Consecutive Integers Calculator

Created by Maciej Kowalski, PhD candidate
Reviewed by Steven Wooding
Last updated: Oct 14, 2022

Welcome to Omni's consecutive integers calculator, where we'll tackle math problems concerning consecutive numbers. For example, we sometimes know the sum of three consecutive integers and would like to find the individual values themselves. We'll also cover two common variants of such exercises, i.e., where we limit ourselves to consecutive odd integers or consecutive even integers.

But before we get to such sophisticated questions, let's answer the fundamental one: what are consecutive integers?

What are consecutive integers?

By definition, integers are the numbers that we can represent without a fractional part. Equivalently, they consist of all positive whole numbers (i.e., 11, 22, 33, and so on), their opposites (i.e., βˆ’1-1, βˆ’2-2, βˆ’3-3, and so on), and zero. However, note that the first sentence means that, say, 4/24/2 is also an integer since we can represent it as a whole number.

The meaning of consecutive is "following each other continuously." We know that we can put all real numbers in order on what we call the number line. Therefore, we can indeed say that one value "follows" another, by which we usually mean that it's larger. However, there are infinitely many numbers between, say, 33 and 44: there's the mixed number 3343\tfrac{3}{4}, the decimal 3.5193.519, or the Ο€\pi from circle calculations. So, what are consecutive numbers?

In fact, what this term usually means is "consecutive integers." True, the simplification is misleading, and mathematicians might even call it conceptually wrong. Nevertheless, we need to embrace it.

So, what are consecutive integers? Using the two definitions above, they are whole numbers following each other. For instance, we know that among integers, 44 comes after 33, and there's nothing else (i.e., no integer) in between. Farther consecutive integers are 55, 66, 77, 88, and so on, while going the other way we have ..., βˆ’1-1, 00, 11, and 22 (note how we put dots before βˆ’1-1 since in this case, the list goes on forever in that direction).

Note how for an arbitrary integer xx, its consecutive numbers take the form x+1x + 1, x+2x + 2, x+3x + 3, and so on. This simple observation will prove crucial in the calculations to come.

The meaning of consecutive numbers seems like a simple concept, don't you think? So let's spice it up a bit and add parity.

Consecutive odd/even integers

We say that an integer (positive or negative) is even if it's divisible by 22 (i.e., equals 00 modulo 22). Otherwise, we say that it's odd. These terms describe what we call the parity of the integer.

πŸ™‹ Parity is an important concept in mathematics as in computer science. Discover with Omni how even and odd matter in your computer with our parity bit calculator!

Sometimes, real-life scenarios (by which we mean math textbooks) want us to only deal with consecutive even integers or consecutive odd integers. Omni's consecutive integers calculator decided to address such life-like problems (i.e., help you get a good grade) and have the option to choose whether you need the values to be even, odd, or whichever.

In calculations, we will need to describe such integers algebraically. In the above section, we used xx, x+1x+1, x+2x + 2, and so on. Observe that we can't just copy that here. Indeed, for x=5x = 5, the three expressions give 55, 66, and 77, while for x=6x = 6 they are 66, 77, and 88. Neither triple lists consecutive even integers nor consecutive odd integers.

At the beginning of this section, we said that even integers are divisible by 22. That means we can represent them in the form 2β‹…x2 \cdot x with xx some integer. Indeed, 4=2β‹…24 = 2 \cdot 2, 18=2β‹…918 = 2 \cdot 9, βˆ’22=2β‹…(βˆ’11)-22 = 2 \cdot(-11), etc.

On the other hand, odd numbers are not divisible by 22, or, equivalently, they give the remainder 11 when divided by 22. Therefore, they differ by 11 from some adjoining even number, so we can represent them in the form 2β‹…x+12 \cdot x + 1. Indeed, 7=2β‹…3+17 = 2 \cdot 3 + 1, 17=2β‹…8+117 = 2\cdot 8 + 1, βˆ’39=2β‹…(βˆ’20)+1-39 = 2 \cdot (-20) + 1, etc.

Remainder, quotient, modulus... What are those, again? Omni's quotient calculator, remainder calculator, and modulo calculator can help you untangle those terms!

Furthermore, observe that consecutive even integers and consecutive odd integers differ by 22. Indeed, examples of such are βˆ’2-2, 00, 22, 44 and βˆ’5-5, βˆ’3-3, βˆ’1-1, 11, respectively. Therefore, if some real-life scenario forces us to write such sequences algebraically, we can use:

  • 2x2x, 2x+22x + 2, 2x+42x + 4, 2x+62x + 6, and so on for consecutive even integers; and
  • 2x+12x + 1, 2x+32x + 3, 2x+52x + 5, 2x+72x + 7, and so on for consecutive odd integers.

And now it's time to use the above representation and apply it to a couple of math problems with consecutive numbers.

The sum/product of consecutive numbers

There are two math problems typical to consecutive integers. Below, we describe them together with some formulas that may help in finding the solution. Note that each can come in the consecutive even/odd integers variant.

  1. The sum of nn consecutive integers is equal to SS. Find these integers.

    Luckily, this one's quite simple, no matter if we're dealing with the sum of three consecutive integers or twenty.

    In essence, we want to find consecutive numbers a1a_1, a2a_2, a3a_3, ..., ana_n such that a1+a2+a3+...+an=Sa_1 + a_2 + a_3 + ... + a_n = S. From the first section, we know a clever way to write these numbers. To be precise, if we denote the smallest by xx, then the next ones will be x+1x + 1, x+2x + 2, x+3x + 3, up to x+(nβˆ’1)x + (n-1). That gives:

x+(x+1)+(x+2)+(x+3)+...+(x+(nβˆ’1))=Sx + (x + 1) + (x + 2) + (x + 3) + ... + (x + (n-1)) = S

If we now use commutativity of addition and simplify, we get:

nβ‹…x+(1+2+3+...+(nβˆ’1))=Sn\cdot x + (1 + 2 + 3 + ... + (n-1)) = S

Next, we observe that the summands in brackets form an arithmetic sequence. Therefore, we can use the formula for the sum of such a sequence and obtain:

nβ‹…x+nβ‹…(nβˆ’1)2=Sn \cdot x + \frac{n\cdot (n - 1)}{2} = S

Lastly, we move the constant term to the right, divide both sides by nn, and get:

x=Sβˆ’nβ‹…(nβˆ’1)2nx = \frac{S - \frac{n \cdot (n - 1)}{2}}{n}

which lets us calculate all the integers in question. However, remember that not all values of SS give an answer. In some cases, the xx at the end may turn out to be a fractional expression. That would mean no integers satisfy our conditions.

  1. The product of nn consecutive integers is equal to PP. Find these integers.

    This time, we're looking for consecutive numbers a1a_1, a2a_2, a3a_3, ..., ana_n such that a1β‹…a2β‹…a3β‹…...β‹…an=Pa_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n = P. Again, we can denote the smallest of these integers by xx and use the notation from the first section to get

xβ‹…(x+1)β‹…(x+2)β‹…(x+3)β‹…...β‹…(x+(nβˆ’1))=Px \cdot (x + 1) \cdot (x + 2) \cdot (x + 3) \cdot ... \cdot (x + (n-1)) = P

And here's when it becomes tricky. If we rewrite the above equation by using the distributive property of multiplication over addition, we'll get a complicated polynomial of order nn. Such things are far from simple, and there's no algorithmic way to solve high-order polynomial equations.

If n=2n = 2, i.e., we have a product of two consecutive, then the above equation looks like this:

xβ‹…(x+1)=Px\cdot(x+1) = P

or, equivalently:

x2+xβˆ’P=0x^2+x-P=0

We got a simple quadratic equation, so we can quickly solve it using the discriminant formula. Recall that such equations often have two solutions. In our case, this means that two pairs of consecutive integers satisfy our assumptions.

πŸ™‹ Use Omni's discriminant calculator and the quadratic formula calculator to find the solutions to a quadratic equation hassle free!

However, if n=3n = 3, then things get more complicated. We get a cubic equation, and although there are still algorithms to solve them, it's not as easy as before. And as we go higher with nn, things get even more troublesome.

The bright side is that we're only interested in integer solutions. In essence, this means that we can try to find the answer by trial-and-error, i.e., check all reasonable integers and see if any of them fit. Omni's consecutive integers calculator does precisely that.

Alright, that seems like enough technicalities for now. We move on to examples that have actual numbers in them, and we'll take the opportunity to explain how to use the consecutive integers calculator.

Example of using the consecutive integers calculator

Suppose that we know that the sum of three consecutive integers is equal to 4242. Let's find those integers, but before we do this ourselves, let's see how Omni's consecutive integers calculator can do the job for us.

We should fill the variable fields in the tool so that they form a sentence describing our problem. First of all, since we have the sum of three consecutive integers, we choose "the sum" from the top list. Next, we're looking for three numbers, so we input 33 in the box below. Then, we want the sum to equal 4242, so we write 4242 in the third line. Lastly, we choose "any" from the bottom list since we don't limit ourselves to only consecutive even or odd integers.

All in all, the consecutive integers calculator's fields should look like so (line by line):

  • I want the sum;

  • Of 33 consecutive integers;

  • To equal 4242;

  • And I allow any integers.

Once we input all the data, the consecutive integers calculator will spit out the answer underneath. But before we reveal it, let's try to find it ourselves.

We begin by translating the problem into an algebraic expression. If we denote the smallest of the integers by xx, then by following instructions from the first section, we can write the task as:

x+(x+1)+(x+2)=42x + (x + 1) + (x + 2) = 42

which, after simplification, gives:

3x+3=423x + 3 = 42

Now we can solve the linear equation the usual way:

3x = 39

x = 13,

3x=39x=13\begin{split} 3x&=39\\ x&=13 \end{split}

which means that our three integers are 1313, 1414, and 1515, and indeed, 13+14+15=4213 + 14 + 15 = 42.

For comparison, let's now try to find two consecutive odd integers whose product equals 6363. Again, we begin by how we can make the consecutive integers calculator do the dirty work.

This time, we're dealing with multiplication, so we choose "the product" from the top list. We have only two integers, so we input 2 into the second line, and we want the product to be 63, so we write that number in the next box. Lastly, we choose "only odd" from the bottom list since we're interested only in such integers.

All in all, the consecutive integers calculator's fields should look the following:

  • I want the product

  • of 2 consecutive integers

  • to equal 63

  • and I allow only odd integers.

Again, once we input all the data, the consecutive integers calculator will spit out the answer underneath. And again, we don't reveal it just yet!

To find the solution ourselves, recall from the second section how we can represent consecutive odd integers algebraically. To be precise, we denote the smaller one by 2x+12x + 1 for xx some integer, so our problem translates to:

(2x+1)β‹…(2x+3)=63(2x + 1) \cdot (2x + 3) = 63

Or:

4x2+8xβˆ’60=04x^2 + 8x - 60 = 0

Simple calculations reveal that the quadratic equation has two solutions: βˆ’5-5 and 33$$. Therefore, we have two pairs of consecutive odd integers that satisfy our assumptions: βˆ’9-9 and βˆ’7-7, and 77 and 99. Indeed, (βˆ’9)β‹…(βˆ’7)=7β‹…9=63(-9)\cdot (-7) = 7 \cdot 9 = 63.

VoilΓ ! Simple yet satisfying calculations, don't you think? Still, as we've mentioned, the formulas may get a bit messy if we decide to multiply too many binomials. Luckily, there's always Omni Calculator to save us the hassle!

FAQ

How do I find consecutive integers?

To find consecutive integers, you need to:

  1. Specify what you need: any consecutive integers or only even/odd ones.
  2. Denote the smallest of them by:
    • x if you allow any integers;
    • 2x if you want only even integers; or
    • 2x + 1 if you want only odd integers.
  3. Write the next integers as:
    • x + 1, x + 2, x + 3, and so on for any integers;
    • 2x + 2, 2x + 4, 2x + 6, and so on for only even integers; or
    • 2x + 3, 2x + 5, 2x + 7, and so on for only odd integers.
  4. If needed, use the representation to describe the integers' properties.
  5. Use the algebraic description to find the integers.
  6. Enjoy your consecutive integers.

How do I solve consecutive integer problems?

To solve consecutive integer problems, you need to:

  1. Specify what you need: any consecutive integers or only even/odd ones.
  2. Denote the smallest of them by:
    • x if you allow any integers;
    • 2x if you want only even integers; or
    • 2x + 1 if you want only odd integers.
  3. Write the next integers as:
    • x + 1, x + 2, x + 3, and so on for any integers;
    • 2x + 2, 2x + 4, 2x + 6, and so on for only even integers; or
    • 2x + 3, 2x + 5, 2x + 7, and so on for only odd integers.
  4. Use the representations to write equations, e.g., for the sum or product.
  5. Solve the equation in the usual way.
  6. Substitute the solution for x to find the integers.
  7. Enjoy the solution to your consecutive integer problem.

How do I find consecutive odd integers?

To find consecutive odd integers, you need to:

  1. Denote the smallest of them by 2x + 1.
  2. Write the next integers as 2x + 1, 2x + 3, 2x + 5, and so on.
  3. If needed, use the representation to describe the integers' properties.
  4. Use the algebraic description to find the integers.
  5. Enjoy your consecutive odd integers.

How do I find three consecutive integers?

To find three consecutive integers, you need to:

  1. Specify what you need: any consecutive integers or only even/odd ones.
  2. Denote the smallest of them by:
    • x if you allow any integers;
    • 2x if you want only even integers; or
    • 2x + 1 if you want only odd integers.
  3. Write the next integers as:
    • x + 1 and x + 2 for any integers;
    • 2x + 2 and 2x + 4 for only even integers; or
    • 2x + 3 and 2x + 5 for only odd integers.
  4. If needed, use the representation to describe the integers' properties.
  5. Use the algebraic description to find the integers.
  6. Enjoy your three consecutive integers.

How do I find two consecutive integers?

To find two consecutive integers, you need to:

  1. Specify what you need: any consecutive integers or only even/odd ones.
  2. Denote the smallest of them by:
    • x if you allow any integers;
    • 2x if you want only even integers; or
    • 2x + 1 if you want only odd integers.
  3. Write the next integer as:
    • x + 1 for any integers;
    • 2x + 2 for only even integers; or
    • 2x + 3 for only odd integers.
  4. If needed, use the representation to describe the integers' properties.
  5. Use the algebraic description to find the integers.
  6. Enjoy your two consecutive integers.

How do I find two consecutive odd integers?

To find two consecutive odd integers, you need to:

  1. Denote the smallest of them by 2x + 1.
  2. Write the next integer as 2x + 3.
  3. If needed, use the representation to describe the integers' properties.
  4. Use the algebraic description to find the integers.
  5. Enjoy your two consecutive odd integers.
Maciej Kowalski, PhD candidate
Data
I want
the sum
of
consecutive integers
to equal
and I allow
any
integers
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