our final answer is C, 9.

\$x^a * y^a = (xy)^a\$

(Note: the exponents must be the same for this rule to apply)

Why is this true? Think about it using real numbers.

If you have \$2^4 * 3^4\$, you have:

\$(2 * 2 * 2 * 2) * (3 * 3 * 3 * 3)\$ => \$(2 * 3) * (2 * 3) * (2 * 3) * (2 * 3)\$

So you have \$(2 * 3)^4\$, or \$6^4\$

Dividing Exponents:

\${x^a}/{x^b} = x^[a-b]\$ (Note: the bases must be the same for this rule to apply)

Why is this true? Think about it using real numbers.

\${2^6}/{2^2}\$ can also be written as:

\${(2 * 2 * 2 * 2 * 2 * 2)}/{(2 * 2)}\$

If you cancel out your bottom 2s, you’re left with \$(2 * 2 * 2 * 2)\$, or \$2^4\$

So \${2^6}/{2^2}\$ => \$2^[6-2]\$ => \$2^4\$

If \$x\$ and \$y\$ are positive integers, which of the following is equivalent to \$(2x)^{3y}-(2x)^y\$?

A. \$(2x)^{2y}\$
B. \$2^y(x^3-x^y)\$
C. \$(2x)^y[(2x)^{2y}-1]\$
D. \$(2x)^y(4x^y-1)\$
E. \$(2x)^y[(2x)^3-1]\$

In this problem, you must distribute out a common element—the \$(2x)^y\$—by dividing it from both pieces of the expression.

This means that you must divide both \$(2x)^{3y}\$ and \$(2x)^y\$ by \$(2x)^y\$. Let's start with the first:

\${(2x)^{3y}}/{(2x)^y}\$

Because this is a division problem that involves exponents with the same base, we say:

\${(2x)^{3y}}/{(2x)^y} = (2x)^[3y - y]\$

So we are left with:

\$(2x)^{2y}\$

Now, for the second part of our equation, we have:

\${(2x)^y}/{(2x)^y}\$

Again, we are dividing exponents that have the same base. So by the same process, we would say:

\${(2x)^y}/{(2x)^y} = (2x)^[y - y] = (2x)^0 = 1\$

(Why 1? Because, as you'll see below, anything raised to the power of 0 = 1)

So our final answer looks like:

\${(2x)^y}{((2x)^{2y} - 1)}\$

Which means our final answer is C.

Taking Exponents to Exponents:

\$(x^a)^b = x^[a * b]\$

Why is this true? Think about it using real numbers.

\$(2^3)^4\$ can also be written as:

\$(2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2)\$

If you count them, 2 is being multiplied by itself 12 times. So \$(2^3)^4 => 2^[3 * 4] => 2^12\$

\$(x^y)^6 = x^12\$, what is the value of \$y\$?

A. 2
B. 4
C. 6
D. 10
E. 12

Because exponents taken to exponents are multiplied together, our problem would look like:

\$y * 6 = 12\$

\$y = 2\$

So our final answer is A, 2.

Distributing Exponents:

\$(x/y)^a = {x^a}/{y^a}\$

Why is this true? Think about it using real numbers.

\$(2/4)^3\$ can be written as:

\$(2/4) * (2/4) * (2/4)\$

\$8/64 = 1/8\$

You could also say \$2^3/4^3\$ => \$8/64\$ => \$1/8\$

\$(xy)^z = x^z * y^z\$

If you are taking a modified base to the power of an exponent, you must distribute that exponent across both the modifier and the base.

\$(3x)^3\$ => \$3^3 * x^3\$

(Note on distributing exponents: you may only distribute exponents with multiplication or division—exponents do not distribute over addition or subtraction. \$(x + y)^a\$ is NOT \$x^a + y^a\$, for example)

Special Exponents:

For the SAT you should know what happens when you have an exponent of 0:

\$x^0=1\$ where \$x\$ is any number except 0

(Why any number but 0? Well 0 to any power other than 0 is 0, because \$0x = 0\$. And any other number to the power of 0 is 1. This makes \$0^0\$ undefined, as it could be both 0 and 1 according to these guidelines.)

Solving an Exponent Question:

Always remember that you can test out exponent rules with real numbers in the same way that we did above. If you are presented with \$(x^2)^3\$ and don’t know whether you are supposed to add or multiply your exponents, replace your x with a real number!

\$(2^2)^3 = (4)^3 = 64\$

Now check if you are supposed to add or multiply your exponents.

\$2^[2+3] = 2^5 = 32\$

\$2^[2 * 3] = 2^6 = 64\$

So you know you’re supposed to multiply when exponents are taken to another exponent.

This also works if you are given something enormous, like \$(x^23)^4\$.

You don’t have to test it out with \$2^23\$! Just use smaller numbers like we did above to figure out the rules of exponents. Then, apply your newfound knowledge to the larger problem.

And the philosophical debate continues.

## Roots

Root questions are common on the SAT, and you should expect to see at least one during your test.

Roots are technically fractional exponents. You are likely most familiar with square roots, however, so you may have never heard a root expressed in terms of exponents before.

A square root asks the question: "What number needs to be multiplied by itself one time in order to equal the number under the root sign?"

So \$√36 = 6\$ because 6 must be multiplied by itself one time to equal 36.

In other words, \$6^2 = 36\$

Another way to write \$√36\$ is to say \$^2√36\$. The 2 at the top of the root sign indicates how many numbers (2 numbers, both the same) are being multiplied together to become 36. (Note: you do not expressly need the 2 at the top of the root sign—a root without an indicator is automatically a square root.)

So \$^3√27 = 3\$ because three numbers, all of which are the same (\$3 * 3 * 3\$), multiplied together equals 27. Or \$3^3 = 27\$.

Fractional Exponents

If you have a number to a fractional exponent, it is just another way of asking you for a root.

So \$16^{1/2} => ^2√16\$

To turn a fractional exponent into a root, the denominator becomes the value to which you take the root.

But what if you have a number other than 1 in the numerator?

\$16^{2/3} => ^3√16^2\$

The denominator becomes the value to which you take the root, and the numerator becomes the exponent to which you take the number under the root sign.

Distributing Roots

\$√xy = √x * √y\$

Just like with exponents, roots can be separated out.

So \$√20\$ => \$√2 * √10\$ or \$√4 * √5\$

\$√x * √y = √xy\$

Because they can be separated, roots can also come together.

So \$√2 * √10\$ => \$√20\$

Reducing Roots

It is common to encounter a problem with a mixed root, where you have an integer multiplied by a root (like \$3√2\$).

Here, \$3√2\$ is reduced to its simplest form, but let's say you had something like this instead:

\$2√12\$

Now \$2√12\$ is NOT as reduced as it can be. In order to reduce it, we must find out if there are any perfect squares that factor into 12. If there are, then we can take them out from under the root sign.

(Note: if there is more than one perfect square that can factor into your number under the root sign, use the largest one.)

12 has several factor pairs. These are:

\$1 * 12\$

\$2 * 6\$

\$3 * 4\$

Well 4 is a perfect square because \$2 * 2 = 4\$. That means that \$√4 = 2\$.

This means that we can take 4 out from under the root sign. Why? Because we know that \$√xy = √x * √y\$.

So \$√12 = √4 * √3\$. And \$√4 = 2\$. So 4 can come out from under the root sign and be replaced by 2 instead.

\$√3\$ is as reduced as we can make it, since it is a prime number.

We are left with \$2√3\$ as the most reduced form of \$√12\$

(Note: you can test to see if this is true on most calculators. \$√12 = 3.4641\$ and \$2 *√3 = 2 * 1.732 = 3.4641\$. The two expressions are identical.)

Now to finish the problem, we must multiply our reduced form of \$√12\$ by 2. Why? Because our original expression was \$2√12\$.

\$2 * 2√3 = 4√3\$

So \$2√12\$ in its most reduced form is \$4√3\$

## Remainders

Questions involving remainders generally show up at least once or twice on any given SAT. A remainder is the amount left over when two numbers do not divide evenly. If you divide 12 by 4, you will not have any remainder (your remainder will be zero). But if you divide 13 by 4, you will have a remainder of 1, because there is 1 left over.

You can think of the division as \$13/4 = 3{1/4}\$. That extra 1 is left over.

Most of you probably haven’t worked with integer remainders since elementary school, as most higher level math classes and questions use decimals to express the remaining amount after a division (for the above example, \$13/4 = 3 \remainder 1\$ or \$3.25\$). But for some situations, decimals simply do not apply.

Joanne’s hens laid a total of 33 eggs. She puts them into cartons that fit 6 eggs each. How many eggs will she have left that do NOT make a full carton of eggs?

\$33/6 = 5 \remainder 3\$. So Joanne can make 5 full baskets with 3 eggs left over.

Some remainder questions may seem incredibly obscure, but they are all quite basic once you understand what is being asked of you.

Which of the following answers could be the remainders, in order, when five positive consecutive integers are divided by 4?

A. 0, 1, 2, 3, 4
B. 2, 3, 0, 1, 2
C. 0, 1, 2, 0, 1
D. 2, 3, 0, 3, 2
E. 2, 3, 4, 3, 2

This question may seem complicated at first, so let’s break it down into pieces.

The question is asking us to find the list of remainders when positive consecutive integers are divided by 4. This means we are NOT looking for the answer plus remainders—we are just trying to find the remainders by themselves.

We will discuss consecutive integers below in the guide, but for now understand that "positive consecutive integers" means positive integers in a row along a number line. So positive consecutive integers increase by 1 continuously. 11, 12, 13, 14, 15, etc. are an example of positive consecutive integers.

We also know that any number divided by 4 can have a maximum remainder of 3. Why? Because if any number could be divided by 4 with a remainder of 4 left over, it means it could be divided by 4 one more time!

For example, \$16/4 = 4 \remainder 0\$ because 4 goes into 16 exactly 4 times. (It is NOT \$3 \remainder 4\$.)

So that automatically lets us get rid of answer choices A and E, as those options both include a 4 for a remainder.

Now we also know that, when positive consecutive integers are divided by any number, the remainders increase by 1 until they hit their highest remainder possible. When that happens, the next integer remainder resets to 0. This is because our smaller number has gone into the larger number an even number of times (which means there is no remainder).

For example, \$10/4 = 2 \remainder 2\$, \$11/4 = 2 \remainder 3\$, \$12/4 = 3 \remainder 0\$, and \$13/4 = 3 \remainder 1\$

Once the highest remainder value is achieved (n - 1, which in this case is 3), the next remainder resets to 0 and then the pattern repeats again from 1.

So we’re looking for a pattern where the remainders go up by 1, reset to 0 after the remainder = 3, and then repeat again from 1.

This means the answer is B, 2, 3, 0, 1, 2

Luckily, Joanne's remaining eggs did not go unloved for long.

## Prime numbers

The SAT loves to test students on prime numbers, so you should expect to see one question per test on prime numbers. Be sure to understand what they are and how to find them.

A prime number is a number that is only divisible by two numbers—itself and 1.

For example, 11 is a prime number because \$1 * 11\$ is its only factor. (11 is not evenly divisible by 2, 3, 4, 5, 6, 7, 8, 9, or 10).

12 is NOT a prime number, because its factors are 1, 2, 3, 4, 6, and 12. It has more factors than just itself and 1.

1 is NOT a prime number, because its only factor is 1.

The only even prime number is 2. Questions about primes come up fairly often on the SAT and understanding that 2 (and only 2!) is a prime number will be invaluable for solving many of these.

A prime number \$x\$ is squared and then added to a different prime number, \$y\$. Which of the following could be the final result?

1. An even number
2. An odd number
3. A positive number

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Now this question relies on your knowledge of both number relationships and primes. You know that any number squared (the number times itself) will be an even number if the original number was even, and an odd number if the original number was odd. Why? Because an even * an even = an even, and an odd * an odd = an odd (\$6 * 6 = 36\$ & \$7 * 7 = 49\$).

Next, we are adding that square to another prime number. You’ll also remember that an even number + an odd number is odd, an odd number + an odd number is even, and an even number + an even number is even.

Knowing that 2 is a prime number, let’s replace x with 2. \$2^2 = 4\$. Now if y is a different prime number (as stipulated in the question), it must be odd, because the only even prime number is 2. So let’s say \$y = 3\$.

\$4 + 3 = 7\$. So the end result is odd.

This means II is correct.

But what if both x and y were odd prime numbers? So let’s say that \$x = 3\$ and \$y = 5\$.

So \$3^2 = 9\$.

\$9 + 5 = 14\$. So the end result is even.

This means I is correct.

Now, for option number III, our results show that it is possible to get a positive number result, since both our results were positive.

This means the final answer is E, I, II, and III

If you forgot that 2 was a prime number, you would have picked D, I and III only, because there would have been no possible way to get an odd number. Remembering that 2 is a prime number is the key to solving this question.

Another typical prime number question on the SAT will ask you to identify how many prime numbers fall in a certain range of numbers.

How many prime numbers are between 30 and 50, inclusive?

A. Two
B. Three
C. Four
D. Five
E. Six

This might seem intimidating or time-consuming, but I promise you do NOT need to memorize a list of prime numbers.

First, eliminate all even numbers from the list, as you know the only even prime number is 2.

Next, eliminate all numbers that end in 5. Any number that ends is 5 or 0 is divisible by 5.

Now your list looks like this:

31, 33, 37, 39, 41, 43, 47, 49

This is much easier to work with, but we need to narrow it down further.

(You could start using your calculator here, or you can do this by hand.) A way to see if a number is divisible by 3 is to add the digits together. If that number is 3 or divisible by 3, then the final result is divisible by 3.

For example, the number 31 is NOT divisible by 3 because \$3 + 1 = 4\$, which is not divisible by 3. However 33 is divisible by 3 because \$3 + 3 = 6\$, which is divisible by 3.

So we can now eliminate 33 (\$3 + 3 = 6\$) and 39 (\$3 + 9 = 12\$) from the list.

We are left with 31, 37, 41, 43, 47, 49.

Now, to make sure you try every necessary potential factor, take the square root of the number you are trying to determine is prime. Any integer equal to or less than the square root will be a potential factor, but you do not have to try any numbers higher.

Why? Well let’s take 36 as an example. Its factors are:

1, 2, 3, 4, 6, 9, 12, 18, and 36. But now look at the factor pairings.

1 & 36

2 & 18

3 & 12

4 & 9

6 & 6

(9 & 4)

(12 & 3)

(18 & 2)

(36 & 1)

After you get past 6, the numbers repeat. If you test out 4, you will know that 9 goes evenly into your larger number—no need to actually test 9 just to get 4 again!

So all numbers less than or equal to a potential prime’s square root are the only potential factors you need to test.

Going back to our list, we have 31, 37, 41, 43, 47, 49.

Well the closest square root to 31 and 37 is 6. We already know that neither 2 nor 3 nor 5 factor evenly into 31 and 37. Neither do 4, or 6. You’re done. Both 31 and 37 must be prime.

As for 41, 43, 47, and 49, the closest square root of these is 7. We already know that neither 2 nor 3 nor 5 factor evenly into 41, 43, 47, or 49.

7 is the exact square root of 49, so we know 49 is NOT a prime.

As for 41, 43, and 47, neither 4 nor 6 nor 7 go into them evenly, so they are all prime.

You are left with 31, 37, 41, 43, and 47.

So your answer is D, there are five prime numbers (31, 37, 41, 43, and 47) between 30 and 50.

Prime numbers, Prime Directive, either way I'm sure we'll live long and prosper.

## Absolute Values

Absolute values are a concept that the SAT loves to use, as it is all too easy for students to make mistakes with absolute values. Expect to see one question on absolute values per test (though very rarely more than one).

An absolute value is a representation of distance along a number line, forward or backwards. This means that an absolute value equation will always have two solutions.

It also means that whatever is in the absolute value sign will be positive, as it represents distance along a number line and there is no such thing as a negative distance.

An equation \$|x + 3| = 14\$, has two solutions:

\$x = 11\$

\$x = -17\$

Why -17? Well \$-17 + 3 = -14\$ and, because it is an absolute value (and therefore a distance), the final answer becomes positive. So \$|-14| = 14\$

When you are presented with an absolute value, instead of doing the math in your head to find the negative and positive solution, rewrite the equation into two different equations.

When presented with the above equation \$|x + 3| = 14\$, take away the absolute value sign and transform it into two equations—one with a positive solution and one with a negative solution.

So \$|x + 3| = 14\$ becomes:

\$x + 3 = 14\$

AND

\$x + 3 = -14\$

Solve for \$x\$

\$x = 11\$ and \$x = -17\$

\$|10 - k| = 3\$

\$|k - 5| = 8\$.

What is a value for \$k\$ that fulfills both equations above?

We know that any given absolute value expression will have two solutions, so we must find the solution that each of these equations shares in common.

For our first absolute value equation, we are trying to find the numbers for \$k\$ that, when subtracted from 10 will give us 3 and -3.

That means our \$k\$ values will be 7 and 13. Why? Because \$10 - 7 = 3\$ and \$10 - 13 = -3\$

Now let's look at our second equation. We know that the two numbers for \$k\$ for \$k - 5\$ must give us both 8 and -8.

This means our \$k\$ values will be 13 and -3. Why? Because \$13 - 5 = 8\$ and \$-3 - 5 = -8\$.

13 shows up as a solution for both problems, which means it is our answer.

So our final answer is 13, this is the number for \$k\$ that can solve both equations.

## Consecutive Numbers

Questions about consecutive numbers may or may not show up on your SAT. If they appear, it will be for a maximum of one question. Regardless, they are still an important concept for you to understand.

Consecutive numbers are numbers that go continuously along the number line with a set distance between each number.

So an example of positive, consecutive numbers would be: 4, 5, 6, 7, 8

An example of negative, consecutive numbers would be: -8, -7, -6, -5, -4

(Notice how the negative integers go from greatest to least—if you remember the basic guide to integers, this is because of how they lie on the number line in relation to 0)

You can write unknown consecutive numbers out algebraically by assigning the first in the series a variable, \$x\$, and then continuing the sequence of adding 1 to each additional number.

The sum of four positive, consecutive integers is 54. What is the first of these integers?

If x is our first, unknown, integer in the sequence, so you can write all four numbers as:

\$x + (x + 1) + (x + 2) + (x + 3) = 54\$

\$4x + 6 = 54\$

\$4x = 48\$

\$x = 12\$

So, because x is our first number in the sequence and \$x= 12\$, the first number in our sequence is 12.

You may also be asked to find consecutive even or consecutive odd integers. This is the same as consecutive integers, only they are going up every other number instead of every number. This means there is a difference of two units between each number in the sequence instead of 1.

An example of positive, consecutive even integers: 8, 10, 12, 14, 16

An example of positive, consecutive odd integers: 15, 17, 19, 21, 23

Both consecutive even or consecutive odd integers can be written out in sequence as:

\$x, x + 2, x + 4, x + 6\$, etc. No matter if the beginning number is even or odd, the numbers in the sequence will always be two units apart.

What is the median number in the sequence of five positive, consecutive odd integers whose sum is 185?

\$x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 185\$

\$5x + 20 = 185\$

\$5x = 165\$

\$x = 33\$

So the first number in the sequence is 33. This means the full sequence is:

33, 35, 37, 39, 41

The median number in the sequence is 37.

Bonus history lesson—Grover Cleveland is the only US president to have ever served two non-consecutive terms.

## Steps to Solving an SAT Integer Question

Because SAT integer questions are so numerous and varied, there is no set way to approach them that is entirely separate from approaching other kinds of SAT math questions. But there are a few techniques that will help you approach your SAT integer questions (and by extension, most questions on SAT math).

#1: Make sure the question requires an integer.

If the question does NOT specify that you are looking for an integer, then any number—including decimals and fractions—are fair game. Always read the question carefully to make sure you are on the right track.

#2: Use real numbers if you forget your integer rules.

Forget whether positive, even consecutive integers should be written as \$x + (x + 1)\$ or \$x + (x + 2)\$? Test it out with real numbers!

14, 16, 18 are consecutive even integers. If \$x = 14\$, \$16 = x + 2\$, and \$18 = x + 4\$.

This works for most all of your integer rules. Forget your exponent rules? Plug in real numbers! Forget whether an even * an even makes an even or an odd? Plug in real numbers!

#3: Keep your work organized.

Like with most SAT math questions, integer questions can seem more complex than they are, or will be presented to you in strange ways. Keep your work well organized and keep track of your values to make sure your answer is exactly what the question is asking for.

Santa is magic and has to double-check his list. So make sure you double-check your work too!

## Test Your Knowledge

1. If \$a^x * a^6 = a^24\$ and \$(a^3)^y = a^15\$, what is the value of \$x + y\$?

A. 9
B. 12
C. 23
D. 30
E. 36

2. If \$48√48 = a√b\$ where \$a\$ and \$b\$ are positive integers and \$a > b\$, which of the following could be a value of \$ab\$?

A. 48
B. 96
C. 192
D. 576
E. 768

3. What is the product of the smallest prime number that is greater than 50 and the greatest prime number that is less than 50?

4. If \$j, k\$, and \$n\$ are consecutive integers such that \$0<j<k<n\$ and the units (ones) digit of the product \$jn\$ is 9, what is the units digit of \$k\$?

A. 0
B. 1
C. 2
D. 3
E. 4

Answers: C, D, 2491, A

1. In this question, we are being asked both to multiply bases with exponents as well as take a base with an exponent to another exponent. Essentially, the question is testing us on whether or not we know our exponent rules.

If we remember our exponent rules, then we know that we must add exponents when we are multiplying two of the same base together.

So \$a^x * a^6 = a^24\$ => \$a^{x + 6} = a^24\$

\$x + 6 = 24\$

\$x = 18\$

We have our value for \$x\$. Now we must find our \$y\$.

We also know that, when taking a base and exponent to another exponent, we must multiply the exponents.

So \$(a^3)^y = a^15\$ => \$a^{3 * y} = a^15\$

\$3 * y = 15\$

\$y = 5\$

In the final step, we must add our \$x\$ and \$y\$ values together:

\$18 + 5 = 23\$

So our final answer is C, 23.

2. We are starting with \$48√48\$ and we know we must reduce it. Why? Because we are told that our first \$48 = a\$ and our second \$48 = b\$ AND that \$a > b\$. Right now our \$a\$ and \$b\$ are equal, but, by reducing the expression, we will be able to find an \$a\$ value that is greater than our \$b\$

So let's find all the factors of 48 to see if there are any perfect squares.

48

\$1 * 48\$

\$2 * 24\$

\$3 * 16\$

\$4 * 12\$

\$6 * 8\$

Two of these pairings have perfect squares. 16 is our largest perfect square, which means that it will be the number we must use to reduce \$48√48\$ down to its most reduced form. Though we are not explicitly asked to find the most reduced form of \$48√48\$, we can start there for now.

So \$48√48 = 48 * √16 * √3\$

\$48 * 4 *√3\$

\$192√3\$

This means that our \$a = 192\$ and our \$b = 3\$, then:

\$ab = 192 * 3 = 576\$

So our final answer is D, 576.

(Special note: you'll notice how we are told to find one possible value for \$ab\$, not necessarily \$ab\$ when \$48√48\$ is at its most reduced. So if our above answer hadn't matched one of our answer options, we would have had to reduce \$48√48\$ only part way.

\$48√48 = 48 * √4 * √12\$

\$48 * 2 * √12\$

\$96√12\$

This would make our \$a = 96\$ and our \$b = 12\$, meaning that our final answer for \$ab\$ would be \$96 * 12 = 1152\$.)

3. This question requires us to be able to figure out which numbers are prime. Let us use the same methods we used during the above section on prime numbers.

All prime numbers other than 2 will be odd and there is no prime number that ends in 5. So let's list the odd numbers (excluding ones that end in 5's) above and below 50.

41, 43, 47, 49, 51, 53, 57, 59

We are trying to find the ones closest to 50 on either side, so let's first test the highest number in the 40's.

49 is the perfect square of 7, which means it is divisible by more than just itself and 1. We can cross 49 off the list.

47 is not divisible by 3 because \$7 + 4 = 11\$ and 11 is not divisible by 3. It is also not divisible by any even number (because an even * an even = an even), by 5, or by 7. This means it must be prime.

(Why did we stop here? Remember that we only have to test potential factors up until the closest square root of the potential prime. \$√47\$ is between \$6^2 = 36\$ and \$7^2 = 49\$, so we tested 7 just to be safe. Once we saw that 7 could not go into 47, we proved that 47 is a prime.)

47 is our largest prime less than 50.

Now let's test the smallest number greater than 50.

51 is odd, but \$5 + 1 = 6\$, which is divisible by 3. That means that 51 is also divisible by 3 and thus cannot be prime.

53 is not divisible by 3 because \$5 + 3 = 8\$, which is not divisible by 3. It is also not divisible by 5 or 7. Therefore it is prime. (Again, we stopped here because the closest square root to 53 is between 7 and 8. And 8 cannot be a prime factor because all of its multiples are even).

This means our smallest prime less than 50 is 47 and our largest is 53. Now we just need to find the product of those two numbers.

\$47 * 53 = 2491\$

Our final answer is 2491.

4. We are told that \$j\$, \$k\$, and \$n\$ are consecutive integers. We also know they are positive (because they are greater than 0) and that they go in ascending order, \$j\$ to \$k\$ to \$n\$.

We are also told that \$jn\$ equals a number with a units digit of 9. So let's find all the ways to get a product of 9 with two numbers.

\$1 * 9\$

\$3 * 3\$

The only way to get any number that ends in 9 (units digit 9) from the product of two numbers is in one of two ways:

#1: Both the original numbers have a units digit of 3

#2: The two original numbers have units digits of 1 and 9, respectively.

Now let's visualize positive consecutive integers. Positive consecutive integers must go up in order with a difference of 1 between each variable. So \$j, k, n\$ could look like any collection of three numbers along a consistent number line.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, etc.

There is no possible way that the units digits of the first and last of three consecutive numbers could both be 3. Why? Because if one had a units digit of 3, the other would have to end in either 1 or 5. Take 13 as an example. If \$j\$ were 13, then \$n\$ would have to be 15. And if \$n\$ were 13, then \$j\$ would have to be 11.

So we know that neither \$j\$ nor \$n\$ has a units digit of 3.

Now let's see if there is a way that we can give \$j\$ and \$n\$ units digits of 1 and 9 (or 9 and 1).

If \$j\$ were given a units digit of 1, \$n\$ would have a units digit of 3. Why? Picture \$j\$ as 11. \$n\$ would have to be 13, and \$11 * 13 = 143\$, which means the units digit of their product is not 9.

But what if \$n\$ was a number with a units digit of 1? \$j\$ would have a units digit of 9. Why? Picture \$n\$ as 11 now. \$j\$ would be 9.

\$9 * 11 = 99\$. The units digit is 9.

And if the last digit of \$j\$ is 9 and the numbers \$j, k, \and n\$ are consecutive, then \$k\$ has to end in 0.

So our final answer is A, 0.

## The Take-Aways

Integers and integer questions can be tricky for some students, as they often involve concepts not tested in high school level math classes (when’s the last time you dealt with integer remainders, for example?). But most integer questions are much simpler than they appear.

If you know your definitions—integers, consecutive integers, absolute values, etc.—and you know how to pay attention to what the question is asking you to find, you’ll be able to solve most any integer question that comes your way.

## What’s Next?

Whew! You’ve done your paces on integers, both basic and advanced. Now that you’ve tackled these foundational topics of the SAT math, make sure you’ve got a solid grasp of all the math topics covered by the SAT math section, so that you can take on the SAT with confidence.

Find yourself running out of time on SAT math? Check out our article on how to buy yourself time and complete your SAT math problems before time’s up.

Feeling overwhelmed? Start by figuring out your ideal score and check out how to improve a low SAT math score.

Already have pretty good scores and looking to get a perfect 800 on SAT Math? Check out our article on how to get a perfect score written by a full SAT scorer.