how to solve problems with division

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How to Do Division

Last Updated: January 14, 2024 Fact Checked

This article was co-authored by Grace Imson, MA and by wikiHow staff writer, Christopher M. Osborne, PhD . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 522,468 times.

Division is one of the 4 major operations in arithmetic, alongside addition, subtraction, and multiplication. In addition to whole numbers, you can divide decimals, fractions, or exponents. You can do long division or, if one of the numbers is a single digit, short division. Start by mastering long division, though, because it is the key to the entire operation.

Long Division

• Sample problem #1 (beginner): 65 ÷ 5 . Place the 5 outside the division bar, and the 65 inside it. It should look like 5厂65 , but with the 65 underneath the horizontal line.
• Sample problem #2 (intermediate): 136 ÷ 3 . Place the 3 outside the division bar, and the 136 inside it. It should look like 3厂136 , but with the 136 underneath the horizontal line.

• In sample problem #1 ( 5厂65 ), 5 is the divisor and 6 is the first digit of the dividend (65). 5 goes into 6 one time, so place a 1 on the top of the divisor bar, aligned above the 6.
• In sample problem #2 ( 3厂136 ), 3 (the divisor) does not go into 1 (the first digit of the dividend) and result in a whole number. In this case, write a 0 above the division bar, aligned above the 1.

• In sample problem #1 ( 5厂65 ), multiply the number above the bar (1) by the divisor (5), which results in 1 x 5 = 5 , and place the answer (5) just below the 6 in 65.
• In sample problem #2 ( 3厂136 ), there is a zero above the division bar, so when you multiply this by 3 (the divisor), your result is zero. Write a zero on a new line just below the 1 in 136.

• In sample problem #1 ( 5厂65 ), subtract the 5 (the multiplication result in the new row) from the 6 right above it (the first digit of the dividend): 6 - 5 = 1 . Place the result (1) in another new row right below the 5.
• In sample problem #2 ( 3厂136 ), subtract 0 (the multiplication result in the new row) from the 1 right above it (the first digit in the dividend). Place the result (1) in another new row right below the 0.

• In sample problem #1 ( 5厂65 ), drop the 5 from 65 down so that it’s beside the 1 that you got from subtracting 5 from 6. This gives you 15 in this row.
• In sample problem #2 ( 3厂136 ), carry down the 3 from 136 and place it beside the 1, giving you 13.

• To continue 5厂65 , divide 5 (the dividend) into the new number (15), and write the result (3, since 15 ÷ 5 = 3 ) to the right of the 1 above the division bar. Then, multiply this 3 above the bar by 5 (the dividend) and write the result (15, since 3 x 5 = 15 ) below the 15 under the division bar. Finally, subtract 15 from 15 and write 0 in a new bottom row.
• Sample problem #1 is now complete, since there are no more digits in the divisor to carry down. Your answer (13) is above the division bar.

• For 3厂136 : Determine how many times 3 goes into 13, and write the answer (4) to the right of the 0 above the division bar. Then, multiply 4 by 3 and write the answer (12) below the 13. Finally, subtract 12 from 13 and write the answer (1) below the 12.

• For 3厂136 : Continue the process for another round. Drop down the 6 from 136, making 16 in the bottom row. Divide 3 into 16, and write the result (5) above the division line. Multiply 5 by 3, and write the result (15) in a new bottom row. Subtract 15 from 16, and write the result (1) in a new bottom row.
• Because there are no more digits to carry down in the dividend, you’re done with the problem and the 1 on the bottom line is the remainder (the amount left over). Write it above the division bar with an “r.” in front of it, so that your final answer reads “45 r.1”.

Short Division

• In order to do short division , your divisor can't have more than one digit.
• Sample problem: 518 ÷ 4 . In this case, the 4 will be outside the division bar, and the 518 inside it.

• In the sample problem, 4 (the divisor) goes into 5 (the first digit of the dividend) 1 time, with a remainder of 1 ( 5 ÷ 4 = 1 r.1 ). Place the quotient, 1, above the long division bar. Place a small, superscript 1 beside the 5, to remind yourself that you had a remainder of 1.
• The 518 under the bar should now look like this: 5 1 18.

• In the sample problem, the number formed by the remainder and the second number of the dividend is 11. The divisor, 4, goes into 11 twice, leaving a remainder of 3 ( 11 ÷ 4 = 2 r.3 ). Write the 2 above the division line (giving you 12) and the 3 as a superscript number beside the 1 in 518.
• The original dividend, 518, should now look like this: 5 1 1 3 8.

• In the sample problem, the next (and final) dividend number is 38—the remainder 3 from the previous step, and the number 8 as the last term of the dividend. The divisor, 4, goes into 38 nine times with a remainder of 2 ( 38 ÷ 4 = 9 r.2 ), because 4 x 9 = 36 , which is 2 short of 38. Write this final remainder (2) above the division bar to complete your answer.
• Therefore, your final answer above the division bar is 129 r.2.

Dividing Fractions

• Your problem might be, for example, 3/4 ÷ 5/8 . For convenience, use horizontal instead of diagonal lines to separate the numerator (top number) and denominator (bottom number) of each fraction.

• In the sample problem, reverse 5/8 so the 8 is on top and the 5 is on the bottom.

• For example: 3/4 x 8/5 .

• In this case, the numerators are 3 and 8, and 3 x 8 = 24 .

• The denominators are 4 and 5 in the sample problem, and 4 x 5 = 20 .

• In the sample problem, then, 3/4 x 8/5 = 24/20 .

• 24: 1, 2, 3, 4 , 6, 8, 12, 24
• 20: 1, 2, 4 , 5, 10, 20
• 24/20 = 6/5 . Therefore, 3/4 ÷ 5/8 = 6/5

• In the sample problem, 5 goes into 6 one time with a remainder of 1. Therefore, the new whole number is 1, the new numerator is 1, and the denominator remains 5.
• As a result, 6/5 = 1 1/5 .

Dividing Exponents

• As a beginner, start with a sample problem in which both numbers with exponents already have the same base—for instance, 3 8 ÷ 3 5 .

• In the sample problem: 8 - 5 = 3 .

• Therefore: 3 8 ÷ 3 5 = 3 3 .

Dividing Decimals

• For the example 65.5 ÷ 0.5 , 0.5 goes outside the division bar, and 65.5 goes inside it.

• In the sample problem, you only need to move the decimal point over one spot for both the divisor and dividend. So, 0.5 becomes 5, and 65.5 becomes 655.
• If, however, the sample problem used 0.5 and 65.55, you’d need to move the decimal point 2 places in 65.55, making it 6555. As a result, you’d also have to move the decimal point in 0.5 2 places. To do this, you’d add a zero to the end and make it 50.

• In the sample problem, the decimal in 655 would appear after the last 5 (as 655.0). So, write the decimal point above the division line right above where that decimal point in 655 would appear.

• Divide 5 into the hundredths digit, 6. You get 1 with a remainder of 1. Place 1 in the hundredths place on top of the long division bar, and subtract 5 from 6 below the number six.
• Your remainder, 1, is left over. Carry the first five in 655 down to create the number 15. Divide 5 into 15 to get 3. Place the three above the long division bar, next to the 1.
• Carry down the last 5. Divide 5 into 5 to get 1, and place the 1 on top of the long division bar. There is no remainder, since 5 goes into 5 evenly.
• The answer is the number above the long division bar (131), so 655 ÷ 5 = 131 . If you pull out a calculator, you’ll see that this is also the answer to the original division problem, 65.5 ÷ 0.5 .

Practice Problems and Answers

Community Q&A

You Might Also Like

• ↑ https://www.mathsisfun.com/long_division.html
• ↑ https://www.k5learning.com/blog/step-step-guide-long-division
• ↑ https://www.bbc.co.uk/bitesize/topics/znmtsbk/articles/zqpddp3
• ↑ http://www.mathsisfun.com/fractions_division.html
• ↑ https://www.ck12.org/arithmetic/divide-fractions/lesson/Quotients-of-Fractions-MSM6/?referrer=concept_details
• ↑ http://www.mathsisfun.com/algebra/variables-exponents-multiply.html
• ↑ https://www.mathsisfun.com/dividing-decimals.html
• ↑ https://www.bbc.co.uk/bitesize/topics/zh7xpv4/articles/zwdc4xs

About This Article

To do simple division, think about how many times one number can go into another number. For example, 6 ÷ 2 is 3, because 3 goes into 6 two times. For larger numbers, it's helpful to spend time reviewing the multiplication tables. To do long division, write the number you want to divide under the division bar, and place the number you want to divide by outside of the bar. For example, if you want to calculate 72 ÷ 3, place 72 under the division bar and 3 outside of it. Then, calculate how many times 3 goes into the first number under the division bar. In this case, you’re calculating how many times 3 goes into 7. The answer is 2, with 1 left over. Write the number 2 above the bar, and the remainder – in this case, 1 – below the 7. Then, if there are any numbers left under the division bar, bring them down to the same row as the remainder. So in this case, you’d write a 2 beside the 1 to get 12. Then, repeat the process: how many times does 3 go into 12? In this example, 3 goes into 12 four times, so you’d write 4 on the line above the problem, beside the other numbers. Therefore, 72 ÷ 3 = 24. If you want to learn how to divide fractions, keep reading the article! Did this summary help you? Yes No

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Long Division

Long Division is a method for dividing large numbers, which breaks the division problem into multiple steps following a sequence. Just like the regular division problems, the dividend is divided by the divisor which gives a result known as the quotient, and sometimes it gives a remainder too. Let us learn how to divide using the long division method , along with long division examples with answers, which include the long division steps in this article.

What is Long Division Method?

Long division is a method for dividing large numbers into steps or parts, breaking the division problem into a sequence of easier steps. It is the most common method used to solve problems based on division . Observe the following long division method to see how to divide step by step and check the divisor, the dividend, the quotient, and the remainder.

The above example also showed us how to do 2 digit by 1 digit division.

Parts of Long Division

While performing long division, we need to know the important parts of long division. The basic parts of long division can be listed as follows:

The following table describes the parts of long division with reference to the example shown above.

How to do Long Division?

Division is one of the four basic mathematical operations, the other three being addition , subtraction , and multiplication . In arithmetic, long division is a standard division algorithm for dividing large numbers, breaking down a division problem into a series of easier steps. Let us learn about the steps that are followed in long division.

Long Division Steps

In order to perform division, we need to understand a few steps. The divisor is separated from the dividend by a right parenthesis 〈)〉 or vertical bar 〈|〉 and the dividend is separated from the quotient by a vinculum (an overbar). Now, let us follow the long division steps given below to understand the process.

• Step 1: Take the first digit of the dividend from the left. Check if this digit is greater than or equal to the divisor.
• Step 2: Then divide it by the divisor and write the answer on top as the quotient.
• Step 3: Subtract the result from the digit and write the difference below.
• Step 4: Bring down the next digit of the dividend (if present).
• Step 5: Repeat the same process.

Let us have a look at the examples given below for a better understanding of the concept. While performing long division, we may come across problems when there is no remainder, while some questions have remainders. So, first, let us learn division in which we get remainders.

Division with Remainders

Case 1: When the first digit of the dividend is equal to or greater than the divisor.

Example: Divide 435 ÷ 4

Solution: The steps of this long division are given below:

• Step 1: Here, the first digit of the dividend is 4 and it is equal to the divisor. So, 4 ÷ 4 = 1. So, 1 is written on top as the first digit of the quotient.
• Step 2: Subtract 4 - 4 = 0. Bring the second digit of the dividend down and place it beside 0.
• Step 3: Now, 3 < 4. Hence, we write 0 as the quotient and bring down the next digit of the dividend and place it beside 3.
• Step 4: So, we have 35 as the new dividend. We can see that 35 > 4 but 35 is not divisible by 4, so we look for the number just less than 35 in the table of 4 . We know that 4 × 8 = 32 which is less than 35 so, we go for it.
• Step 5: Write 8 in the quotient. Subtract: 35 - 32 = 3.
• Step 6: Now, 3 < 4. Thus, 3 is the remainder and 108 is the quotient.

Case 2: When the first digit of the dividend is less than the divisor.

Example: Divide 735 ÷ 9

Solution: Let us divide this using the following steps.

• Step 1: Since the first digit of the dividend is less than the divisor, put zero as the quotient and bring down the next digit of the dividend. Now consider the first 2 digits to proceed with the division.
• Step 2: 73 is not divisible by 9 but we know that 9 × 8 = 72 so, we go for it.
• Step 3: Write 8 in the quotient and subtract 73 - 72 = 1.
• Step 4: Bring down 5. The number to be considered now is 15.
• Step 5: Since 15 is not divisible by 9 but we know that 9 × 1 = 9, so, we take 9.
• Step 6: Subtract: 15 - 9 = 6. Write 1 in the quotient.
• Step 7: Now, 6 < 9. Thus, remainder = 6 and quotient = 81.

Case 3: This is a case of long division without a remainder.

Division without Remainder

Example: Divide 900 ÷ 5

Solution: Let us see how to divide step by step.

• Step 1: We will consider the first digit of the dividend and divide it by 5. Here it will be 9 ÷ 5.
• Step 2: Now, 9 is not divisible by 5 but 5 × 1 = 5, so, write 1 as the first digit in the quotient.
• Step 3: Write 5 below 9 and subtract 9 - 5 = 4.
• Step 4: Since 4 < 5, we will bring down 0 from the dividend to make it 40.
• Step 5: 40 is divisible by 5 and we know that 5 × 8 = 40, so, write 8 in the quotient.
• Step 6: Write 40 below 40 and subtract 40 - 40 = 0.
• Step 7: Bring down the next 0 from the dividend. Since 5 × 0 = 0, we write 0 as the remaining quotient.
• Step 9: Therefore, the quotient = 180 and there is no remainder left after the division, that is, remainder = 0.

Long division problems also include problems related to long division by a 2 digit number, long division polynomials and long division with decimals. Let us get an an idea about these in the following sections.

Long Division by a 2 Digit Number

Long division by a 2 digit number means dividing a number by a 2-digit number . For long division by a 2 digit number , we consider both the digits of the divisor and check for the divisibility of the first two digits of the dividend.

For example, if we need to divide 7248 by 24, we can do it using the long division steps. Let us see how to divide step by step.

• Step 1: Since it is a long division by a 2 digit number, we will check for the divisibility of the first two digits of the dividend. The first 2 digits of the dividend are 72 and it is greater than the divisor, so, we will proceed with the division.
• Step 2: Using the multiplication table of 24, we know that 24 × 3 = 72. So we write 3 in the quotient and 72 below the dividend to subtract these. Subtract 72 - 72 = 0.
• Step 3: Bring down the next number from the dividend, that is, 4. The number to be considered now is 4.
• Step 4: Since 4 is smaller than 24, we will put 0 as the next quotient, since 24 × 0 = 0 and write 0 below 4 to subtract 4 - 0 = 4
• Step 5: Bring down the next number from the dividend, that is, 8 and place it next to this 4. The number to be considered now is 48.
• Step 6: Using the multiplication table of 24, we know that 24 × 2 = 48. So we write 2 in the quotient and 48 below the dividend to subtract these. Subtract 48 - 48 = 0. Thus, remainder = 0 and quotient = 302. This means, 7248 ÷ 24 = 302.
• Long Division of Polynomials

When there are no common factors between the numerator and the denominator , or if you can't find the factors, you can use the long division process to simplify the expression. For more details about long division polynomials, visit the Dividing Polynomials page.

Long Division with Decimals

Long division with decimals can be easily done just like the normal division. We just need to keep in mind the decimals and keep copying them as they come. For more details about long division with decimals, visit the Dividing Decimals page.

How to Divide Decimals by Whole Numbers?

When we need to divide decimals by whole numbers, we follow the same procedure of long division and place the decimal in the quotient whenever it comes. Let us understand this with the help of an example.

Example: Divide 36.9 ÷ 3

• Step 1: Here, the first digit of the dividend is 3 and it is equal to the divisor. So, 3 ÷ 3 = 1. So, 1 is written on top as the first digit of the quotient and we write the product 3 below the dividend 3.
• Step 2: Subtract 3 - 3 = 0. Bring the second digit of the dividend down and place it beside 0, that is, 6
• Step 3: Using the multiplication table of 3, we know that 3 × 2 = 6. So we write 2 in the quotient and 6 below the dividend to subtract these. Subtract 6 - 6 = 0.
• Step 4 : Now comes the decimal point in the dividend. So, place a decimal in the quotient after 12 and continue with the normal division.
• Step 5: Bring down the next number from the dividend, that is, 9. The number to be considered now is 9.
• Step 6: Using the multiplication table of 3, we know that 3 × 3 = 9. So we write 3 in the quotient and 9 below the dividend to subtract these. Subtract 9 - 9 = 0. Thus, remainder = 0 and quotient = 12.3. This means, 36.9 ÷ 3 = 12.3

Long Division Tips and Tricks:

Given below are a few important tips and tricks that would help you while working with long division:

• The remainder is always smaller than the divisor.
• For division, the divisor cannot be 0.
• The division is repeated subtraction, so we can check our quotient by repeated subtractions as well.
• We can verify the quotient and the remainder of the division using the division formula : Dividend = (Divisor × Quotient) + Remainder.
• If the remainder is 0, then we can check our quotient by multiplying it with the divisor. If the product is equal to the dividend, then the quotient is correct.

☛ Related Articles

• Long Division Formula
• Long Division with Remainders Worksheets
• Long Division Without Remainders Worksheets
• Long Division with 2-digit Divisors Worksheets
• Long Division Calculator

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Long Division Examples with Answers

Example 1: Ron planted 75 trees equally in 3 rows. Use long division to find out how many trees did he plant in each row?

The total number of trees planted by Ron = 75. The number of rows = 3. To find the number of trees in each row, we need to divide 75 by 3 because there is an equal number of trees in each of the three rows. Let us also observe how to do 2 digit by 1 digit division here.

Therefore, the number of trees in each row = 25 trees.

Example 2: $4000 needs to be distributed among 25 men for the work completed by them at a construction site. Calculate the amount given to each man.

The total amount is $4000. The number of men at work = 25. We need to calculate the amount given to each man. To do so, we have to divide 4000 by 25 using the long division method. Let us see how to work with long division by a 2 digit number and also see how to do long division step by step.

Each man will be given $160. Therefore, $160 is the amount given to each man.

Example 3: State true or false with respect to long division.

a.) In the case of long division of numbers, the remainder is always smaller than the divisor.

b.) We can verify the quotient and the remainder of the division using the division formula: Dividend = (Divisor × Quotient) + Remainder.

a.) True, in the case of long division of numbers, the remainder is always smaller than the divisor.

b.) True, we can verify the quotient and the remainder of the division using the division formula: Dividend = (Divisor × Quotient) + Remainder.

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Practice Questions on Long Division

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FAQs on Long Division

What is long division in math.

Long division is a process to divide large numbers in a convenient way. The number which is divided into smaller groups is known as a dividend, the number by which we divide it is called the divisor, the value received after doing the division is the quotient, and the number left after the division is called the remainder.

The following steps explain the process of long division. This procedure is explained with examples above on this page.

• Write the dividend and the divisor in their respective positions.
• Take the first digit of the dividend from the left.
• If this digit is greater than or equal to the divisor, then divide it by the divisor and write the answer on top as the quotient.
• Write the product below the dividend and subtract the result from the dividend to get the difference. If this difference is less than the divisor, and there are no numbers left in the dividend, then this is considered to be the remainder and the division is done. However, if there are more digits in the dividend to be carried down, we continue with the same process until there are no more digits left in the dividend.

What are the Steps of Long Division?

Given below are the 5 main steps of long division. For example, let us see how we divide 52 by 2.

• Step 1: Consider the first digit of the dividend which is 5 in this example. Here, 5 > 2. We know that 5 is not divisible by 2.
• Step 2: We know that 2 × 2 = 4, so, we write 2 as the quotient.
• Step 3: 5 - 4 = 1 and 1 < 2 (After writing the product 4 below the dividend, we subtract them).
• Step 4: 1 < 2, so we bring down 2 from the dividend and we get 12 as the new dividend now.
• Step 5: Repeat the process till the time you get a remainder less than the divisor. 12 is divisible by 2 as 2 × 6 = 12, so we write 6 in the quotient, and 12 - 12 = 0 (remainder).

Therefore, the quotient is 26 and the remainder is 0.

How to do Long Division with 2 Digits?

For long division with 2 digits, we consider both the digits of the divisor and check for the divisibility of the first two digits of the dividend. If the first 2 digits of the dividend are less than the divisor, then consider the first three digits of the dividend. Proceed with the division in the same way as we divide regular numbers. This procedure is explained with examples above on this page under the heading of 'Long Division by a 2 Digit Number'.

What is the Long Division of Polynomials?

In algebra , the long division of polynomials is an algorithm to divide a polynomial by another polynomial of the same or the lower degree. For example, (4x 2 - 5x - 21) is a polynomial that can be divided by (x - 3) following some defined rules, which will result in 4x + 7 as the quotient.

How to do Long Division with Decimals?

The long division with decimals is performed in the same way as the normal division. This procedure is explained with examples above on this page under the heading of 'How to Divide Decimals by Whole Numbers'? For more details, visit the page about dividing decimals . The basic steps of long division with decimals are given below.

• Write the division in the standard form.
• Start by dividing the whole number part by the divisor.
• Place the decimal point in the quotient above the decimal point of the dividend.
• Bring down the digits on the tenths place, i.e., the digit after the decimal.
• Divide and bring down the other digit in sequence.
• Divide until all the digits of the dividend are over and a number less than the divisor or 0 is obtained in the remainder.

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How to Do Long Division: Step-by-Step Instructions

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In math, few skills are as practical as knowing how to do long division . It's the art of breaking down complex problems into manageable steps, making it an essential tool for students and adults alike.

This operation has many practical uses in our daily lives. For instance, imagine you have a bag of 2,436 candies and want to share them equally among 4 friends. Long division helps you determine that each friend gets 609 candies, ensuring everyone gets their fair share.

Let's dive into the fundamentals of long division and learn about other everyday situations where we can put it to use.

What Is Long Division?

How to do long division in simple steps, long division method: an apple example, using long division in everyday life, how to divide a decimal point by a whole number, practice problems and answers.

Long division is a handy way to divide big numbers by smaller ones, helping us figure out how many times one number fits into another. It turns a tricky math problem into easier steps.

When we do long division, we work with four main parts:

• the big number we want to divide (called the " dividend ")
• the smaller number we're dividing by (the " divisor ")
• the answer to our division (the " quotient ")
• sometimes a little bit left over (the " remainder ")

Long Division vs. Short Division

Short and long division are both methods to divide numbers, but they differ in complexity. The short-division method is a quick way to find the answer when dividing simple numbers. For example, say you want to divide 36 by 6. You write it as 36 ÷ 6, using a division sign, and quickly get the answer, which is 6.

Long division is used for bigger, more complicated numbers, typically two or more digits. This method involves several steps, like writing out the numbers neatly and carefully.

Let's dive into long division with a clear example. We'll use 845 ÷ 3 to walk through this step-by-step process:

• Set up the problem. Write the dividend (845) under the division bar and the divisor (3) outside the bar.
• Divide. Look at the first digit of the dividend (8). How many times does 3 go into 8? Twice, because 3 x 2 = 6, and that's the closest we can get without going over. Write the 2 above the division bar, over the 8.
• Multiply. Multiply the quotient (2) by the divisor (3). (2 x 3 = 6). Write 6 under the 8.
• Subtract. Subtract 6 from 8 to get 2. Draw a line under the 6, subtract, and write 2 below the line.
• Bring down the next digit. Now, bring down the next digit of the dividend, which is 4, to sit next to the 2, making 24.
• Repeat the steps. 3 goes into 24 eight times (3 x 8 = 24), so write 8 above the bar next to the 2. Subtract 24 from 24 to get 0. Now, follow the same process you used in steps 1 through 5 and bring down the last digit, which is 5, to form 05. The number 3 goes into 5 once (3 x 1 = 3), leaving a remainder of 2. Write the 1 above the bar and the remainder 2 below after subtracting 3 from 5.
• The final answer with a remainder. You've divided 845 by 3 to get a final answer of 281 with a remainder of 2.
• Convert the remainder to decimal form. Depending on how far along you are in learning long division, this may be your final answer. If you've progressed to decimals, you will add .0 to 845 and put a decimal point above the division bar, right after the 1. Bring 0 down to form 20. The number 3 goes into 20 six times (3 x 6 = 18). Write 6 after the decimal point above the division bar. Normally, you would continue adding another 0 after 845. until there is no remainder, but since 20 – 18 = 2, you would be repeating this process infinitely because 3 does not divide evenly into 845. Instead, you will draw a horizontal line over the 6 in 281.6 to indicate that it is a repeating decimal. A calculator would show the answer as 281.666667 to indicate that the repeating decimal rounds up.

Now let's use a practical example to work through the long division process.

Imagine you just went apple picking and came home with a massive haul of delicious fruit. In your kitchen, you have 456 apples, and you want to share them equally among 3 baskets to give to your friends, so you're dividing 3 by 456 (456 ÷ 3).

To figure out how many apples go into each basket, you'd tackle the division problem step by step.

• 3 goes into the first digit (4) once, so you write 1 above the division bar, above the 4 in 456. Then you show the subtraction: 4 – 3 = 1.
• Bring down the next digit (5) to form 15. 3 goes into 15 five times (3 x 5 = 15), so you write 5 above the division bar, above the 5 in 456. Then you show the subtraction: 15 – 15 = 0.
• Bring down the final digit (6) to form 06. 3 goes into 6 twice (3 x 2 = 6), so you write 2 above the division bar, above the 2 in 456. Then you show the subtraction: 6 – 6 = 0.
• Since there is no remainder left to divide, you quotient is now written atop the division bar: 152. You will need to place 152 apples in each of the 3 baskets to evenly distribute the 456 apples.

Long division also pops up in real-life situations . Think about when you need to divide something, like pizza or cake, into equal parts.

Want to cut a large recipe in half or figure out how many days are left till summer vacation? Long division can help with that. It's a great way to help us figure out those splits and manage resources better.

And, of course, practicing long division sharpens our problem-solving skills . It teaches us to tackle big problems step by step, breaking them down into smaller, more manageable pieces. This approach is super helpful in math and figuring out all sorts of challenges we might face.

So, long division is more than just a bunch of steps we follow. It's a key that unlocks a lot of doors in the world of math and beyond, helping us understand and connect different concepts and apply them in all sorts of ways.

Dividing decimals by whole numbers is useful in our everyday lives. For instance, if you're splitting a sum of money equally among a certain number of people, you'll need to divide the total (a decimal) by the number of people (a whole number) to determine how much each person gets.

Dividing a decimal point (decimal number) by a whole number is similar to regular division, but you must be mindful of the placement of the decimal point. Here's how to do it:

Example : Divide 0.5 by 5.

• Set up the problem. Begin by setting up the division, with 0.5 as the dividend (the number you're dividing, which will be under the division bar) and 5 as the divisor (the number you're dividing by, which will be to the left of the division bar).
• Begin dividing. 5 goes into the first digit of the dividend 0 times, so you'll write 0 above the division bar, above the 0 in 0.5, and place a decimal point after the 0 you just wrote. It should be directly above the decimal point in the dividend.
• Bring down the next digit. Bring down the 5 to form 05 (you do not bring the decimal down). 5 goes into 5 once (5 x 1 = 5), so you'll write 1 above the division bar, above the 5 in 0.5.
• Show the final answer. When you show the subtraction (5 – 5 = 0), you'll have no remainder. This means the number above the division bar is your final answer: 0.1.

Let's put our long division skills to the test with some word problems. Tackle these problems one step at a time, and don't rush. If you get stuck, pause and review the steps. Remember, practice makes perfect, and every problem is an opportunity to improve your long-division skills.

1. Emma has 672 pieces of candy to share equally among her 4 friends. How many pieces of candy does each friend get?

Solution : To find out, divide 672 by 4. Start with the first part of 672, which is 6, and see how many times 4 can fit into it. It fits 1 time, leaving us with 2. Bringing down the 7 turns it into 27, which 4 fits into 6 times, leaving us with 3. Finally, bringing down the 2 to join the remaining 3 makes 32, which 4 divides into 8 times. So, each friend gets 168 pieces of candy.

2. A teacher has 945 stickers to distribute equally in 5 of her classes. How many stickers does each class get?

Solution : We'll divide 945 by 5. Looking at 9 first, 5 goes into it 1 time. With 4 leftover, we bring down the 4 from 945 to get 44, which 5 divides into 8 times with another 4 leftover. Lastly, bringing down the 5 to the remaining 4 makes 45, which 5 divides into 9 times. Therefore, each class receives 189 stickers.

3. A library has 2,310 books to be placed equally on 6 shelves. How many books will each shelf contain?

Solution : Divide 2,310 by 6. Starting with 23, 6 goes into it 3 times with 5 leftover. After subtracting, we bring down the 1 to get 51, which 6 divides into 8 times with 3 leftover. Bringing down the 0 to the remaining 3 gives us 30, which 6 divides into 5 times. So, each shelf will have 385 books.

This article was updated in conjunction with AI technology, then fact-checked and edited by a HowStuffWorks editor.

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Division in Math

Division is a math superpower that breaks down a whole — whether you’re cutting a pizza or divvying up some candy!

Author Christina Levandowski

Expert Reviewer Jill Padfield

Published: August 24, 2023

• Key takeaways
• Division is an opposite game – If you multiply numbers, you can “undo” them using division. It’s multiplication’s opposite function! 
• There’s a few signs to look for – There are three main symbols for division.
• You won’t always get “even Stevens” – Sometimes, you’ll have a little left over. That leftover number is known as the “remainder.”

Table of contents

What is division?

Common symbols and terminology, properties of division, how to divide in 6 easy steps, what is long division, working with remainders.

• Let’s practice together!

Practice problems

Division is one of the most important math skills you’ll practice, helping you to undo multiplication problems or break off parts of a “whole.” We know it looks complicated, but it really isn’t! You just need to know what signs to look for that tell you when division is needed. 

Like addition and subtraction, division uses a few special terms and symbols. Knowing these can help you to work out your problems quickly and correctly. 

We know it sounds complicated right now — but with a little practice and this handy guide, you’ll be flying through your math homework in no time!

Division is a process in math that lets you break down a number into multiple, equal parts. Sometimes, you can cut everything down into whole number parts, and, sometimes, you’ll be left with a little leftover, giving you a decimal or fraction for an answer rather than a whole number. 

You’ll often see division problems vertically, like this:

It can also be written horizontally: 10 ÷ 2, as 10/2 , or using a division bar: 2 ⟌ 10.

No matter how you see it, though, the use for it is always the same. You’re breaking down a number or quantity into smaller pieces. 

Let’s take a look at some key terms that’ll help you build your division skills.

Division is a simple mathematical operation, but there are still a few terms to know to help you find the correct solution. 

Here are the terms you need to know to solve division equations with ease:

➗ — This is known as a division sign, and it tells you that a number needs to be broken down into multiple pieces. 

⟌ — This is the division bar, and it also means to divide. On the outside of the bar, you’ll see the number determining how many pieces are needed from the whole (the divisor), and the dividend on the inside, which is what you’ll be dividing. The answer goes on the top of the bar. 

∕ — This is known as the division slash. Generally, the divisor comes first, and the dividend will appear second.

Important vocabulary

• Divisor – The divisor is the number that is determining how many pieces are needed from the whole. For example: in 15 ÷ 3, three would be the divisor. It’s also the number located outside of the bracket when you see a division bar.
• Dividend – The dividend is the number that’s being divided, and it’s found inside the division bar.
• Quotient – The quotient is your answer, which goes after the equals (=) sign or on the top of the division bar.
• Remainder – In some cases, you’ll have a remainder — which means that the divisor can’t be equally divided into the dividend. The remainder is written to the side of your equation next to the division bar.

Anytime you see the word “property” in math, know that it’s just a rule to remember as you work through your groups of problems. Here are some of the most important properties of division that you need to know: 

• The Division By 1 Property:  If a number is divided by 1, the quotient will always be the original number. 
• The Division By Itself Property: If a number is divided by itself, the quotient will always be 1. 
• The Division By 0 Property: If a number is divided by 0, it’s “undefined” and cannot be solved. 
• The Division Of 0 By (Any) Number Property: If a 0 value is divided by any number, you’ll have 0 as your quotient.

Knowing these helpful properties can help you to do basic operations (like division) confidently. Remember — these are division facts, so these properties will always be true…no matter what problem you’re working to find the quotient to!

Now that you know the terms and properties of your division operation, it’s time to practice your skills. Let’s work the problem below together. 

1. Prepare your equation

We know that the problem above can feel overwhelming — so we want to take this moment to remind you that what we’re doing is breaking down a number into smaller numbers (or smaller groups of numbers). 

First things first, we have to prepare the equation. Feel free to keep it horizontal,  write it vertically, or use a division bar if you’d like. Use whatever method you feel comfortable with. 

Remember: The dividend (15) belongs inside the division bar if you choose to use that method. 

2. Start with the first digit of the dividend from the left

As we begin to divide, we need to start from the first digit from the left (in this case, 1) and ask ourselves: Does the divisor (3) go into 1 at least once? 

The answer here is “no,” so we will then evaluate the first AND second integer (making 15) as a dividend. 

We ask again: Does the divisor (3) go into 15 at least once? 

Now, the answer is “yes” — we just have to count how many times 3 can go into 15, starting our division process.* 

*NOTE: You can do this by using basic arithmetic operations (such as multiplication) to “undo” the problem (i.e., 3 x ? = 15) or counting by threes until you reach 15. 

In our case, 3 goes into 15 a total of five times.

3. Divide it by the divisor and write the answer on top as the quotient

Now that we know that 15 ÷ 3 = 5, it’s time to write it into our equation. Go ahead and write 5 behind the equals sign or standing tall at the top of your division bar. 

4. Subtract the product of the divisor and the digit written in the quotient from the first digit of the dividend

Now, we have to check our work. We have to ask ourselves: What is 5 x 3? Does it equal our dividend? If it does, you’re golden — you’ve done it! 

Do the multiplication, and then subtract your product to ensure that there’s no other steps remaining (like you’d see in the case of a remainder). 

In our example, 15 – 15 = 0…so no remainder or further action is needed.

5. Bring down the next digit in the dividend (if possible)

In other problems, if you did have a three or four digit dividend, you might need to bring down the next digit in the dividend, and determine if your divisor divides that number cleanly. 

You would then repeat the division process, putting your answer over the third or “next” place above the division bar as part of the quotient. 

Next, yo would repeat step 4 to determine if more steps in the division process are needed.

In our example, we don’t have to do this, so we will leave it as is. Good work!

Congratulations! You just broke a large number down into equal, separate parts. It’s time to repeat the process for your other problems. 

Long division is a form of division that’s used to break down larger numbers and will generally repeat steps 1-6 above at least three or more times. 

We’ll work on that stuff later — for now, let’s just focus on mastering the basics!

What happens when you wind up with a little extra left over, you might ask? While it can look pretty scary, it’s simple to solve.

To do this, you’ll repeat steps one through five above until you get a number that cannot continue to be divided evenly. At this point, you’ll do a few additional steps:

• Determine how many times the divisor goes in to the product of your current answer and the divisor. This won’t be a clean number, and that’s OKAY — that’s what your remainder process is for.
• Complete the subtraction steps. After you get your number, complete the subtraction steps and write your answer below the subtraction bar.
• For example: In the case of 16 ➗ 3, we would write the quotient as: 5R1.

When you see that there’s zero left over, or if there is no way for the divisor to divide into the dividend, that means that your problem is solved!

Let’s practice together

• We ask: “How many times can 6 go into 2?” 
• 6 is greater than 2, so we will not be able to put a number over the 2. We then consider, “How many times can 6 go into 20?”
• Well, this is a bit of a challenge! 6 does not go into 20 evenly. 6 x 3= 18, and 6 x 4= 24. So, 6 can go into 20 three times, but it won’t go evenly.
• So, we add the 3 over the 0, above the division bar.
• We put the product of 6 x 3 (our divisor x our quotient) under the dividend and subtract to determine if the a remainder in our difference. 
• There is a remainder of 2. We write our quotient as: 3R2 .

• We know that our divisor is going to be 1, and our dividend (the number being divided) is 5. We identify them, and we put them properly into a division bar. 
• We ask: “How many times can 1 go into 5?” 
• Instead of working the problem counting or using multiplication, we remember the Division By 1 Property. 
• We put 5 at the top of our division bar, since any integer that is divided by 1 will always be itself. 
• There is no remainder for these types of Division By 1 Property problems. We can move on to the next problem.

• We know that our divisor is going to be 2, and our dividend (the number being divided) is 0. We identify them, and we put them properly into a division bar. 
• We ask, “How many times can 2 go into 0?” 
• Instead of working the problem counting or using multiplication, we remember the Division Of 0 By (Any) Number Property. 
• We put 0 at the top of our division bar, since any integer that attempts to divide 0 as a dividend will always result in a quotient of zero. 
• There is no remainder for these types of Division Of 0 By (Any) Number Property problems. We move on to the next problem.

Ready to give it a go?

You’ve done great so far — and you’re well on your way to mastering the art of division. Don’t be afraid to keep trying and make mistakes. 

Practice makes perfect, so we’ve given you a few more problems to practice as you work to perfect your skills. Remember: You can always scroll up to walk through the tutorials and refresh yourself on the terms, placement, and properties you’ll need to solve these correctly. 

By the end of this session, we’re confident that you’ll be ready to claim that A+ on your next math test. You can do it!

Click to reveal the answer.

The answer is 2 .

The answer is 1R6 .

The answer is 4 . 

Parent Guide

The answer is 2.

How did we get here? 

• We identify 4 as the dividend and 2 as the divisor, and place them in the division bar. 
• We ask: “How many times can 2 go into 4?” We determine this using the “count by twos” method, which shows us that 2 goes into 4 a total of two times. 
• We put 2 at the top of our division bar as the quotient, and multiply it by our divisor (2). We then subtract the product of our multiplication from the number to get an answer of 0, which shows us that there is no remainder. You’re done!

The answer is 1R6.

• We identify 8 as our divisor and 14 as our dividend, and place them in the division bar. 
• We ask: “How many times can 8 go into 14?”, as 8 will not go into 1. We determine this using the “count by eights” method, which shows us that 8 goes into 14 just once. 
• We write a 1 in the quotient place above the 4 under the division bar. We then multiply 1 x 8 to get a product of 8, which is placed below the 14 under the division bar. 
• Now, we do the math and subtract 8 from 14. We’ll get 6 as our difference. 
• We then write our quotient as 1R6.

The answer is 4. 

How did we get here?

• We identify 5 as our divisor and 20 as our dividend, and place them in the division bar. 
• We ask: “How many times can 5 go into 20,” as 5 will not go into 2 at all. We determine this using the “count by fives” method, which shows us that 5 can go into 20 cleanly four times. 
• We place a “4” in our quotient place, and multiply 4 x 5 to get a product of 20. This is written under the division bar as a subtraction problem. 
• We subtract 20 – 20, resulting in a difference of 0. 
• This means that 4 is our final quotient with no remainder.

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FAQs about math strategies for kids

We understand that diving into new information can sometimes be overwhelming, and questions often arise. That’s why we’ve meticulously crafted these FAQs, based on real questions from students and parents. We’ve got you covered!

Division is the mathematical process that breaks down a big value into smaller values. 

There are plenty of times you’ll use division in your everyday life. Some of the most common ways might be to break up an even quantity of something, determining how much of an ingredient to use, or grouping up items for use. 

Division is the inverse of multiplication. This means that it naturally undoes any sort of operation that’s done with multiplication. 

The three main parts of division are the divisor, dividend, and quotient. 

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Christina has written for hundreds of clients from small businesses to Indeed.com. She has extensive experience working with marketing strategy and social media marketing, and has her own business creating assets for clients in the space. She enjoys being an entrepreneur and has also started pursuing investment opportunities as time permits.

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Step by Step Guide for Long Division

What is long division.

Long division is a way to solve division problems with large numbers. Basically, these are division problems you cannot do in your head.

Getting started

One of the problems students have with long division problems is remembering all the steps. Here’s a trick to mastering long division. Use the acronym DMSB , which stands for:

D = Divide M = Multiply S = Subtract B = Bring down

This sequence of letters can be hard to remember, so think of the acronym in the context of a family:  

Dad, Mother, Sister, Brother .

Write D M S B in the corner of your worksheet to remember the sequence you’re about to use.

How to write it down

First, you have to write down the problem in long division format. A typical division problem looks like this:

Dividend ÷ Divisor = Quotient

To write this down in long division format it looks like this:

Let’s try a fairly simple example:

Now, let’s write that problem down in the long division format:

We’re ready to start using the acronym: D M S B

Step 1: D for Divide

How many times will 5 go into 65? That’s too hard to work out in your head, so let’s break it down into smaller steps.

The first problem you’ll work out in this equation is how many times can you divide 5 into 6. The answer is 1. So you put 1 on the quotient line.

Step 2: M for Multiply

You multiply your answer from step 1 and your divisor: 1 x 5 = 5. You write 5 under the 6.

Step 3: S for Subtract

Next you subtract. In this case it will be 6 – 5 = 1.

Step 4: B for Bring down

The last step in the sequence is to bring down the next number from the dividend, which in this case is 5. You write the 5 next to the 1, making the number 15.

Now you start all over again:

How many times can you divide 5 into 15. The answer is 3. So you put 3 on the quotient line.

You multiply your answer from step 1 and your divisor: 3 x 5 = 15. Write this underneath the 15.

Now we subtract 15 from 15. 15 – 15 = 0.

There is no need for step 4. We have finished the problem.

Once you have the answer, do the problem in reverse using multiplication (5 x 13 = 65) to make sure your answer is correct.

K5 Learning has a number of free long division worksheets for grade 4 , grade 5 and grade 6 . Check them out in our math worksheet center .

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Long Division – Definition, Steps, Method, Examples

Definition of long division, steps to carry out long division, fun facts about long division, solved examples on long division , practice problems on long division, frequently asked questions on long division.

In Math, l ong division is the mathematical method for dividing large numbers into smaller groups or parts. It helps to break down a problem into simple and easy steps . The long divisions have dividends, divisors, quotients , and remainders. In a long division problem, the dividend is the large number that is divided by another number called the divisor . The quotient is the result of the division, and the excess quantity that cannot be divided is called the remainder . 

Example of Long Division 

Here is an example that will help us understand this concept:

As 75 is not a multiple of 4, it is not divisible by 4 and will leave a remainder at the end. 

Related Worksheets

Symbol of Division

To show that two numbers are divided, we can add a division sign ‘÷’ between them. So, for example, if we have to show 36 divided by 6, we can write it as 36 ÷ 6. 

We can also show it in a fraction form as 366. 

There are five steps to solve every long division problem with ease. 

Let’s have a look at the examples given below for a better understanding of the concept.

Dividing Decimals Using A Long Division

Long division can also be used to divide decimal numbers into equal groups . It follows the same steps as that of long division, namely, – divide, multiply, subtract , bring down and repeat or find the remainder. 

Here’s an example of long division with decimals .

• If the dividend is 0, the quotient will always be zero.
• The remainder is always less than the divisor.
• Multiplying the quotient with the divisor and adding the remainder will give the dividend. 

(Divisor x Quotient) + Remainder = Dividend 

• When the remainder is 0, the dividend is the product of divisor and quotient.

Divisor x Quotient = Dividend, when remainder = 0

Long Division is an instrument that allows you to divide large numbers into multiple smaller groups or parts. When we divide a dividend with a divisor, the quotient obtained is the number of groups that can be made, and the remainder identifies how many elements or numbers that will be left ungrouped. To understand this concept even better, check out the wide range of interesting exercises available on SplashLearn and become a long division wiz! 

Question 1:  Divide 726 by 4  

Hence, the quotient of the problem is 181, and the remainder is 2.

Question 2: What is the remainder when we divide 248 by 8?

Hence, the quotient of the problem is 31, and the remainder is 0. 

Question 3: Laila’s mother collected all family pictures and wanted to place all of them in an album. If each page of an album can contain 9 pictures, how many pages of the album will she need if there are 285 pictures?

The quotient of the problem is 31, and the remainder is 6. So, she needs 31 + 1 (for the remaining six pictures) or 32 pages. 

Long Division

Attend this Quiz & Test your knowledge.

On dividing 426 by 4, we get the remainder as

Find the dividend if the divisor is 8, quotient is 71 and remainder is 4., find the digit that can replace a in the following division problem..

Can the divisor be 0?

No, division by 0 is not defined in Math generally.

What is the Long Division Method?

In Math, long division is the mathematical method for dividing large numbers into multiple smaller groups or parts. The number which we divide into small groups is called the dividend, the number by which we divide is called the divisor. It helps to break down a problem into simple and easy steps. 

How can we verify the quotient and remainder of a division problem?

​​We can verify the quotient and the remainder of the division using the division formula : (Divisor x Quotient) + Remainder = Dividend 

What is the difference between long division and short division?

The short division is great for dividing larger numbers by one-digit numbers, whereas the long division is handy for dividing large numbers by numbers with two or more digits. 

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Long Division Calculator – with Steps to Solve

Enter the divisor and dividend below to calculate the quotient and remainder using long division. The results and steps to solve it are shown below.

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How to Do Long Division with Remainders

Parts of a long division problem, steps to calculate a long division problem, how to get the quotient and remainder as a decimal, how to do long division without division, frequently asked questions.

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Learning long division is a crucial milestone in understanding essential math skills and a rite of passage to completing elementary school. It strikes fear in elementary school students and parents alike.

A recent study found that the understanding of long division and fractions in elementary school is directly linked to the student’s ability to learn and understand algebra later in school. [1]

Have no fear!

Learning long division can be easy, and in just a few easy steps, you can solve any long division problem. Follow along as we break it down, but first, we need to cover the anatomy of a long division problem.

There are a few parts to a long division problem, as shown in the image above.

The dividend is the number being divided and appears to the right and under the division line.

The divisor is the number being divided by and appears to the left of the division line.

The quotient is the solution and is shown above the dividend over the division line. Often in long division, the quotient is referred to as just the whole number part of the solution.

The remainder is the remaining part of the solution, or what’s leftover, that doesn’t fit evenly into the quotient.

There are a few main steps to solving a long division problem: divide, multiply, subtract, bringing the number down, and repeating the process.

Step One: Set up the Expression

The first step in solving a long division problem is to draw the equation that needs to be solved. If the problem is already in long division form, then skip along to step two.

If it’s not, this is how to draw the long division problem.

Start by drawing a vertical bar to separate the divisor and dividend and an overbar to separate the dividend and quotient.

Place the dividend to the right of the vertical bar and under the overbar. Place the divisor to the left of the vertical bar.

For example , to divide 75 by 4, the long division problem should look like this:

Step Two: Divide

With the long division problem drawn, start by dividing the first digit in the dividend by the divisor.

You can also think about this as counting the number of times the divisor will evenly fit into this digit in the dividend.

If the divisor does not fit into the first digit an even number of times, drop the remainder or decimal portion of the result and write the whole number portion of the result in the quotient above the overline directly above the digit in the dividend.

For example , the divisor “4” goes evenly into the first digit of the dividend “7” one time, so a “1” can be added to the quotient above the 7.

Step Three: Multiply

The next step is to multiply the divisor by the digit just added to the quotient. Write the result below the digit in the dividend.

This step forms the part of the expression for the next step.

Continuing with our example, multiplying the divisor “4” by “1”, which we found in the previous step, equals “4”. So, add a “4” below the first digit in the dividend.

Step Four: Subtract

Now, add a minus sign “-” before the number added in the previous step and draw a line below it to form a subtraction expression.

Continuing the example above, add a “-” before the “4” and a subtraction line below it.

Now that you have created a subtraction problem, it’s time to solve it.

To solve, subtract “7” minus “4”, which equals “3”, so write a “3” below the subtraction line.

Note: if the resulting value of the subtraction problem is greater than the divisor, then you made a mistake in step 2 and should double-check your work.

If the long division problem has a dividend that is a single digit, then hooray, you’re done! The remaining number that is the result of the subtraction problem is the remainder , and the number above the dividend is the whole number quotient.

If more digits are remaining in the dividend, then proceed to the next step.

Step Five: Pull Down the Next Number

At this point in the process, it’s time to operate on the next number in the dividend. To do this, pull down the next digit in the dividend and place it directly to the right of the result from the subtraction problem above.

The next digit in the dividend is “5”. So, pull “5” down and write it next to the “3” found in the previous step.

Step Six: Repeat

At this point, you might be wondering where to go from here. Repeat steps two to five until all the digits in the dividend have been pulled down, divided, multiplied, and subtracted.

When dividing, use the result of the subtraction problem combined with the pulled-down digit as the dividend and divide the divisor into it.

Continuing the examples above, divide the result of the subtraction problem and the pulled-down digit by the divisor. Thus, the next step is to divide 35 by 4. The result is “8”, so add “8” to the quotient.

Next, multiply the quotient digit “8” by the divisor “4”, which equals 32. Add “32” to the long division problem and place a negative sign in front of it.

Next, repeat the subtraction process, subtracting 32 from 35, which equals 3. Add a “3” below the subtraction line. Since there are no longer any remaining digits in the dividend, this is the remainder portion of the solution.

Therefore, 75 divided by 4 is 18 with a remainder of 3. As you practice these steps, use the calculator above to confirm your answer and validate your steps solving long division problems.

If you’ve gotten this far, then you should have a good idea of how to solve a long division problem, but you might be stuck if you need to get the quotient as a decimal rather than a whole number with a remainder.

To calculate the quotient in decimal form, follow the steps above the get the whole number and remainder.

Next, divide the remainder by the divisor to get the remainder as a decimal. Finally, add the decimal to the quotient to get the quotient in decimal form.

For example , 75 ÷ 4 is 18 with a remainder of 3.

Divide 3 by 4 to get the decimal 0.75. 3 ÷ 4 = 0.75

Then, add 0.75 to 18 to get the quotient as a decimal. 0.75 + 18 = 18.75

Thus, the decimal form of 75 ÷ 4 equals 18.75.

While it defeats the purpose of actually learning how to do long division, there is technically a way to perform long division without actually doing any division. The way to do this is as follows.

Set up the long division expression the exact same way as you would normally.

Step Two: Repeatedly Subtract the Divisor

Now, subtract the divisor from the dividend. Afterward, subtract the divisor again from the remaining value. Do this repeatedly until the remaining value is less than the divisor.

Step Three: Count the Number of Subtractions

Finally, to find the quotient, simply count the number of times you subtracted the divisor. This is the whole number portion of the quotient, and the final remaining value is the remainder.

Note: While this method of solving long division problems may seem easier, it is often very impractical to do so. For example, in the above example of 75 divided by 4, you would need to repeat the subtraction 18 times!

Therefore, traditional long division is the vastly superior method.

Why is long division important?

Long division is important not just because it is a tool that allows us to solve difficult division problems, but because it helps to teach logical thinking that will prepare students to excel in solving future mathematical problems.

Why do we still teach long division?

We still teach long division because it teaches students how to think logically, a valuable skill that is shown not just to improve future understanding of algebraic concepts, but also to help solve problems in all aspects of their lives.

How do you check a long division answer?

Just like subtraction is the opposite of addition, multiplication is the opposite of division. Therefore, to check a long division answer, multiply the quotient by the divisor, and if it equals the dividend, then the answer is correct!

Can you do long division on a calculator?

While a calculator can solve division problems, it will not list out the steps used in evaluating a long division problem, and will therefore not improve your understanding of how to perform long division.

Recommended Math Resources

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• Hexadecimal Calculator & Converter Right Arrow Icon
• Triangle Area Calculator Right Arrow Icon
• Right Triangle Calculator Right Arrow Icon
• Carnegie Mellon University, Press Release: Carnegie Mellon-Led Research Team Finds Knowledge Of Fractions and Long Division Predicts Long-Term Math Success, https://www.cmu.edu/news/stories/archives/2012/june/june15_mathsuccess.html

Division Word Problems (1-step word problems)

In these lessons, we will learn to solve word problems involving division.

Related Pages 2-Step Division Word Problems More Word Problems More Singapore Math Lessons

Here are some examples of division word problems that can be solved in one step. We will illustrate how block diagrams or tape diagrams can be used to help you to visualize the division word problems in terms of the information given and the data that needs to be found.

We use division or multiplication when the problem involves equal parts of a whole. The following diagram shows how to use division to find unknown size of parts or groups or to find unknown number of parts or groups. Scroll down the page for examples and solutions.

Example: There are 160 grade 3 students in a school. The students are to be equally divided into 5 classes. How many students do we have in each class?

160 ÷ 5 = 32

We have 32 students in each class.

Example: Melissa made 326 cupcakes. She packed 4 cupcakes into each box. How many boxes of cupcakes did she pack? How many cupcakes were left unpacked?

326 ÷ 4 = 81 remainder 2

She packed 81 boxes of cupcakes.

2 cupcakes were left unpacked.

How to solve multiplication and division problems by drawing a diagram? Division: Finding the Number in Each Group

Example: Victor opened a bag of pretzels and counted 56. He gave each of 7 friends an equal number of pretzels. How many pretzels did each friend receive?

Use Tape Diagrams to solve Word Problems with Unknown Number of Groups

Example: A vet gives the dogs in her office 4 bones each. She used 24 bones. How many dogs got bones?

Tape Diagrams - Unknown Size of Groups How to make a tape diagram to solve a word problem with an unknown size of groups?

Example: Mrs. Silverglat has 21 pretzels. She gives seven students am equal amount of pretzels. How many pretzels does each student get?

Practice solving the following multiplication and division word problems.

• Dan went to the market on Friday. He bought two tomatoes. On Sunday, he bought six times as many. How many tomatoes did he buy on Sunday?
• In July, a construction company built 360 miles of road. In February, the company laid down 60 miles of road. How many times more road did the company complete in July?
• Linh ran 21 miles. Linh ran three times as far as Sophie. How far did Sophie run?
• Molly’s bedroom is 220 square feet. Molly’s dining room is five times the size of her bedroom. How large is her dining room?

Using Division Tape Diagrams to find Unknown Number of Groups

Example: After playing Belmont, the 24 Islander players traveled to South Boston. This time they went by car and 3 players rode in each care. How many cars did they need?

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Unit 5: Division

About this unit.

Do you like breaking things apart? Then you're going to love learning about division! We'll use place value, area models, and estimation techniques to make dividing by 1-digit numbers a breeze. In this unit, you will even get the inside scoop on what to do when there's a pesky remainder left over.

• Division: FAQ (Opens a modal)
• Estimating division that results in non-whole numbers (Opens a modal)
• Understanding remainders (Opens a modal)
• Introduction to remainders (Opens a modal)
• Interpreting remainders (Opens a modal)
• Estimate to divide by 1-digit numbers Get 3 of 4 questions to level up!
• Interpret remainders Get 3 of 4 questions to level up!
• Divide with remainders (2-digit by 1-digit) Get 3 of 4 questions to level up!

Divide multiples of 10, 100, and 1,000 by 1-digit numbers

• Quotients that are multiples of 10 (Opens a modal)
• Divide multiples of 10, 100, and 1,000 by 1-digit numbers Get 5 of 7 questions to level up!

Division with place value

• Division using place value (Opens a modal)
• Divide using place value Get 3 of 4 questions to level up!

Division with area models

• Division with area models (Opens a modal)
• Create division equations with area models Get 3 of 4 questions to level up!
• Divide by 1-digit numbers with area models Get 3 of 4 questions to level up!

Estimate quotients

• Estimating quotients (Opens a modal)
• Estimate quotients (3- and 4-digit divided by 1-digit) Get 3 of 4 questions to level up!

Multi-digit division with partial quotients

• Introduction to division with partial quotients (no remainder) (Opens a modal)
• Division with partial quotients (remainder) (Opens a modal)
• Intro to long division (no remainders) (Opens a modal)
• Divide multi-digit numbers by 2, 3, 4, and 5 (remainders) Get 5 of 7 questions to level up!
• Divide multi-digit numbers by 6, 7, 8, and 9 (remainders) Get 3 of 4 questions to level up!

Multiplication, division word problems

• Division word problem: field goals (Opens a modal)
• Multiplication word problem: pizza (Opens a modal)
• Multiplication and division word problems Get 5 of 7 questions to level up!

Multi-step word problems

• 2-step estimation word problem (Opens a modal)
• 2-step estimation word problems Get 3 of 4 questions to level up!
• Represent multi-step word problems using equations Get 3 of 4 questions to level up!
• Multi-step word problems with whole numbers Get 3 of 4 questions to level up!

Long Division Calculator

Division is one of the basic arithmetic operations, the others being multiplication (the inverse of division), addition, and subtraction. The arithmetic operations are ways that numbers can be combined in order to make new numbers. Division can be thought of as the number of times a given number goes into another number. For example, 2 goes into 8 4 times, so 8 divided by 4 equals 2.

Division can be denoted in a few different ways. Using the example above:

8 ÷ 4 = 2

In order to more effectively discuss division, it is important to understand the different parts of a division problem.

Components of division

Generally, a division problem has three main parts: the dividend, divisor, and quotient. The number being divided is the dividend, the number that divides the dividend is the divisor, and the quotient is the result:

One way to think of the dividend is that it is the total number of objects available. The divisor is the desired number of groups of objects, and the quotient is the number of objects within each group. Thus, assuming that there are 8 people and the intent is to divide them into 4 groups, division indicates that each group would consist of 2 people. In this case, the number of people can be divided evenly between each group, but this is not always the case. There are two ways to divide numbers when the result won't be even. One way is to divide with a remainder, meaning that the division problem is carried out such that the quotient is an integer, and the leftover number is a remainder. For example, 9 cannot be evenly divided by 4. Instead, knowing that 8 ÷ 4 = 2, this can be used to determine that 9 ÷ 4 = 2 R1. In other words, 9 divided by 4 equals 2, with a remainder of 1. Long division can be used either to find a quotient with a remainder, or to find an exact decimal value.

How to perform long division?

To perform long division, first identify the dividend and divisor. To divide 100 by 7, where 100 is the dividend and 7 is the divisor, set up the long division problem by writing the dividend under a radicand, with the divisor to the left (divisorvdividend), then use the steps described below:

This is the stopping point if the goal is to find a quotient with a remainder. In this case, the quotient is 014 or 14, and the remainder is 2. Thus, the solution to the division problem is:

100 ÷ 7 = 14 R2

To continue the long division problem to find an exact value, continue the same process above, adding a decimal point after the quotient, and adding 0s to form new dividends until an exact solution is found, or until the quotient to a desired number of decimal places is determined.

Solving the SAT problem using spiking neural P systems with coloured spikes and division rules

• Research Paper
• Open access
• Published: 23 May 2024

Cite this article

You have full access to this open access article

• Prithwineel Paul 1   na1 &
• Petr Sosík 1   na1  

Spiking neural P systems (SNPS) are variants of the third-generation neural networks. In the last few decades, different variants of SNPS models have been introduced. In most of the SNPS models, spikes are represented using an alphabet with just one letter. In this paper, we use a deterministic SNPS model with coloured spikes (i.e. the alphabet representing spikes contains multiple letters), together with neuron division rules to demonstrate an efficient solution to the SAT problem. As a result, we provide a simpler construction with significantly less class resources to solve the SAT problem in comparison to previously reported results using SNPSs.

Avoid common mistakes on your manuscript.

1 Introduction

Membrane computing is a well-known natural computing model. The computing models in membrane computing are inspired by the working of biological cells. In the last decades, many researchers have constructed different variants of membrane computing models inspired by different biological phenomena [ 1 ]. One such type of cell inspiring many computational models is the neuron. The structure and function of biological neurons communicating via sending impulses (spikes) was the main motivation behind the construction of a popular variant of membrane computing model known as the spiking neural P system (SNPS) [ 2 , 3 ]. Since SNNs (i.e. spiking neural networks) [ 4 ] belong to third-generation neural networks, SNPS models can also be considered third-generation neural networks.

Membrane computing models are mainly categorised into three types, i.e. (1) cell like; (2) tissue like; (3) neural like. One of the most popular directions of research in membrane computing is solving computationally hard problems using different variants of P systems. A comprehensive survey on the use of variants of membrane computing models for solving NP-hard problems, i.e. NP-complete (SAT, SUBSET-SUM) and PSPACE-complete problems (QSAT, Q3SAT) can be found in [ 5 , 6 ]. For instance, SNPS with pre-computed resources has been used to solve QSAT, Q3SAT problems [ 7 ], SUBSET-SUM problem [ 8 , 9 , 10 ], SAT & 3SAT problems [ 10 , 11 ] in a polynomial or even linear time. SNPS with neuron division and budding can solve the SAT problem in a polynomial time with respect to the number of literals n and the number of clauses m . Moreover, these SNP systems can solve the problem in a deterministic manner [ 12 , 13 ]. Similarly, an SNPS with budding rules [ 14 ] and an SNPS with division rules [ 15 ] can solve the SAT problem in polynomial time and a deterministic manner. Furthermore, SNPSs with structural plasticity and time-free SNPS models can solve the SUBSET-SUM problem in a feasible time [ 16 , 17 ]. In [ 18 ], it has been proved that SNPS with astrocytes producing calcium can solve the SUBSET-SUM problem in a polynomial time. Recently, a linear time uniform solution for the Boolean SAT problem using self-adapting SNPS with refractory period and propagation delay is derived in [ 19 ]. Finally, SNPS models have been used to solve maximal independent set selection problems from distributed computing [ 20 ].

In this paper, we construct a new variant of SNPS model, i.e. spiking neural P system with coloured spikes and division rules. The idea of coloured spikes was introduced by Song et al. [ 21 ] to simplify the construction of a SNPS solving complex tasks. Division rules were introduced for the first time in [ 12 , 22 ] to solve the SAT problem using SNPS with division rules and budding. Both the spiking rules and the division rules used here to solve the SAT problem are deterministic in nature. In this paper, we show that with this combined variant, we can solve the SAT problem efficiently using a lower number of steps as well as less amount of other resources.

The paper is organized in the following manner: in Sect.  2 , we discuss the structure and function of the SNP system with coloured spikes and neuron division. In Sect.  3 , we describe a solution to the SAT problem using this variant of SNPS. In Sect.  4 , we give a brief comparison of descriptional and computational complexity of two other SNPS models with division rules, in the amount of resources necessary to solve the SAT problem. Section 5 is conclusive in nature.

2 Spiking neural P system with coloured spikes and neuron division

In this section, we introduce the new variant of SNPS model, i.e. spiking neural P systems with coloured spikes and neuron division, which we mentioned in Sect.  1 . More precisely, this variant combines properties of two existing SNPS models, i.e. SNPS with neuron division [ 12 ] and SNPS with coloured spikes [ 21 ]. Division rules are used to obtain an exponential workspace in polynomial time to apply the strategy of trading space for time, while the use of coloured spikes allows for a simpler and more efficient construction of the SNPS. In the sequel we denote by L ( E ) the language associated to E ,  where E is a regular expression over an alphabet S .

Definition 1

A spiking neural P system with coloured spikes and neuron division of degree \(m \ge 1\) is a construct of the form \(\Pi = (S, H, syn, \sigma _1, \sigma _2, \ldots , \sigma _m, R,\) in ,  out ) where

\(m \ge 1\) represents the initial degree, i.e. the number of neurons initially present in the system;

\(S = \{a_1, a_2, \ldots , a_g\}, g \in \mathbb {N}\) is the alphabet of spikes of different colours;

H is the set containing labels of the neurons;

\(syn \subseteq H \times H\) represents the synapse dictionary between the neurons where \((i, i) \notin syn\) for \(i \in H\) .

\(\sigma _1, \sigma _2, \ldots , \sigma _m\) are neurons initially present in the system, with \(\{1,\dots ,m\}\subseteq H,\) where each neuron \(\sigma _i = \langle n_1^{i}, n_2^{i}, \ldots , n_g^{i}\rangle ,\) for \(1 \le i \le m,\) contains initially \(n_{j}^{i} \ge 0\) spikes of type \(a_j\) \((1 \le j \le g)\) ;

R represents the set containing the rules of the system \(\Pi\) . Each neuron labelled \(i\in H\) contains rules denoted by \([r]_i.\) The rules in R are divided into three categories:

Spiking rule: \([E / a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g} \rightarrow a_1^{p_1} a_2^{p_2} \ldots a_g^{p_g}; d]_i\) where \(i \in H\) , E is a regular expression over S ; \(n_j \ge p_j \ge 0\) \((1 \le j \le g)\) ; \(d \ge 0\) is called delay. Furthermore, \(p_j>0\) for at least one j ,  \(1 \le j \le g.\)

Forgetting rule: \([a_1^{t_1} a_2^{t_2} \ldots a_n^{t_n} \rightarrow \lambda ]_i\) where \(i \in H\) , and \(a_1^{t_1} a_2^{t_2} \ldots a_n^{t_n} \notin L(E)\) for each regular expression E associated with any spiking rule present in the neuron i .

Neuron division rule: \([E]_i \rightarrow []_j\,||\,[]_k\) where E is a regular expression over S and \(i, j, k \in H\) .

in and out represent the input and output neurons, respectively.

A spiking rule \([E / a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g} \rightarrow a_1^{p_1} a_2^{p_2} \ldots a_g^{p_g}; d]_i\) is applicable when the neuron \(\sigma _i\) contains spikes \(a_1^{c_1} a_2^{c_2} \ldots a_n^{c_n} \in L(E)\) and \(c_j \ge n_j\) \((1 \le j \le g)\) . After application of the rule, \(n_j\) copies of the spike \(a_j\) \((1\le j\le g)\) are consumed while \((c_j - n_j)\) copies remains inside the neuron \(\sigma _i\) . Furthermore, \(p_j\) spikes \((1\le j\le g)\) are sent to all neurons \(\sigma _i\) is connected to. These are either neurons \(\sigma _k\) such that \((i,k)\in syn,\) or neurons to which \(\sigma _i\) inherited synapses during neuron division, as described bellow. If the delay \(d = 0\) , then the spikes leave the neuron i immediately. However, if \(d \ge 1\) and the rule is applied at time t , then the neuron i will be closed at step \(t, t+1, t+2, \ldots , t+ d -1\) . During this phase, the neuron cannot receive any spike from outside nor can apply any rule. At step \((t + d)\) , the neuron spikes and it can also receive spikes from other neurons and apply a rule.

The spiking rule \([E / a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g} \rightarrow a_1^{p_1} a_2^{p_2} \ldots a_g^{p_g}; d]_i\) is simply written as \([a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g} \rightarrow a_1^{p_1} a_2^{p_2} \ldots a_g^{p_g}; d]_i\) if \(E = a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g}\) . If \(d = 0\) , then it becomes \([E/a_1^{n_1} a_2^{n_2} \ldots a_g^{n_g} \rightarrow a_1^{p_1} a_2^{p_2} \ldots a_g^{p_g}]_i\) .

A forgetting rule \([a_1^{t_1} a_2^{t_2} \ldots a_n^{t_n} \rightarrow \lambda ]_i\) is applicable when the neuron \(\sigma _i\) contains exactly the spikes \(a_1^{t_1} a_2^{t_2} \ldots a_n^{t_n}\) . Then all these spikes are consumed in the neuron \(\sigma _i\) .

A division rule \([E]_i \rightarrow []_j\,||\,[]_k\) is applicable if a neuron \(\sigma _i\) contains spikes \(a_1^{c_1} a_2^{c_2} \ldots a_n^{c_n} \in L(E)\) . Then after application of the rule, two new neurons labelled j and k are created from \(\sigma _i\) and the spikes are consumed. Child neurons contain developmental rules from R labelled j and k ,  respectively, and initially they do not contain any spikes. Also, child neurons inherit synaptic connections of the parent neuron. If there exists any connection \((\sigma _t, \sigma _i)\) (i.e. \(\sigma _t\) is connected to \(\sigma _i\) ), then after division there exist connections \((\sigma _t, \sigma _j)\) and \((\sigma _t, \sigma _k)\) . Similarly, if there exists a connection \((\sigma _i, \sigma _t)\) , then \((\sigma _j, \sigma _t)\) and \((\sigma _k, \sigma _t)\) will exist after division. Moreover, the child neurons will create synaptic connections which are provided by the synapse dictionary.

The configuration of the system provides information about the number of spikes present in the neurons, the synaptic connections with the other neurons and whether the neuron is open/closed (i.e. whether it can receive/send spikes). The SNPS models work as a parallel distributive computing model where the rules present in all neurons are applied in parallel synchronized manner and the configuration moves to the next configuration. Each computation starts from the initial configuration and stops at a final configuration (i.e. when no rules are further applicable in any of the neurons). If a neuron has more than one rule which is applicable in a given step, then one of these rules is non-deterministically chosen. However, in this work, such a situation never occurs and rules in the neurons are applied in a deterministic manner.

Since already the introductory variant of the SNPS presented in [ 2 ] is computationally universal in the Turing sense, and as our variant of the SNPS is an extension of this basic model, its computational universality follows by the results presented already in [ 2 ]. In the next section, we use the deterministic SNPS with coloured spikes and neuron division to obtain a uniform solution to the SAT problem in a linear time.

3 A solution to the SAT problem

Many decision problems have been solved using spiking neural P systems in uniform as well as semi-uniform ways [ 7 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 ]. Let M be a decision problem. If the problem M is solved in a semi-uniform manner, then for each instance I of M , a spiking neural P system \(\Pi _{I, M}\) is constructed in polynomial time (with respect to the size of the instance I ). The structure and initial configuration of the system \(\Pi _{I, M}\) depends on the instance I . Moreover, the system \(\Pi _{I, M}\) halts (or it may spike a specified number of spikes within a time interval) if and only if I is a positive instance of the decision problem M . A uniform solution of M contains the family \(\{\Pi _{M}(n)\}_{n \in \mathbb {N}}\) of SNP systems where for each instance \(I \in M\) of size n , polynomial number of spikes based on n is introduced into a specified input neuron of \(\Pi _{I, M}\) . Moreover, the system \(\Pi _{I, M}\) halts if and only if I is a positive instance of M . The uniform solutions are strictly associated with the structure of the problem instead of being associated with only one instance of the decision problem. This feature makes the uniform solutions to decision problems more preferable than the semi-uniform solutions. Also, in order to obtain a semi-uniform solution, the SNP system does not require any specific input neuron. However, in the case of uniform solutions, the system must have a specified input neuron(s) which receives the description of an instance of the decision problem in the form of a spike train.

The SAT problem (or the Boolean satisfiability problem) [ 23 ] is a well-known NP-complete decision problem. Each instance is a formula in propositional logic with variables obtaining values TRUTH or FALSE . The SAT decision problem is the problem of determining whether there exists an assignment of truth values to the variables such that the whole formula evaluates to TRUTH .

Let us consider an instance represented by the formula \(\gamma _{n, m} = C_1 \wedge C_2 \wedge \cdots \wedge C_m\) in the conjunctive normal form, where \(C_i\) \((1 \le i \le m)\) represent the clauses. Each clause is a disjunction of literals of the form \(x_j\) or \(\lnot x_j\) , where \(x_j\) are logical variables, \(1\le j \le n.\) Moreover, the class of SAT instances with n variables and m clauses is denoted by SAT ( n ,  m ). So \(\gamma _{n,m} \in SAT(n,m)\) . In order to solve the SAT problem, at first we have to encode the instance \(\gamma _{n,m}\) using spikes in the SNPS, so that the encoded instance could be sent to the input neuron. In this work, we consider the SNPS model with coloured spikes, i.e. different variables are encoded by different types of spikes. The encoding of the instance \(\gamma _{n,m}\) is as follows:

In addition to \(\alpha _{i,j},\) the encoding of the instance contains other auxiliary spikes. The term \(a^{n+1}\) is added at the beginning in order to give the system a necessary initial period during which it generates an exponential workspace with \(2^n\) neurons. The encoding of each clause is separated by \(a_c\) and the end of the encoding is identified by \(a_f\) .

The spiking neural P system with coloured spikes and division rules solving the instances in SAT ( n ,  m ) is described below.

3.1 The SNPS description

The structure of the SNPS with coloured spikes and division rules is as follows: \(\Pi _{n,m} = (S, H, syn, \sigma _{1_0}, \sigma _{1_0'}, \sigma _{n+1}, \sigma _{n+2}, \sigma _{n+3}, \sigma _{n+4}, \sigma _{n+5}, R, in, out)\) where

\(S = \{ a_i, a_i'\ |\ 1\le i\le n\} \cup \{a, a_s, a_c, a_f\}.\)

\(H = \{i, i', i_0, i_0' \ |\ i = 1, 2, \ldots , n \}\) \(\cup\) \(\{ n+1, n+2, n+3, n+4, n+5\}\) \(\cup\) \(\{in, out\}\) \(\cup\) \(\{t_i, f_i\ |\ i = 1, 2, \ldots , n \}\) .

\(syn = \{(i, t_i), (i', f_i)\ |\ i = 1, 2, \ldots , n\}\) \(\cup\) \(\{ (n+2, n+1), (n+2, n+3), (n+3, n+2), (n+4, n+2), (n+4, n+3) \}\) \(\cup\) \(\{ (in, 1_0), (in, 1_0'), (in, n+5), (n+5, t_1), (n+5, f_1) \}\) \(\cup\) \(\{(t_1, out), (f_1, out) \}.\)

the initial configuration of the system contains neurons with labels in ,  out ,  \(1_0, 1_0', n+1, n+2, n+3, n+4, n+5\) . The neurons with labels \(1_0, 1_0', n+2, n+3\) and \(n+4\) contain the spike a ,  and the remaining neurons contain no spike.

Rules in R are divided into four modules: (1) generating module; (2) input module; (3) checking module; (4) output module.

Rules in the generating module:

\([a]_{i_0} \rightarrow []_{i}\,||\,[]_{(i+1)_0}; i = 1, 2, \ldots , n-2\)

\([a]_{(n-1)_0} \rightarrow []_{n-1}\,||\,[]_{n}\)

\([a]_{i_0'} \rightarrow []_{i'}\,||\,[]_{(i+1)_0'}; i = 1, 2, \ldots , n-2\)

\([a]_{(n-1)_0'} \rightarrow []_{(n-1)'}\,||\,[]_{n'}\)

\([a]_{t_i} \rightarrow []_{t_{i+1}}\,||\,[]_{f_{i+1}}; i = 1, 2, \ldots , n-1\) ;

\([a]_{f_i} \rightarrow []_{t_{i+1}}\,||\,[]_{f_{i+1}}; i = 1, 2, \ldots , n-1\) ;

\([a]_{n+1} \rightarrow []_{t_1}\,||\,[]_{f_1}\)

\([a \rightarrow a]_{n+2}\)

\([a^{2} \rightarrow \lambda ]_{n+2}\)

\([a \rightarrow a]_{n+3}\)

\([a^{2} \rightarrow \lambda ]_{n+3}\)

\([a \rightarrow a; n+1]_{n+4}\)

\([S^* / a_c \rightarrow a_c]_{n+5}\)

\([S^* / a_f \rightarrow a_f]_{n+5}.\)

Rules in the input module:

\([a_i \rightarrow a_i]_{in}; i = 1, 2, \ldots , n\)

\([a_i' \rightarrow a_i']_{in}; i = 1, 2, \ldots , n\)

\([a \rightarrow a]_{in}\) ;

\([a_c \rightarrow a_c]_{in}\) ;

\([a_f \rightarrow a_f]_{in}\) ;

\([S^{*} / a_i \rightarrow a]_i; i = 1, 2, \ldots , n\)

\([S^{*} / a_i' \rightarrow a]_{i'}; i = 1, 2, \ldots , n\)

Note: the last two rules are not seen in Fig. 1 as the neurons \(\sigma _i\) , \(i = 1, 2, \ldots , n,\) appear during the generating phase.

Rules in the checking module:

\([a_saa / a \rightarrow a_s]_{t_n}\)

\([a_sa_c \rightarrow \lambda ]_{t_n}\)

\([a_sa_ca / a_ca \rightarrow a_s]_{t_n}\)

\([a_sa_f \rightarrow a]_{t_n}\)

\([a_saa / a \rightarrow a_s]_{f_n}\)

\([a_sa_c \rightarrow \lambda ]_{f_n}\)

\([a_sa_ca / a_ca \rightarrow a_s]_{f_n}\)

\([a_sa_f \rightarrow a]_{f_n}.\)

Rules in the output module:

\([a_s^{+} a^{+} / a \rightarrow a]_{out}.\)

The initial structure of the SNPS with coloured spikes and division rules solving an instance of the SAT problem using contains 9 neurons (see Fig. 1 ). Computation of the SNPS is divided into four stages: (1) generating; (2) input; (3) checking; (4) output.

Initial structure of the SNPS solving the SAT problem

In the generating stage, the division rules are used to create an exponential number of neurons which are further used during the input and checking stages. In the input stage, the input neuron receives the encoded instance of the SAT problem. This stage overlaps with the checking stage during which the SNPS verifies whether any assignment of values of the variables \(x_1, x_2, \ldots , x_n\) satisfies all the clauses \(C_i\) \((1 \le i \le m)\) present in the proposition formula \(\gamma _{n,m}\) . Finally, if the output neuron spikes, it confirms that the formula \(\gamma _{n,m}\) is satisfiable.

3.2 Generating stage

Neurons \(1_0\) and \(1_0'\) contain initially the spike a and the rule \([a]_{1_0} \rightarrow []_{1}\,||\,[]_{2_0}\) and \([a]_{1_0'} \rightarrow []_{1'}\,||\,[]_{2_0'},\) respectively. Thus, the neuron \(\sigma _{1_0}\) creates two neurons with labels 1 and \(2_0,\) and the neuron \(\sigma _{1_0'}\) creates two neurons with labels \(1'\) and \(2_0'\) at time \(t=1\) .

The input neuron has synaptic connections to the neurons with labels \(1_0\) and \(1_0',\) which are inherited by the neurons \(1, 1',\) \(2_0\) and \(2_0'.\) At times \(i=1,2,\dots ,n+1,\) the input neuron receives the spike a from the input spike train and at times \(i=2,3,\dots ,n-1\) it sends the spike to neurons \(i_0\) and \(i_0'\) \((2\le i\le n-1).\) At times \(i=n,\) \(n+1\) and \(n+2,\) the input neuron sends spikes, too, but no neurons with labels \(i_0\) and \(i_0'\) \((n\le i\le n+2\) exist.)

Neurons \(i_0\) and \(i_0'\) \((2\le i\le n-2)\) contain the rule \([a]_{i_0} \rightarrow []_{i}\,||\,[]_{(i+1)_0}\) and \([a]_{i_0'} \rightarrow []_{i'}\,||\,[]_{(i+1)_0'},\) i.e. the neuron \(\sigma _{i_0}\) creates two neurons with labels i and \((i+1)_0,\) and the neuron \(\sigma _{i_0'}\) creates two neurons with labels \(i'\) and \((i+1)_0'\) \((2 \le i \le n-1).\)

Finally, at step \(t=n-1,\) neurons with labels \((n-1)_0\) and \((n-1)_0'\) divide and create neurons with labels \(n-1,\) n ,  \((n-1)'\) and \(n'.\) So after application of all these division rules, there is a layer of neurons \(\sigma _i,\) \(\sigma _{i'}\) \((1\le i\le n)\) shown in Fig. 7 , and the input neuron has synaptic connections to all of them.

Simultaneously, the neurons with labels \(n+1, n+2, n+3\) and \(n+4\) create subsequently \(2^{n}\) neurons which will be used in the checking stage. The circuit controlling the generating stage is depicted in Fig.  2 . Initially, the neurons \(n+2, n+3\) and \(n+4\) contain one spike. At time \(t = 1\) , \(\sigma _{n+2}\) and \(\sigma _{n+3}\) spike and \(\sigma _{n+1}\) receives a spike at time \(t = 2\) . At \(t = 2\) , the rule \([a]_{n+1} \rightarrow []_{t_1}\,||\,[]_{f_1}\) is applied (Fig.  3 ). So at \(t = 3\) , two new neurons with labels \(t_1\) and \(f_1\) are created (Fig.  4 ).

Details of the generating module producing the exponential workspace of \(2^n\) neurons in the first n computational steps of the SNPS

Structure of the SNPS at time \(t = 2\)

Structure of the SNPS at time \(t = 3\)

Note that the neuron \(\sigma _1\) (resp. \(\sigma _{1'}\) ) is connected to \(\sigma _{t_1}\) (resp. \(\sigma _{f_1}\) ) using the synaptic connections from synapse dictionary. Moreover, \(\sigma _{t_1}\) and \(\sigma _{f_1}\) are connected to the output neuron with the synaptic connections from the synapse directory (see Fig.  4 ). The neuron \(\sigma _{t_1}\) contains the rule \([a]_{t_1} \rightarrow []_{t_2}\,||\,[]_{f_2}\) and neuron \(\sigma _{f_1}\) contains \([a]_{f_1} \rightarrow []_{t_2}\,||\,[]_{f_2}\) .

At the same time \(t=3,\) neurons \(\sigma _{t_1}\) and \(\sigma _{f_1}\) receive the spike a from \(\sigma _{n+2}\) . Hence, at \(t = 4\) , each of \(\sigma _{t_1}\) and \(\sigma _{f_1}\) is divided into \(\sigma _{t_2}\) and \(\sigma _{f_2}\) . Again, these neurons receive one spike from \(\sigma _{(n+2)}\) and they are divided at the next step. This process will continue until \(t = n+2\) . Newly created neurons \(t_i, f_i\) \((1\le i\le n)\) will also get further synapses due to the synapse dictionary, as described above. After \((n+2)\) steps, we will show a system in Fig.  7 .

Finally, the rule \(a \rightarrow a; n+1\) in \(\sigma _{n+4}\) is applied at time \(t = n+2\) and thus both \(\sigma _{n+2}\) and \(\sigma _{n+3}\) receive one spike at \(t = n+3\) . Next, the rule \(a^{2} \rightarrow \lambda\) is applied and both spikes in \(\sigma _{n+2}\) and \(\sigma _{n+3}\) are consumed. So no spikes remain inside them and they are inactive from now on.

3.3 Input stage

Recall that the input neuron receives the encoding of an instance of the SAT problem with n variables and m clauses, i.e. \(\gamma _{n,m}\) , where

The encoding of each clause is ended by \(a_c\) and after the input is completely read, it ends with \(a_f\) . The initial buffer \(a^{n+1}\) is used to delay the input of the clause by \(n+1\) steps, giving the SNPS enough time to generate \(2^{n}\) checking neurons. The input neuron in has the following rules: (1) \(a \rightarrow a;\) (2) \(a_i \rightarrow a_i\ (1 \le i \le n);\) (3) \(a_i' \rightarrow a_i'\ (1 \le i \le n);\) (4) \(a_c \rightarrow a_c;\) (5) \(a_f \rightarrow a_f\) . Initially, the input neuron is empty and when it receives \(a, a_i, a_i', a_c\) or \(a_f\) as input, it spikes and sends the same spike to neurons with labels \(i, i'\) \((1 \le i \le n)\) and \(n+5\) .

At time \(t = n + 2\) , the input neuron receives \(\alpha _{1,1}\) as input. Since the input neuron contains the rules \([a_i \rightarrow a_i]_{in}; [a_i' \rightarrow a_i']_{in}\) \((1 \le i \le n)\) , it will spike immediately. The neurons with label i contain the rule \([S^{*} / a_i \rightarrow a]_i\) \((1 \le i \le n)\) and neurons with label \(i'\) contain the rule \([S^{*} / a_i' \rightarrow a]_{i'}\) \((1 \le i \le n)\) . These rules are activated upon the existence of the literal \(x_i\) (resp. \(\lnot x_i\) ) in a clause. If \(\sigma _i\) spikes using the rule \([S^{*} / a_i \rightarrow a]_i\) , it signals that the literal \(x_i\) is present in a clause. Similarly, the use of the rule \([S^{*} / a_i' \rightarrow a]_{i'}\) in \(\sigma _{i'}\) signals the presence of \(\lnot x_i\) in a clause.

3.4 Checking stage

After the generating stage, the input neuron is connected to neurons \(\sigma _i\) and \(\sigma _{i'}\ (1 \le i \le n)\) . These neurons, in turn, are connected to the checking layer consisting of \(2^n\) neurons labelled \(t_n\) or \(f_n,\) see Fig. 7 . Each of the checking neurons \(\sigma _{t_n}\) or \(\sigma _{f_n}\) has exactly n incoming synapses from the input neurons \(\sigma _i\) or \(\sigma _{i'}\ (1 \le i \le n)\) . These synapses represent one of the \(2^{n}\) possible assignments of ‘TRUTH’ or ‘FALSE’ to the n variables of the formula. The structure of these synapses was created by either inherited synapses or by the synapse dictionary during the generating stage. The incoming synapses are indicated under checking neurons in Figs.  4 , 5 , 6 , and 7 by expressions in parentheses. Let us consider the assignment \(t_1 t_2 f_3 \ldots f_n\) (i.e. the first two variables have ‘TRUTH’ and the remaining ones have ‘FALSE’). The corresponding checking neuron has synapses from neurons \(\sigma _1,\) \(\sigma _2,\) \(\sigma _3',\) \(\dots ,\) \(\sigma _n',\) in the input module.

Structure of the SNPS at time \(t = 4\)

Structure of the SNPS at time \(t = 5\)

Structure of the SNPS at time \(t = n+2\)

The input module receives encoded clauses one by one. Spiking of the neuron \(\sigma _i\) in the input module signals the presence of the literal \(x_i\) in the clause, and spiking of the neuron \(\sigma _{i'}\) signals the presence of the literal \(\lnot x_i.\) Therefore, each of the neurons \(\sigma _{t_n}\) or \(\sigma _{f_n}\) obtains one or more spike a if its assignment satisfies the clause. These neurons contain rules \((1) a_s a a / a \rightarrow a_s;\ (2) a_s a_c \rightarrow \lambda ;\ (3) a_s a_c a / a_c a \rightarrow a_s;\ (4) a_s a_f \rightarrow a\) .

Initially, neurons in the checking module contain the spike \(a_s.\) If it receives the spike a from the input module, no rule can be applied. When another a spike comes, the rule \(a_s aa / a \rightarrow a_s\) is applied and only one spike a remains in the neuron. When the encoding of the clause is read completely, the spike \(a_c\) is received from \(\sigma _{n+5}\) . If the assignment of a checking neuron satisfies the clause, the neuron now has the spikes \(a_s a_c a\) and the rule \(a_s a_c a / a_c a \rightarrow a_s\) is applied. Otherwise, the neuron contains the spike \(a_s a_c\) and they are consumed by the rule \(a_s a_c \rightarrow \lambda\) . Since the neuron looses the spike \(a_s\) , no further computation will take place inside it.

3.5 Output stage

The output neuron can receive spikes \(a_s\) due to the application of the rules \(a_saa / a \rightarrow a_s\) and \(a_sa_ca / a_ca \rightarrow a_s\) in checking neurons during the checking stage. However, these spikes are ignored. Finally, when the formula is completely read, the checking neurons receive the spike \(a_f\) from \(\sigma _{n+5}\) . If any of them still has the spike \(a_s\) (meaning that its assignment satisfies all clauses), it spikes using the rule \(a_sa_f \rightarrow a\) and the spike a is sent to the output neuron. Next, the rule \(a_s^{+} a^{+} / a \rightarrow a\) is applied in the output neuron, confirming that the formula is satisfiable.

4 Discussion

In this section, we compare parameters of our solution to the SAT problem using a SNPS with coloured spikes and division rules with two other published papers [ 13 , 15 ] presenting similar solutions to the SAT with SNP systems. Both papers use neuron division rules, the authors of [ 13 ] use also neuron dissolution rules. We compare five parameters of descriptional complexity of the used SNP systems and also the running time complexity. The results are summarized in the table below assuming a solution to an instance of SAT ( n ,  m ),  i.e. with m clauses and n variables.

A more detailed analysis of the running time shows that the generation stage in our paper requires \((n+1)\) steps when the exponential workspace is created. The total number of steps required for the following input stage reading the encoded formula via the input neuron is \(m(n+1) + 1.\) , The checking stage largely overlaps with the input one and requires two more steps to complete. Finally the output stage required only one step.

It follows that all descriptional complexity parameters used in our model have significantly lower values than in the two compared papers. More specifically, the number of neurons, number of neuron labels, size of the synapse dictionary and number of rules used in our solution are significantly lower than in previous solutions and we are able to achieve this result using only 5 initial spikes. Especially, the number of rules is linear to n ,  while in the previously reported solutions it is quadratic or even exponential. Note that the coloured spikes largely help to organize the work of the system and allow for these simplifications.

The only exception is the running time which is \(\mathcal {O}(n+m)\) in [ 13 ], while our SNPS runs in time \(\mathcal {O}(nm).\) The explanation is simple: the authors of [ 13 ] use m input neurons and a special encoding of the formula where each clause is sent in parallel to the n designated input neuron in just one step. Therefore, the input stage in [ 13 ] takes only \(\mathcal {O}(m)\) steps. We conjecture that such an input encoding can be employed also in our construction, diminishing the total running time to \(\mathcal {O}(n+m).\)

5 Conclusion

We presented a deterministic spiking neural P system with coloured spikes and division rules which has been used to solve the SAT problem in linear time. We have shown that our model uses significantly less resources than those reported in [ 13 , 15 ] to solve the SAT problem. It is fair to note that we use a linear number of different spikes with respect to the number n of variables in SAT, while the two mentioned papers use just one type of spike. However, most types of spikes in our construction is used just to encode the input formula. We could use a similar input encoding and an input module with multiple input neurons as that in [ 13 ] with only a few changes which would lower the number of different spikes (= colours) in our model to 5. Furthermore, this input module would restrict our running time to \(\mathcal {O}(n+m).\) This is left for future research. As another possible future research direction one could focus on an efficient solution to PSPACE-complete problems using a similar SNPS model as that discussed in this paper.

Data availability

No datasets were generated or analysed during the current study.

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Acknowledgements

This work was supported by the Silesian University in Opava under the Student Funding Plan, project SGS/9/2024.

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Prithwineel Paul and Petr Sosík have contributed equally to this work.

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Paul, P., Sosík, P. Solving the SAT problem using spiking neural P systems with coloured spikes and division rules. J Membr Comput (2024). https://doi.org/10.1007/s41965-024-00153-0

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