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Sample examples of division with remainder. Dividing natural numbers with a remainder: rule, examples of solutions

Read the topic of the lesson: “Division with a remainder.” What do you already know about this topic?

Can you distribute 8 plums equally on two plates (Fig. 1)?

Rice. 1. Illustration for example

You can put 4 plums in each plate (Fig. 2).

Rice. 2. Illustration for example

The action we performed can be written like this.

8: 2 = 4

Do you think it is possible to divide 8 plums equally onto 3 plates (Fig. 3)?

Rice. 3. Illustration for example

Let's act like this. First, put one plum in each plate, then a second plum. We will have 2 plums left, but 3 plates. This means that we cannot distribute them evenly further. We put 2 plums in each plate, and we had 2 plums left (Fig. 4).

Rice. 4. Illustration for example

Let's continue observing.

Read the numbers. Among the given numbers, find those that are divisible by 3.

11, 12, 13, 14, 15, 16, 17, 18, 19

Test yourself.

The remaining numbers (11, 13, 14, 16, 17, 19) are not divisible by 3, or they say "shared with the remainder."

Let's find the value of the quotient.

Let's find out how many times 3 is contained in the number 17 (Fig. 5).

Rice. 5. Illustration for example

We see that 3 ovals fit 5 times and 2 ovals remain.

The completed action can be written like this.

17: 3 = 5 (remaining 2)

You can also write it in a column (Fig. 6)

Rice. 6. Illustration for example

Look at the pictures. Explain the captions to these figures (Fig. 7).

Rice. 7. Illustration for example

Let's look at the first picture (Fig. 8).

Rice. 8. Illustration for example

We see that 15 ovals were divided into 2. 2 were repeated 7 times, with the remainder being 1 oval.

Let's look at the second picture (Fig. 9).

Rice. 9. Illustration for example

In this figure, 15 squares were divided into 4. 4 were repeated 3 times, with the remainder being 3 squares.

Let's look at the third picture (Fig. 10).

Rice. 10. Illustration for example

We can say that 15 ovals were divided into 3. 3 were repeated 5 times equally. In such cases the remainder is said to be 0.

Let's do the division.

We divide seven squares into three. We get two groups, and one square remains. Let's write down the solution (Fig. 11).

Rice. 11. Illustration for example

Let's do the division.

Let's find out how many times four are contained in the number 10. We see that the number 10 contains four times 2 times and 2 squares remain. Let's write down the solution (Fig. 12).

Rice. 12. Illustration for example

Let's do the division.

Let's find out how many times two are contained in the number 11. We see that in the number 11 two are contained 5 times and 1 square remains. Let's write down the solution (Fig. 13).

Rice. 13. Illustration for example

Let's draw a conclusion. Dividing with a remainder means finding out how many times the divisor is contained in the dividend and how many units are left.

Division with a remainder can also be performed on the number line.

On the number line we mark segments of 3 divisions and see that there are three divisions three times and one division remains (Fig. 14).

Rice. 14. Illustration for example

Let's write down the solution.

10: 3 = 3 (remaining 1)

Let's do the division.

On the number line we mark segments of 3 divisions and see that there are three divisions three times and two divisions remain (Fig. 15).

Rice. 15. Illustration for example

Let's write down the solution.

11: 3 = 3 (remaining 2)

Let's do the division.

On the number line we mark segments of 3 divisions and see that we got exactly 4 times, there is no remainder (Fig. 16).

Rice. 16. Illustration for example

Let's write down the solution.

12: 3 = 4

Today in the lesson we got acquainted with division with a remainder, learned how to perform the named action using a drawing and number beam, practiced solving examples on the topic of the lesson.

References

M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 1. - M.: “Enlightenment”, 2012.
M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 2. - M.: “Enlightenment”, 2012.
M.I. Moro. Math lessons: Methodical recommendations for the teacher. 3rd grade. - M.: Education, 2012.
Regulatory document. Monitoring and evaluation of learning outcomes. - M.: “Enlightenment”, 2011.
"School of Russia": Programs for primary school. - M.: “Enlightenment”, 2011.
S.I. Volkova. Mathematics: Test work. 3rd grade. - M.: Education, 2012.
V.N. Rudnitskaya. Tests. - M.: “Exam”, 2012.

Nsportal.ru ().
Prosv.ru ().
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Homework

1. Write down the numbers that are divisible by 2 without a remainder.

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19

2. Perform division with a remainder using a picture.

3. Perform division with a remainder using the number line.

4. Create an assignment for your friends on the topic of the lesson.

Many numbers cannot be divided completely; when dividing, there is often a remainder other than zero. In this article we will look at division methods natural numbers with the remainder and consider their application in detail using examples.

Let's start with dividing natural numbers with a remainder, then consider division using sequential subtraction. Finally, let's finish with an analysis of the method for selecting an incomplete quotient. We present an algorithm for division with a remainder for the most general case and show how to check the result of dividing natural numbers with a remainder.

This is one of the most convenient ways of division. It is described in detail in a separate article devoted to dividing natural numbers by a column. Here we will not present the entire theory again, but will concentrate specifically on the case of division with a remainder.

We will give an example solution, since it is easiest to understand the essence of the method in practice.

Example 1. How to divide natural numbers with a remainder?

Divide the natural number 273844 by the natural number 97.

We divide by a column and write:

Result: the partial quotient of division is 2823, and the remainder is 13.

Dividing numbers with a remainder through sequential subtraction

To find the partial quotient and remainder, you can resort to sequential subtraction of the divisor from the dividend. This method is not always advisable, but in some cases it is very convenient to use. Let's look at the example again.

Example 2. Division with a remainder through sequential subtraction.

Let us have 7 apples. We need to put these 7 apples into bags of 3 apples. In other words, 7 divided by 3.

Take 3 apples from the initial quantity and put them in one bag. We will have 7 - 3 = 4 apples left. Now, from the remaining apples, we again take 3 pieces and put them in another bag. There are 4 - 3 = 1 apple left.

1 apple is the remainder of the division, since at this stage we can no longer form another package with three apples and the division is essentially completed. Division result:

7 ÷ 3 = 2 (remainder 1)

This means that the number 3 seems to fit into the number 7 twice, and one is the remainder, less than 3.

Let's look at another example. This time, we will present only mathematical calculations, without resorting to analogies.

Example 3. Division with a remainder through sequential subtraction.

Let's calculate: 145 ÷ 46.

The number 99 is greater than 46, so we continue to sequentially subtract the divisor:

We repeat this operation again:

As a result, we needed to sequentially subtract the divisor from the dividend 3 times before we got the remainder - the result of the subtraction, which is less than the divisor. In our case, the remainder is the number 7.

145 ÷ 46 = 3 (remainder 7).

The sequential subtraction method is not suitable when the dividend is less than the divisor. In this case, you can immediately write down the answer: the incomplete quotient is equal to zero, and the remainder is equal to the dividend itself.

If a< b , то a ÷ b = 0 (остаток a) .

For example:

12 ÷ 36 = 0 (remainder 12) 47 ÷ 88 = 0 (remainder 47)

Also regarding the method of sequential subtraction, it should be noted that it is convenient only in cases where the entire division operation is reduced to a small number of subtractions. If the dividend is many times larger than the divisor, using this method will be impractical and involves a lot of cumbersome calculations.

Method for selecting an incomplete quotient

When dividing natural numbers with a remainder, you can calculate the result by selecting an incomplete quotient. We will show you how to conduct the selection process and what it is based on.

First, we determine among which numbers we need to look for the incomplete quotient. From the very definition of the division process, it is clear that the incomplete quotient is equal to zero, or is one of the natural numbers 1, 2, 3, etc.

Secondly, we will establish the connection between the divisor, the dividend, the incomplete quotient and the remainder. Consider the equation d = a - b · c. Here d is the remainder of the division, a is the dividend, b is the divisor, c is the incomplete quotient.

Thirdly, let's not forget that the remainder is always less than the divisor.

Now let's look at the selection process itself. The dividend a and the divisor b are known to us from the very beginning. As an incomplete quotient c we will successively take numbers from the series 0, 1, 2, 3, etc. Applying the formula d = a - b · c and calculating the resulting value with a divisor, we will end the process when the remainder d is less than the divisor b. The number taken as c at this step will be an incomplete quotient.

Let's look at the application of this method using an example.

Example 4. Division with remainder by selection method

Divide 267 by 21.

a = 267 ; b = 21. Let's select an incomplete quotient.

We use the formula d = a - b · c and we will sequentially iterate over c, giving it the values 0, 1, 2, 3, etc.

If c = 0, we have: d = a - b · c = 267 - 21 · 0 = 267. The number 267 is greater than 21, so we continue the substitution.

With c = 1 we have: d = a - b · c = 267 - 21 · 1 = 246. Because 246 > 21, repeat the process again.

With c = 2 we have: d = a - b · c = 267 - 21 · 2 = 267 - 42 = 225; 225 > 21.

With c = 3 we have: d = a - b · c = 267 - 21 · 3 = 267 - 63 = 204 ; 204 > 21.

With c = 12 we have: d = a - b c = 267 - 21 12 = 267 - 252 = 15; 15< 21 .

Algorithm for dividing natural numbers with remainder

When the above methods for selecting incomplete quotient and sequential subtraction require too cumbersome calculations, the following method is used for division with a remainder. Let's consider an algorithm for dividing a natural number a by a number b with a remainder.

Let us remember that in the case when a< b, неполное частное равно нулю, а остаток равен делимомому a . Мы будем рассматривать случай, когда a >b.

Let us formulate three questions and answer them:

What is known there?
What do we need to find?
How will we do this?

Initially, the dividend and divisor are known: a and b.

You need to find the partial quotient c and the remainder d.

Let us present a formula that specifies the relationship between the dividend, divisor, partial quotient and remainder. a = b c + d. It is this relationship that we will take as the basis for the algorithm for dividing natural numbers with a remainder. The dividend a must be represented as a sum a = b · c + d, then we will find the required quantities.

The division algorithm, thanks to which we represent a as the sum a = b · c + d, is very similar to the algorithm for dividing natural numbers without a remainder. Below are the steps of the algorithm using the example of dividing the number 899 by 47.

1. First of all, we look at the dividend and the divisor. We find out and remember how many digits the number in the dividend is written for more number in the divider. In our specific example The dividend has three digits, and the divisor has two.

Let's remember this number.

2. To the right of the divisor notation, add the number of zeros determined by the difference between the number of digits in the dividend and the divisor. In our case, we need to add one zero. If the written number is greater than the dividend, then you need to subtract one from the number remembered in the first paragraph.

In our example, we add a zero to the right of 47. Since 470< 899 , запомненное в предыдущем пункте число не нужно уменьшать на единицу. Таким образом, число 1 так и остается у нас в памяти.

3. To the right of the number 1 we assign the number of zeros, equal to the number, defined in the previous paragraph. In our example, adding one zero to one, we get the number 10. As a result of this action, we received a working unit of discharge, which we will work with further.

4. We will sequentially multiply the divisor by 1, 2, 3. . etc. units of the working digit until we get a number that is greater than or equal to the dividend.

The working category in our example is tens. After multiplying the divisor by one unit of the working digit, we get 470.

470 < 899 , поэтому умножаем на еще одну единицу рабочего разряда. Получаем: 47 · 20 = 940 ; 940 > 899 .

The number we got at the penultimate step (470 = 47 10) is the first of the terms we were looking for.

5. Find the difference between the dividend and the first term found. If the resulting number is greater than the divisor, then we proceed to finding the second term.

We repeat steps 1 - 5, but take the number obtained here as the dividend. If we again get a number greater than the divisor, repeat steps 1 - 5 again in a circle, but with the new number as the dividend. We continue until the number obtained here is less than the divisor. Let's move on to the final stage. Looking ahead, let's say that the last number received will be equal to the remainder.

Let's look at an example. 899 - 470 = 429, 429 > 47. We repeat steps 1 - 5 of the algorithm with the number 429 taken as the dividend.

1. The number 429 has one more digit than the number 47. Let's remember the difference - the number 1.

2. Add one zero to the right of the dividend. We get the number 470. Since 470 > 429, subtract 1 from the number 1 remembered in the previous paragraph and get 1 - 1 = 0. Remember 0.

3. Since in the previous paragraph we received the number 0 and remembered it, we do not need to add a single zero to the one on the right. Thus, the working digit is units

4. Consistently multiply the divisor 47 by 1, 2, 3. . etc. We will not give detailed calculations, but pay attention to the final result: 47 9 = 423< 429 , 47 · 10 = 470 >429. Thus, the second term we are looking for is 47 · 9 = 423.

5. The difference between 429 and 423 is equal to the number 6. Since 6< 47 , это третье, и последнее искомое слагаемое. Перейдем к завершающему этапу алгоритма деления столбиком.

6. The goal of the previous steps was to represent the dividend as the sum of several terms. For our example, we got 899 = 470 + 423 + 6. Remember that 470 = 47 10, 423 = 47 9. Let's rewrite the equality:

899 = 47 10 + 47 9 + 6

Let us apply the distributive property of multiplication.

899 = 47 10 + 47 9 + 6 = 47 (10 + 9) + 6

899 = 47 19 + 6.

Thus, we presented the dividend in the form of the previously given formula a = b · c + d.

The required unknowns: incomplete quotient c = 19, remainder d = 6.

Of course, when deciding practical examples there is no need to describe all the actions in such detail. Let's show this:

Example 5. Dividing natural numbers with a remainder

Let's divide the numbers 42252 and 68.

We use an algorithm. The first five steps give the first term - the number 40800 = 68 600.

We repeat the first five steps of the algorithm again with the number 1452 = 42252 - 40800 and get the second term 1360 = 68 20

For the third time we go through the steps of the aglorhythm, but with the new number 92 = 1452 - 1360. The third term is equal to 68 = 68 1. The remainder is 24 = 92 - 68.

As a result we get:

42252 = 40800 + 1360 + 68 + 24 = 68 600 + 68 20 + 68 1 + 24 = 68 (600 + 20 + 1) + 24 = 68 621 + 24

The partial quotient is 621, the remainder is 24.

Dividing natural numbers with a remainder. Checking the result

Dividing natural numbers with a remainder, especially for large numbers, is a rather labor-intensive and cumbersome process. Anyone can make a mistake in calculations. That is why checking the division result will help you understand whether you did everything correctly. Checking the result of dividing natural numbers with a remainder is performed in two stages.

At the first stage, we check whether the remainder is greater than the divisor. If not, then everything is fine. Otherwise, we can conclude that something went wrong.

Important!

The remainder is always less than the divisor!

At the second stage, the validity of the equality a = b · c + d is checked. If the equality turns out to be true after substituting the values, then the division was performed without errors.

Example 6. Checking the result of dividing natural numbers with a remainder.

Let's check if it is true that 506 ÷ 28 = 17 (remainder 30).

Compare the remainder and the divisor: 30 > 28.

This means that the division was performed incorrectly.

Example 7. Checking the result of dividing natural numbers with a remainder.

The student divided 121 by 13 and received as a result an incomplete quotient of 9 with a remainder of 5. Did he do the right thing?

To find out, first compare the remainder and the divisor: 5< 13 .

The first check point has been passed, let's move on to the second.

Let's write the formula a = b · c + d. a = 121 ; b = 13 ; c = 9 ; d = 5.

Substitute the values and compare the results

13 9 + 5 = 117 + 5 = 122; 121 ≠ 122

This means that somewhere an error crept into the student’s calculations.

Example 8. Checking the result of dividing natural numbers with a remainder.

The student performed laboratory work in physics. During execution, he needed to divide 5998 by 111. As a result, he got the number 54 with a remainder of 4. Is everything calculated correctly?

Let's check! The remainder 4 is less than the divisor 111, so we move on to the second stage of verification.

We use the formula a = b · c + d, where a = 5998; b = 111 ; c = 54 ; d = 4.

After substitution, we have:

5998 = 111 54 + 4 = 5994 + 4 = 5998.

The equality is correct, which means the division is correct.

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How to teach a child division? The simplest method is learn long division. This is much easier than carrying out calculations in your head; it helps you avoid getting confused, not “losing” the numbers, and developing a mental scheme that will work automatically in the future.

How is it carried out?

Division with a remainder is a method in which a number cannot be divided into exactly several parts. As a result of this mathematical operation, in addition to the whole part, an indivisible piece remains.

Let's give a simple example how to divide with remainder:

There is a jar for 5 liters of water and 2 jars of 2 liters each. When water is poured from a five-liter jar into two-liter jars, 1 liter of unused water will remain in the five-liter jar. This is the remainder. In digital form it looks like this:

5:2=2 rest (1). Where is 1 from? 2x2=4, 5-4=1.

Now let's look at the order of division into a column with a remainder. This visually simplifies the calculation process and helps not to lose numbers.

The algorithm determines the location of all elements and the sequence of actions by which the calculation is performed. As an example, let's divide 17 by 5.

Main stages:

Correct entry. Dividend (17) – located on the left side. To the right of the dividend, write the divisor (5). A vertical line is drawn between them (indicating the division sign), and then, from this line, a horizontal line is drawn, emphasizing the divisor. The main features are indicated in orange.
Search for the whole. Next, the first and simplest calculation is carried out - how many divisors fit into the dividend. Let's use the multiplication table and check in order: 5*1=5 - fits, 5*2=10 - fits, 5*3=15 - fits, 5*4=20 - doesn't fit. Five times four is more than seventeen, which means the fourth five does not fit. Let's go back to three. A 17 liter jar will fit 3 five liter jars. We write the result in the form: 3 is written under the line, under the divisor. 3 is an incomplete quotient.
Definition of remainder. 3*5=15. We write 15 under the dividend. We draw a line (indicated by the “=” sign). Subtract the resulting number from the dividend: 17-15=2. We write the result below the line - in a column (hence the name of the algorithm). 2 is the remainder.

Pay attention! When dividing in this way, the remainder must always be less than the divisor.

When the divisor is greater than the dividend

Difficulty arises when the divisor is larger than the dividend. Decimals they are not yet studied in the 3rd grade curriculum, but, following the logic, the answer should be written in the form of a fraction - at best a decimal, at worst - a simple one. But (!) in addition to the program, the calculation method limited by the task: it is necessary not to divide, but to find the remainder! some of them are not! How to solve such a problem?

Pay attention! There is a rule for cases when the divisor is greater than the dividend: the partial quotient is equal to 0, the remainder is equal to the dividend.

How to divide the number 5 by the number 6, highlighting the remainder? How many 6-liter cans will fit into a 5-liter jar? , because 6 is greater than 5.

The assignment requires filling 5 liters - not a single one has been filled. This means that all 5 remain. Answer: partial quotient = 0, remainder = 5.

Division begins to be studied in the third grade of school. By this time, students should already be able to do the division of two-digit numbers by single-digit ones.

Solve the problem: 18 sweets need to be distributed to five children. How many candies will be left?

Examples:

We find the incomplete quotient: 3*1=3, 3*2=6, 3*3=9, 3*4=12, 3*5=15. 5 – overkill. Let's go back to 4.

Remainder: 3*4=12, 14-12=2.

Answer: incomplete quotient 4, 2 left.

You may ask why when divided by 2, the remainder is either 1 or 0. According to the multiplication table, between digits that are multiples of two there is a difference of one.

Another task: 3 pies must be divided into two.

Divide 4 pies between two.

Divide 5 pies between two.

Working with multi-digit numbers

The 4th grade program offers a more complex process of division with increasing calculated numbers. If in the third grade calculations were carried out on the basis of a basic multiplication table ranging from 1 to 10, then fourth graders carry out calculations with multi-digit numbers over 100.

It is most convenient to perform this action in a column, since the incomplete quotient will also be a two-digit number (in most cases), and the column algorithm makes calculations easier and makes them more visual.

Let's divide multi-digit numbers to double digits: 386:25

This example differs from the previous ones in the number of calculation levels, although the calculations are carried out according to the same principle as before. Let's take a closer look:

386 is the dividend, 25 is the divisor. It is necessary to find the incomplete quotient and select the remainder.

First level

Divider - two-digit number. The dividend is three-digit. We select the first two left digits of the dividend - this is 38. We compare them with the divisor. Is 38 more than 25? Yes, that means 38 can be divided by 25. How many whole 25 are in 38?

25*1=25, 25*2=50. 50 is more than 38, let's go back one step.

Answer - 1. Write the unit to the zone not completely private.

38-25=13. Write the number 13 below the line.

Second level

Is 13 more than 25? No - that means you can “lower” the number 6 down by adding it next to 13, on the right. It turned out to be 136. Is 136 more than 25? Yes - that means you can subtract it. How many times can 25 fit into 136?

25*1=25, 25*2=50, 25*3=75, 25*4=100, 25*5=125, 256*=150. 150 is more than 136 – we go back one step. We write the number 5 in the incomplete quotient zone, to the right of one.

Calculate the remainder:

136-125=11. Write it below the line. Is 11 more than 25? No - division cannot be carried out. Does the dividend have digits left? No - there is nothing more to share. The calculations are completed.

Answer: the partial quotient is 15, the remainder is 11.

What if such a division is proposed, when the two-digit divisor is greater than the first two digits of the multi-digit dividend? In this case, the third (fourth, fifth and subsequent) digit of the dividend takes part in the calculations immediately.

Let's give examples for division with three- and four-digit numbers:

75 is a two-digit number. 386 – three-digit. Compare the first two digits on the left with the divisor. 38 is more than 75? No - division cannot be carried out. We take all 3 numbers. Is 386 more than 75? Yes, division can be done. We carry out calculations.

75*1=75, 75*2=150, 75*3=225, 75*4=300, 75*5= 375, 75*6=450. 450 is more than 386 – we go back a step. We write 5 in the incomplete quotient zone.

Find the remainder: 386-375=11. 11 is more than 75? No. Are there any digits left for the dividend? No. The calculations are completed.

Answer: partial quotient = 5, remainder - 11.

Let's check: is 11 more than 35? No - division cannot be carried out. Let's substitute the third number - 119 is more than 35? Yes, we can carry out the action.

35*1=35, 35*2=70, 35*3=105, 35*4=140. 140 is more than 119 – we go back one step. We write 3 in the incomplete balance zone.

Find the remainder: 119-105=14. Is 14 over 35? No. Are there any digits left for the dividend? No. The calculations are completed.

Answer: incomplete quotient = 3, 14 left.

Let's check: is 11 greater than 99? No, we substitute another number. Is 119 more than 99? Yes - let's start the calculations.

11<99, 119>99.

99*1=99, 99*2=198 – overkill. We write 1 in the incomplete quotient.

Find the remainder: 119-99=20. 20<99. Опускаем 5. 205>99. Let's calculate.

99*1=99, 99*2=198, 99*3=297. Too much. We write 2 in the incomplete quotient.

Find the remainder: 205-198=7.

Answer: partial quotient = 12, remainder - 7.

Division with remainder - examples

Learning to divide by column with a remainder

Conclusion

This is how calculations are carried out. If you are careful and follow the rules, then there will be nothing complicated here. Every student can learn to count with a column, because it is fast and convenient.

We will move on from the general idea of dividing natural numbers with a remainder, and in this article we will understand the principles by which this action is carried out. At all division with remainder has a lot in common with dividing natural numbers without a remainder, so we will often refer to the material in this article.

First, let's look at dividing natural numbers with a remainder. Next we will show how you can find the result of dividing natural numbers with a remainder by performing sequential subtraction. After this, we will move on to the method of selecting an incomplete quotient, not forgetting to give examples with detailed description solutions. Next, we will write an algorithm that allows us to divide natural numbers with a remainder in the general case. At the end of the article, we will show how to check the result of dividing natural numbers with a remainder.

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Dividing natural numbers with a remainder

One of the most convenient ways to divide natural numbers with a remainder is long division. In the article Dividing Natural Numbers by Columns, we discussed this division method in great detail. We will not repeat ourselves here, but simply give the solution to one example.

Example.

Divide with the remainder of the natural number 273,844 by the natural number 97.

Solution.

Let's do the division by column:

Thus, the partial quotient of 273,844 divided by 97 is 2,823, and the remainder is 13.

Answer:

273,844:97=2,823 (rest. 13) .

Dividing natural numbers with a remainder through sequential subtraction

You can find the partial quotient and remainder when dividing natural numbers by sequentially subtracting the divisor.

The essence of this approach is simple: sets with the required number of elements are sequentially formed from the elements of the existing set until this is possible, the number of resulting sets gives the incomplete quotient, and the number of remaining elements in the original set is the remainder of the division.

Let's give an example.

Example.

Let's say we need to divide 7 by 3.

Solution.

Let's imagine that we need to put 7 apples into bags of 3 apples. From the original number of apples, we take 3 pieces and put them in the first bag. In this case, due to the meaning of subtracting natural numbers, we are left with 7−3=4 apples. From these we again take 3 pieces and put them in the second bag. After this we are left with 4−3=1 apple. It is clear that this is where the process ends (we cannot form another package with the required number of apples, since the remaining number of apples 1 is less than the quantity 3 we need). As a result, we have two bags with the required number of apples and one apple left.

Then, due to the meaning of dividing natural numbers with a remainder, we can say that we got the following result 7:3=2 (rest. 1).

Answer:

7:3=2 (rest. 1) .

Let's consider the solution to another example, and we will only give mathematical calculations.

Example.

Divide the natural number 145 by 46 using sequential subtraction.

Solution.

145−46=99 (if necessary, refer to the article subtraction of natural numbers). Since 99 is greater than 46, we subtract the divisor a second time: 99−46=53. Since 53>46, we subtract the divisor a third time: 53−46=7. Since 7 is less than 46, we will not be able to carry out the subtraction again, that is, this ends the process of sequential subtraction.

As a result, we needed to successively subtract the divisor 46 from the dividend 145 3 times, after which we got the remainder 7. Thus, 145:46=3 (remaining 7).

Answer:

145:46=3 (remaining 7) .

It should be noted that if the dividend is less than the divisor, then we will not be able to carry out sequential subtraction. Yes, this is not necessary, since in this case we can immediately write the answer. In this case, the partial quotient is equal to zero, and the remainder is equal to the dividend. That is, if a

It must also be said that dividing natural numbers with a remainder using the method considered is good only when a small number of successive subtractions are required to obtain the result.

Selection of incomplete quotient

When dividing given natural numbers a and b with a remainder, the partial quotient c can be found. Now we will show what the selection process is based on and how it should proceed.

First, let's decide among which numbers to look for the incomplete quotient. When we talked about the meaning of dividing natural numbers with a remainder, we found out that an incomplete quotient can be either zero or a natural number, that is, one of the numbers 0, 1, 2, 3, ... Thus, the required incomplete quotient is one of the written numbers, and we just have to go through them to determine which number the partial quotient is.

Next, we will need an equation of the form d=a−b·c, which specifies , as well as the fact that the remainder is always less than the divisor (we also mentioned this when we talked about the meaning of dividing natural numbers with a remainder).

Now we can proceed directly to the description of the process of selecting an incomplete quotient. The dividend a and the divisor b are known to us initially; as an incomplete quotient c, we successively take the numbers 0, 1, 2, 3, ..., each time calculating the value d=a−b·c and comparing it with the divisor. This process ends as soon as the resulting value is less than the divisor. In this case, the number c at this step is the desired incomplete quotient, and the value d=a−b·c is the remainder of the division.

It remains to analyze the process of selecting an incomplete quotient using an example.

Example.

Divide with the remainder of the natural number 267 by 21.

Solution.

Let's select an incomplete quotient. In our example, a=267, b=21. We will sequentially assign c the values 0, 1, 2, 3, ..., calculating at each step the value d=a−b·c and comparing it with the divisor 21.

At c=0 we have d=a−b·c=267−21·0=267−0=267(first multiplication of natural numbers is performed, and then subtraction, this is written in the article). The resulting number is greater than 21 (if necessary, study the material in the article comparing natural numbers). Therefore, we continue the selection process.

At c=1 we have d=a−b·c=267−21·1=267−21=246. Since 246>21, we continue the process.

At c=2 we get d=a−b·c=267−21·2=267−42=225. Since 225>21, we move on.

At c=3 we have d=a−b·c=267−21·3=267−63=204. Since 204>21, we continue the selection.

At c=12 we get d=a−b·c=267−21·12=267−252=15. We received the number 15, which is less than 21, so the process can be considered complete. We selected the incomplete quotient c=12, with the remainder d equal to 15.

Answer:

267:21=12 (rest. 15).

Algorithm for dividing natural numbers with a remainder, examples, solutions

In this section, we will consider an algorithm that allows division with a remainder of a natural number a by a natural number b in cases where the method of sequential subtraction (and the method of selecting an incomplete quotient) requires too much large quantity computing operations.

Let us immediately note that if the dividend a is less than the divisor b, then we know both the partial quotient and the remainder: for a b.

Before we describe in detail all the steps of the algorithm for dividing natural numbers with a remainder, we will answer three questions: what do we initially know, what do we need to find, and based on what considerations will we do this? Initially, we know the dividend a and the divisor b. We need to find the partial quotient c and the remainder d. The equality a=b·c+d specifies the relationship between the dividend, divisor, partial quotient and remainder. From the written equality it follows that if we present the dividend a as a sum b·c+d, in which d is less than b (since the remainder is always less than the divisor), then we will see both the incomplete quotient c and the remainder d.

All that remains is to figure out how to represent the dividend a as a sum b·c+d. The algorithm for doing this is very similar to the algorithm for dividing natural numbers without a remainder. We will describe all the steps, and at the same time we will solve the example for greater clarity. Divide 899 by 47.

The first five points of the algorithm will allow you to represent the dividend as the sum of several terms. It should be noted that the actions from these points are repeated cyclically again and again until all the terms that add up to the dividend are found. In the final sixth point, the resulting sum is converted to the form b·c+d (if the resulting sum no longer has this form), from where the required incomplete quotient and remainder become visible.

So, let's start representing the dividend 899 as the sum of several terms.

First, we calculate how much more the number of digits in the dividend is greater than the number of digits in the divisor, and remember this number.

In our example, the dividend has 3 digits (899 is a three-digit number), and the divisor has two digits (47 is a two-digit number), therefore, the dividend has one more digit, and we remember the number 1.

Now in the divisor entry on the right we add the numbers 0 in the amount determined by the number obtained in the previous paragraph. Moreover, if the written number is greater than the dividend, then you need to subtract 1 from the number remembered in the previous paragraph.

Let's return to our example. In the notation of the divisor 47, we add one digit 0 to the right, and we get the number 470. Since 470<899 , то запомненное в предыдущем пункте число НЕ нужно уменьшать на 1 . Таким образом, у нас в памяти остается число 1 .

After this, to the number 1 on the right we assign the numbers 0 in an amount determined by the number memorized in the previous paragraph. In this case, we get a unit of digit, which we will work with further.

In our example, we assign 1 digit 0 to the number 1, and we get the number 10, that is, we will work with the tens place.

Now we successively multiply the divisor by 1, 2, 3, ... units of the working digit until we get a number greater than or equal to the dividend.

We found out that in our example the working digit is the tens digit. Therefore, we first multiply the divisor by one unit in the tens place, that is, multiply 47 by 10, we get 47 10 = 470. The resulting number 470 is less than the dividend 899, so we proceed to multiplying the divisor by two units in the tens place, that is, we multiply 47 by 20. We have 47·20=940. We got a number that is greater than 899.

The number obtained at the penultimate step during sequential multiplication is the first of the required terms.

In the example being analyzed, the required term is the number 470 (this number is equal to the product 47·100, we will use this equality later).

After this, we find the difference between the dividend and the first term found. If the resulting number is greater than the divisor, then we proceed to find the second term. To do this, we repeat all the described steps of the algorithm, but now we take the number obtained here as the dividend. If at this point we again obtain a number greater than the divisor, then we proceed to find the third term, once again repeating the steps of the algorithm, taking the resulting number as the dividend. And so we proceed further, finding the fourth, fifth and subsequent terms until the number obtained at this point is less than the divisor. As soon as this happens, we take the number obtained here as the last term we are looking for (looking ahead, let’s say that it is equal to the remainder), and move on to the final stage.

Let's return to our example. At this step we have 899−470=429. Since 429>47, we take this number as the dividend and repeat all stages of the algorithm with it.

The number 429 has one more digit than the number 47, so remember the number 1.

Now in the notation of the dividend on the right we add one digit 0, we get the number 470, which is greater than the number 429. Therefore, from the number 1 remembered in the previous paragraph, we subtract 1, we get the number 0, which we remember.

Since in the previous paragraph we remembered the number 0, then to the number 1 there is no need to assign a single digit 0 to the right. In this case, we have the number 1, that is, the working digit is the ones digit.

Now we sequentially multiply the divisor 47 by 1, 2, 3, ... We will not dwell on this in detail. Let's just say that 47·9=423<429 , а 47·10=470>429. The second term we are looking for is the number 423 (which is equal to 47 9, which we will use further).

The difference between 429 and 423 is 6. This number is less than the divisor 47, so it is the third (and last) term we are looking for. Now we can move on to the final stage.

Well, here we come to final stage. All previous actions were aimed at presenting the dividend as the sum of several terms. Now the resulting sum remains to be converted to the form b·c+d. The distributive property of multiplication relative to addition will help us cope with this task. After this, the required incomplete quotient and remainder will become visible.

In our example, the dividend 899 is equal to the sum of three terms 470, 423 and 6. The sum 470+423+6 can be rewritten as 47·10+47·9+6 (remember, we paid attention to the equalities 470=47·10 and 423=47·9). Now we apply the property of multiplying a natural number by a sum, and we get 47·10+47·9+6= 47·(10+9)+6= 47·19+6. Thus, the dividend is transformed to the form we need 899=47·19+6, from which the incomplete quotient 19 and the remainder 6 can be easily found.

So, 899:47=19 (rest. 6).

Of course, when solving examples, you will not describe in such detail the process of division with a remainder.

Instructions

First, test your child's multiplication skills. If a child does not know the multiplication table firmly, then he may also have problems with division. Then, when explaining division, you can be allowed to peek at the cheat sheet, but you still have to learn the table.

Write the dividend and divisor using a vertical separator bar. Under the divisor you will write down the answer - the quotient, separating it with a horizontal line. Take the first digit of 372 and ask your child how many times the number six “fits” in three. That's right, not at all.

Then take two numbers - 37. For clarity, you can highlight them with a corner. Repeat the question again - how many times the number six is contained in 37. To count quickly, it will come in handy. Put the answer together: 6*4 = 24 – not at all similar; 6*5 = 30 – close to 37. But 37-30 = 7 – six will “fit” again. Finally, 6*6 = 36, 37-36 = 1 – suitable. The first digit of the quotient found is 6. Write it under the divisor.

Write 36 under the number 37 and draw a line. For clarity, you can use the sign in the recording. Under the line, put the remainder - 1. Now “descend” the next digit of the number, two, to one - it turns out to be 12. Explain to the child that numbers always “descend” one at a time. Ask again how many “sixes” there are in 12. The answer is 2, this time without a remainder. Write the second digit of the quotient next to the first. Final result – 62.

Also consider the case of division in detail. For example, 167/6 = 27, remainder 5. Most likely, your offspring is about simple fractions haven't heard anything yet. But if he asks questions, what to do next with the remainder can be explained using the example of apples. 167 apples were divided among six people. Everyone got 27 pieces, and five apples remained undivided. You can also divide them by cutting each into six slices and distributing them equally. Each person got one slice from each apple - 1/6. And since there were five apples, each one had five slices - 5/6. That is, the result can be written like this: 27 5/6.