Decision with the residue decision. Division of natural numbers with the residue: rules, examples and solutions

Before we describe in detail all the steps of the division algorithm of natural numbers with the residue, will answer three questions: what is initially known to us that we need to find and on the basis of what considerations will we do this? Initially, we know Delimi A and divider b. We need to find an incomplete private C and a residue d. Equality A \u003d B · C + D sets the relationship between divisible, divisor, incomplete private and residue. From the recorded equality it follows that if we are presenting divisible A in the form of the amount B · C + D, in which D is less than B (as the residue is always smaller than the divider), then we will see and incomplete private C and the residue d.

It remains only to figure out how delimi A is in the form of the amount B · C + D. The algorithm that allows this is very similar to the algorithm of dividing natural numbers without a residue. We describe all the steps, and at the same time we will keep the example of an example for greater clarity. We divide 899 per 47.

The first five points of the algorithm will be allowed to submit divisible as a sum of several terms. It should be noted that the actions from these items are cyclically repeated again and again until all the terms governing in the amount of divisible will be found. In the final sixth paragraph, the amount obtained is converted to the form B · C + D (if the amount received will no longer have such a form), where the desired incomplete private and residue becomes visible.

So, proceed to the submission of dividery 899 as a sum of several terms.

First, calculate how much the number of signs in the divide record is larger than the number of signs in the record of the divider, and remember this number.

In our example, in the records of divide 3 sign (899 - three-digit number), and in the record of the divider - two signs (47 - two-digit number), therefore, in the record of divisible to one sign more, and we remember the number 1.

Now in the records of the divider on the right finish the figures 0 in the amount determined by the number obtained in the previous paragraph. At the same time, if the recorded number is more divisible, then the number is remembered in the previous paragraph.

Return to our example. In the record of the divider 47 add one digit 0, and we obtain 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 that, to the figure 1, you attribute numbers 0 in an amount determined by the number that is stored in the previous paragraph. At the same time we get a unit of discharge, with which we will work further.

In our example, to the figure 1 we attribute 1 digit 0, while we get the number 10, that is, we will work with the discharge of dozens.

Now consistently multiply the divider on 1, 2, 3, ... the units of the working discharge until the moment we receive a number, more or equal to divide.

We found out that in our example, the work discharge is the discharge of tens. Therefore, we first multiply a divider on one unit of discharge of dozens, that is, we multiply 47 by 10, we obtain 47 · 10 \u003d 470. The resulting number 470 is less divided 899, so we turn to the multiplication of the divider into two units of the discharge of dozens, that is, 47 multiply by 20. We have 47 · 20 \u003d 940. We got a number that is more than 899.

The number obtained in the penultimate step with consistent multiplication is the first of the desired terms.

In a disassembled example, the desired term is the number 470 (this number is equal to the product 47 · 100, this is equality we use later).

After that, we find the difference between the divisible and the first found category. If the resulting number is more divisor, then proceed to finding the second terms. To do this, we repeat all the steps of the algorithm, but the number received here is already accepted as divisible. If the number is again obtained at this point, then we proceed to find the third terms, once again repeating the steps of the algorithm, adopting the resulting number as divisible. And so act further, finding the fourth, fifth and subsequent terms, while the number received at this point will not be less than the divider. As soon as this happened, the number received here is taken as the last desired terms (running forward, let's say that it is equal to the residue), and go to the final stage.

Return to our example. At this step, we have 899-470 \u003d 429. Since 429\u003e 47, we take this number as divisible and repeat with it all the stages of the algorithm.

In the record of the number 429 per one sign, more than in the number of Numbers 47, therefore, remember the number 1.

Now in the records of the divide, we finish one digit 0, we obtain the number 470, which is greater than the number 429. Therefore, from the one-stored in the previous paragraph 1, we subtract 1, we obtain a number 0, which I remember.

Since in the previous paragraph, we remembered the number 0, then to the figure 1 it is not necessary to right to attribute any numbers 0. At the same time, we have a number 1, that is, the work discharge is the discharge of units.

Now consistently multiply divider 47 per 1, 2, 3, ... we will not stop at this in detail. Let's just say that 47 · 9 \u003d 423<429 , а 47·10=470>429. The second desired term is the number 423 (which is 47 · 9, which we use on).

The difference between 429 and 423 is 6. This number is less than divider 47, so it is the third (and the last) of the desired terms. Now we can move to the final stage.

Well, we came to final stage. All previous actions were aimed at submitting a divider in the form of a sum of several terms. Now the resulting amount remains to convert b · c + d. With this task, we will help to cope with the distribution property of multiplication relative to addition. After that, the desired incomplete private and residue will be seen.

In our example, Deli-899 is equal to the sum of the three terms 470, 423 and 6. The sum of 470 + 423 + 6 can be rewritten in the form 47 · 10 + 47 · 9 + 6 (remember, we paid attention to the equality 470 \u003d 47 · 10 and 423 \u003d 47 · 9). Now apply the property of multiplying a natural number to the amount, while we obtain 47 · 10 + 47 · 9 + 6 \u003d 47 · (10 + 9) + 6 \u003d 47 · 19 + 6. Thus, divisible was transformed to the above form 899 \u003d 47 · 19 + 6, from where it is easily incomplete private 19 and a residue 6.

So, 899: 47 \u003d 19 (OST. 6).

Of course, when solving examples, you will not describe in such detail the division of the remission.

Read the lesson: "Decision with the residue." What do you already know on this topic?

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

Fig. 1. Illustration for example

In each plate, you can put 4 plums (Fig. 2).

Fig. 2. Illustration for example

The action we performed can be written so.

8: 2 = 4

What do you think it is possible to put on 8 drops to decompose on 3 plates (Fig. 3)?

Fig. 3. Illustration for example

We will act like this. First, put on each plate on one plum, then on the second plum. We will have 2 plums, but 3 plates. So, we can't decompose further. We put in each plate of 2 plums, and 2 plums left (Fig. 4).

Fig. 4. Illustration for example

Continue observation.

Read numbers. Among these numbers, find those that are divided by 3.

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

Check yourself.

The remaining numbers (11, 13, 14, 16, 17, 19) are not divided into 3, or they say "Share with the rest."

Find the value of private.

We learn how many times 3 are contained among 17 (Fig. 5).

Fig. 5. Illustration for example

We see that it fits 3 oval 5 times and 2 oval remained.

The performed action can be written so.

17: 3 \u003d 5 (OST. 2)

Can be recorded in the column (Fig. 6)

Fig. 6. Illustration for example

Consider drawings. Explain the signatures to these drawings (Fig. 7).

Fig. 7. Illustration for example

Consider the first drawing (Fig. 8).

Fig. 8. Illustration for example

We see that 15 ovals were divided by 2. on 2, it repeated 7 times, in the residue - 1 oval.

Consider the second drawing (Fig. 9).

Fig. 9. Illustration for example

In this figure, 15 squares were divided by 4. 4, it was repeated 3 times, in the residue - 3 squares.

Consider the third drawing (Fig. 10).

Fig. 10. Illustration for example

It can be said that 15 ovals were divided by 3. 3 by 3 repeatedly 5 times equally. In such cases, they say that the residue is 0.

Perform division.

Seven squares split three. We get two groups, and one square will remain. We write down the decision (Fig. 11).

Fig. 11. Illustration for example

Perform division.

We learn how many times four are among 10. We see that among 10 four four contains 2 times and 2 squares remain. We write down the solution (Fig. 12).

Fig. 12. Illustration for example

Perform division.

We learn how many times two are contained among 11. We see that among 11 two two contains 5 times and 1 square remains. We write down the solution (Fig. 13).

Fig. 13. Illustration for example

Make a conclusion. Divide with the residue - it means to know how many times the divider is contained in division and how many units will remain.

The division with the residue can be performed on the numeric ray.

On the numeric ray, we note the segments of 3 divisions and see that three divisions turned out to be three times and one division was left (Fig. 14).

Fig. 14. Illustration for example

We write down the decision.

10: 3 \u003d 3 (OST.1)

Perform division.

On the numerical beam, we note the segments of 3 divisions and see that three divisions turned out to be three times and two divisions remained (Fig. 15).

Fig. 15. Illustration for example

We write down the decision.

11: 3 \u003d 3 (OST.2)

Perform division.

On the numeric ray, we note the segments of 3 divisions and see that we got exactly 4 times, the residue is absent (Fig. 16).

Fig. 16. Illustration for example

We write down the decision.

12: 3 = 4

Today, at the lesson, we met the division with the residue, learned to perform the named effect using the drawing and the numeric beam, were trained in solving examples on the topic of the lesson.

Bibliography

1. M.I. Moro, MA Bantova and others. Mathematics: Tutorial. Grade 3: In 2 parts, part 1. - M.: Enlightenment, 2012.
2. M.I. Moro, MA Bantova and others. Mathematics: Tutorial. Grade 3: in 2 parts, part 2. - M.: "Education", 2012.
3. M.I. Moro. Mathematics lessons: Guidelines for teacher. Grade 3. - M.: Enlightenment, 2012.
4. Regulatory document. Control and evaluation of learning outcomes. - M.: "Enlightenment", 2011.
5. "School of Russia": programs for elementary school. - M.: "Enlightenment", 2011.
6. S.I. Volkov. Mathematics: Checking. Grade 3. - M.: Enlightenment, 2012.
7. V.N. Rudnitskaya. Tests. - M.: Exam, 2012.

1. Nsportal.ru ().
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3. Do.gendocs.ru ().

Homework

1. Drink numbers that are divided into 2 without a residue.

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

2. Take a division with the remnant with the drawing.

3. Fit separation with the remnant with a numeric beam.

4. Make an assignment for your comrades on the subject of the lesson.

Division with the rest - This is the division of one number to another, in which the residue is not equal to zero.

The division is not always possible, since there are cases when one number is not divided into another. For example, the number 11 is not divided into 3, as there is no such natural number, when multiplying 3 would be 11.

When the division cannot be done, it is not possible to divide everything divide, but only its largest part that can be divided into a divider. IN this example The largest part of the division, which can be divided into 3 - this is 9 (as a result, we obtain 3), the remaining smaller part of the divide - 2 is not divided into 3.

Speaking of division 11 at 3, 11 is still called divisible, 3 - divider, the result of the division - the number 3 is called incomplete private, and number 2 - residue from division. The division itself in this case is called division with the residue.

Incomplete private called the greatest numberwhich when multiplying the divisor gives a product that does not exceed the divide. The difference between divisible and this product is called the residue. The residue is always less than a divider, otherwise it could also be divided into a divider.

The division with the residue can be recorded like this:

11: 3 \u003d 3 (residue 2)

If, when dividing one natural number to another in the residue, it turns out 0, it is said that the first number is divided by the second aimed. For example, 4 is divided into 2 aimed. Number 5 is not divided into 2 aimed. The word is usually lowered for brevity and say: this number is divided into another, for example: 4 is divided by 2, and 5 is not divided into 2.

Verification of division with the residue

Check the result of dividing with the residue can be in the following way: Incomplete private multiplied to the divider (or vice versa) and add the residue to the resulting product. If the result is a division equal to division, the division with the residue is done correctly:

11: 3 \u003d 3 (residue 2)

In this article we carefully consider division with the rest. Let's start by S. general view about this action, then find out the meaning of the division of natural numbers with the residueAnd we introduce the necessary terms. Then outline the range of tasks solved by dividing natural numbers with the residue. In conclusion, we will dwell on all sorts of links between divisible, divisory, incomplete private and rescue residue.

Navigating page.

Answer:

Delimiya is 79.

It should also be noted that checking the result of dividing natural numbers with the residue is carried out by checking the justice of the equality obtained A \u003d B · C + D.

Finding the residue, if you know Delimi, divider and incomplete private

In terms of its meaning, the residue D is the number of elements that remains in the original set after the exclusion from it elements B times according to C elements. Consequently, due to the meaning of multiplying the natural numbers and the meaning of subtracting natural numbers, equality is fair d \u003d a-b · c . In this way, the residue D from the division of the natural number A on the natural number B is equal to the difference of divide A and the product of the divider B on incomplete private C.

The resulting bond D \u003d A-B · C allows you to find a residue when divisible, divider and incomplete private. Consider the solution of the example.

How to teach a child to divide? The easiest method is - to learn the division of the column. It is much easier than to carry out the calculations in the mind, it helps not get confused, not "losing" numbers and develop a mental scheme that will continue to work automatically.

In contact with

How to run

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

Let's give a simple example how to share with the residue:

There is a bank for 5 liters of water and 2 cans of 2 liters. When, from five liter cans, water is transfused into two-liters, 1 liter of not used water will remain in five liters. This is the balance. In a digital version it looks like this:

5: 2 \u003d 2 OST (1). Where did 1 come from? 2x2 \u003d 4, 5-4 \u003d 1.

Now consider the order of division into a column with the residue. This visually facilitates the calculation process and helps not losing numbers.

The algorithm determines the location of all elements and the sequence of actions on which the calculation is performed. As an example, we divide 17 to 5.

Main steps:

1. Proper recording. Delimi (17) - located on the left side. The right of the dividera write the divider (5). Between them carry out a vertical line (denoting a fission sign), and then, from this line, they spend horizontal, emphasizing the divider. The main features are marked with orange.
2. Search for a whole. Further, the first and simple calculation is carried out - how many divisters fit in Delim. We use the multiplication table and check in order: 5 * 1 \u003d 5 - it is placed, 5 * 2 \u003d 10 - it is placed, 5 * 3 \u003d 15 - placed, 5 * 4 \u003d 20 - not placed. Five times four - more than seventeen, it means that the fourth five does not fit. Return to three. In the 17 liter jar, 3 five-liters will fit. Record the result in shape: 3 Write under the line, under the divider. 3 is incomplete private.
3. Determination of the balance. 3 * 5 \u003d 15. 15 Write under divisible. Let's get a line (indicates the sign "\u003d"). We subtract from the dividated number: 17-15 \u003d 2. We write down the result below below the line - in the column (hence the name of the algorithm). 2 is the residue.

Note! When dividing this way, the residue should always be less than the divider.

When the divider is more divible

Cause difficulty cases when the divider turns out more divisible. Decimal fractions The program for the 3rd grade is not yet studied, but by following logic, the answer must be recorded in the form of a fraction - at best decimal, at worst - simple. But (!) In addition to the program, the calculation methodology limits the task: It is necessary not to divide, and find the rest! Part of them is not! How to solve such a task?

Note! There is a rule for cases when the divider is more divisible: incomplete private equal to 0, the residue is divisible.

How to divide the number 5 by number 6, highlighting the residue? How many 6-liter cans get into five-liter? because 6 more than 5.

On the task, you must fill out 5 liters - not one is filled. So, all 5. Answer: incomplete private \u003d 0, residue \u003d 5.

The division is beginning to study in the third school class. By this time, the disciples should already be allowed to make the division of two-digit numbers to unambiguous.

Decide the task: 18 candies need to be given five children. How many candies will remain?

Examples:

We find incomplete private: 3 * 1 \u003d 3, 3 * 2 \u003d 6, 3 * 3 \u003d 9, 3 * 4 \u003d 12, 3 * 5 \u003d 15. 5 - Bust. Returning to 4.

The residue: 3 * 4 \u003d 12, 14-12 \u003d 2.

Answer: incomplete private 4, 2 left.

You may ask why when dividing 2, the residue is either equal to 1 or 0. on the multiplication table, between numbers, multiple two there is a difference in one.

Another task: 3 pies should be divided for two.

4 Pies split for two.

5 pies split for two.

Work with multivalued numbers

The program for grade 4 offers a more complex process of dividing with an increase in the calculated numbers. If in the third class, the calculations were carried out on the basis of the database multiplication table ranging from 1 to 10, then the quarter-graders of the calculation are carried out with multi-valued numbers more than 100.

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

Slimming multivalued numbers on double-digit: 386:25

This example differs from the previous amount of calculation levels, although the calculations are carried out by the same principle as before. Consider a Read more:

386 - Delimi, 25 - divider. It is necessary to find incomplete private and allocate the residue.

First level

Divider is a two-digit number. Delimi - three-digit. We allocate the first two left numbers from the dividera - it is 38. Compare them with a divider. 38 more 25? Yes, it means 38 can be divided into 25. How many as a whole 25 are in 38?

25 * 1 \u003d 25, 25 * 2 \u003d 50. 50 more than 38, come back one step back.

Answer - 1. Record a unit into the zone not full of private.

38-25 \u003d 13. Write the number 13 below the feature.

Second level

13 more 25? No - it means you can "omit" the number 6 down, adding it next to 13, right. It turned out 136. 136 more than 25? Yes, it means you can subtract it. How many times 25 fit in 136?

25 * 1 \u003d 25, 25 * 2 \u003d 50, 25 * 3 \u003d 75, 25 * 4 \u003d 100, 25 * 5 \u003d 125, 256 * \u003d 150. 150 More 136 - We return back to one step. We write 5 into the incomplete private zone, to the right of one.

Calculate the residue:

136-125 \u003d 11. Written under the feature. 11 more 25? No - the division is impossible. Delimo left figures? No - there is nothing more to share. Calculations are completed.

Answer: Incomplete private equal to 15, in the residue 11.

And if such a division is proposed when the two-digit divider more first Two digits of multivalued division? In this case, the third (fourth, fifth and subsequent) division figure takes part in the calculations immediately.

We give examples For division with three- and four-digit numbers:

75 - two-digit number. 386 - three-digit. Compare the first two digits on the left with the divider. 38 more 75? No - the division is impossible. Take all 3 digits. 386 more than 75? Yes - the division can be done. Calculate.

75 * 1 \u003d 75, 75 * 2 \u003d 150, 75 * 3 \u003d 225, 75 * 4 \u003d 300, 75 * 5 \u003d 375, 75 * 6 \u003d 450. 450 More 386 - We return a step back. We write 5 to the incomplete private area.

We find the residue: 386-375 \u003d 11. 11 more than 75? Not. Still left figures from divide? Not. Calculations are completed.

Answer: Incomplete private \u003d 5, in the residue - 11.

Perform check: 11 more than 35? No - the division is impossible. We substitute the third number - 119 more than 35? Yes - we can conduct action.

35 * 1 \u003d 35, 35 * 2 \u003d 70, 35 * 3 \u003d 105, 35 * 4 \u003d 140. 140 More 119 - We return one step back. We write 3 in the incomplete balance zone.

We find the residue: 119-105 \u003d 14. 14 more than 35? Not. Delivered numbers left? Not. Calculations are completed.

Answer: Incomplete private \u003d 3, left - 14.

Check: 11 more than 99? No - we substitute another figure. 119 more than 99? Yes - start calculations.

11<99, 119>99.

99 * 1 \u003d 99, 99 * 2 \u003d 198 - bust. We write 1 to incomplete private.

We find the residue: 119-99 \u003d 20. twenty<99. Опускаем 5. 205>99. Calculate.

99 * 1 \u003d 99, 99 * 2 \u003d 198, 99 * 3 \u003d 297. Bruep. We write 2 in incomplete private.

We find the residue: 205-198 \u003d 7.

Answer: Incomplete private \u003d 12, residue - 7.

Decision with the residue - examples

Learning to divide in a column with the residue

Output

Thus, calculations are carried out. If you are attentive and execute the rules, then nothing complicated here. Each schoolboy can learn to be considered a column, because it is quick and convenient.