What does grade 3 do in math? Division with remainder, examples and problems - that's what is taught in the lessons. Division with remainder and the algorithm for such calculations will be discussed in the article.
Peculiarities
Consider the topics included in the program that is being studied by grade 3. Division with remainder is highlighted in a special section of mathematics. What is it about? If the dividend is not evenly divisible by the divisor, then the remainder remains. For example, divide 21 by 6. It turns out 3, but the remainder is 3.
In cases when the remainder is equal to zero during the division of natural numbers, they say that the whole division has been performed. For example, if 25 is to be divided by 5, the result is 5. The remainder is zero.
Solution examples
In order to perform division with a remainder, a specific record is used.
Here are examples in mathematics (grade 3). Long division with a remainder can be omitted. It is enough to write in a line: 13: 4 = 3 (remainder 1) or 17: 5 = 3 (remainder 2).
Let's take a closer look at everything. For example, dividing 17 by three results in an integer number of five and a remainder of two. What is the order of solving such an example for division with remainder? First, you need to find the maximum number up to 17, which can be divided by three without a remainder. The largest will be 15.
Further, 15 is divided by the number three, the result of the action will be the number five. Now we subtract the number we found from the dividend, that is, subtract 15 from 17, we get two. A mandatory action is to reconcile the divisor and the remainder. After verification, the response of the action taken must be recorded. 17: 3 = 15 (remainder 2).
If the remainder is greater than the divisor, the action was performed incorrectly. It is according to this algorithm that the 3rd class performs division with the remainder. The examples are first analyzed by the teacher on the blackboard, then the children are invited to test their knowledge by conducting independent work.
Multiplication example
One of the most difficult issues faced by grade 3 is division with remainder. Examples can be tricky, especially when additional columnar calculations are required.
Let's say you want to divide 190 by 27 to get the minimum balance. Let's try to solve the problem using multiplication.
Let's choose a number that, when multiplied, will give the figure as close as possible to the number 190. If we multiply 27 by 6, we get the number 162. Subtract the number 162 from 190, and the remainder will be 28. It turned out to be larger than the original divisor. Therefore, the number six is not suitable for our example as a factor. Let's continue the solution of the example, taking the number 7 for multiplication.
Multiplying 27 by 7, we get the product 189. Next, we will check the correctness of the solution, for this we subtract the result from 190, that is, subtract the number 189. The remainder will be 1, which is clearly less than 27. This is how complex expressions are solved in school (grade 3, division with remainder). Examples always involve recording a response. The whole mathematical expression can be formatted like this: 190: 27 = 7 (remainder 1). Similar calculations can be performed in a column.
This is exactly how class 3 performs division with a remainder. The examples given above will help you understand the algorithm for solving such problems.
Conclusion
In order for the students primary grades the correct computational skills were formed, the teacher during the lessons in mathematics is obliged to pay attention to the explanation of the algorithm of the child's actions when solving problems for division with the remainder.
According to the new federal state educational standards, special attention is paid to an individual approach to learning. The teacher must select tasks for each child, taking into account his individual abilities. At each stage of teaching the rules of division with the remainder, the teacher must carry out intermediate control. It allows him to identify the main problems that arise with the assimilation of the material for each student, timely correct knowledge and skills, eliminate emerging problems, and obtain the desired result.
How to teach a child to divide? The easiest method is learn long division... It is much easier than doing calculations in your mind, it helps you not to get confused, not to "lose" the numbers and to develop a mental scheme that will work automatically in the future.
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How is
Remaining division is a way in which a number cannot be divided into exactly 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 divide with the remainder:
There is a can for 5 liters of water and 2 cans for 2 liters. When water is poured from a five-liter jar into two-liter jars, 1 liter of unused water will remain in a five-liter jar. This is the remainder. Digitally, it looks like this:
5: 2 = 2 rest (1). Where does 1 come from? 2x2 = 4, 5-4 = 1.
Now let's look at the order of division into a long division. This visually facilitates 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 steps:
- Correct entry. Dividend (17) - located by left side... To the right of the dividend, write the divisor (5). A vertical line is drawn between them (denotes a division sign), and then, from this line, a horizontal line is drawn, emphasizing the divider. The main features are highlighted in orange.
- Search for the whole. Next, the first and simplest calculation is carried out - how many dividers fit in 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 that the fourth five does not fit. Back to three. T 17 liter jar fit 3 five-liter. We write the result in the form: 3 we write under the line, under the divisor. 3 is an incomplete quotient.
- Determination of the remainder. 3 * 5 = 15. We write 15 under the dividend. We draw the line (denotes the "=" sign). Subtract the resulting number from the dividend: 17-15 = 2. We write the result below under the line - in a column (hence the name of the algorithm). 2 is the remainder.
Note! When dividing this way, the remainder must always be less than the divisor.
When the divisor is greater than the dividend
Difficulties arise when the divisor is larger than the dividend. Decimal fractions in the program for grade 3 are not yet studied, but, following the logic, the answer must be written in the form of a fraction - in best case decimal, at worst simple. But (!) In addition to the program, the calculation method limits the task: it is necessary not to divide, but to find the remainder! part of it is not! How to solve this problem?
Note! There is a rule for cases when the divisor is greater than the dividend: the incomplete quotient is 0, the remainder is equal to the dividend.
How do you divide the number 5 by the number 6, highlighting the remainder? How many 6-liter cans will fit in a 5-liter one? because 6 is greater than 5.
On assignment, it is necessary to fill 5 liters - none are filled. This means that all 5 remain. Answer: incomplete quotient = 0, remainder = 5.
Division begins to study in the third grade of the school. By this time, students should already, which allows them to divide two-digit numbers by single-digit numbers.
Solve the problem: Give 18 candies to five children. How many candies are left?
Examples:
We find the incomplete quotient: 3 * 1 = 3, 3 * 2 = 6, 3 * 3 = 9, 3 * 4 = 12, 3 * 5 = 15. 5 - brute force. Back to 4.
Remainder: 3 * 4 = 12, 14-12 = 2.
Answer: incomplete quotient 4, 2 left.
You may ask why, when dividing by 2, the remainder is either 1 or 0. According to the multiplication table, between numbers that are multiples of two there is a difference of one.
One more task: 3 pies must be divided by two.
Divide 4 patties for two.
Divide 5 pies for two.
Working with multidigit numbers
The 4th grade program offers more difficult process carrying out division with an increase in the calculated numbers. If in the third grade the calculations were carried out on the basis of the basic multiplication table in the range from 1 to 10, then the fourth graders carry out calculations with multi-digit numbers more than 100.
This action is most convenient to perform 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 more intuitive.
Divide multi-digit numbers to two-digit numbers: 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 isolate the remainder.
First level
The divisor is a two-digit number. The dividend is three-digit. Select the first two left digits from the dividend - this is 38. Compare them with the divisor. 38 is more than 25? Yes, so 38 can be divided by 25. How many whole 25 are included in 38?
25 * 1 = 25, 25 * 2 = 50. 50 is more than 38, go back one step.
The answer is 1. We write the unit to the zone not complete private.
38-25 = 13. We write down the number 13 under the line.
Second level
13 is more than 25? No - it means you can "lower" the number 6 down, adding it next to 13, on the right. It turned out 136. 136 is more than 25? Yes - so you can subtract it. How many times does 25 fit in 136?
25 * 1 = 25, 25 * 2 = 50, 25 * 3 = 75, 25 * 4 = 100, 25 * 5 = 125, 256 * = 150. 150 more than 136 - go back one step. We write the number 5 in the incomplete private area, to the right of one.
We calculate the remainder:
136-125 = 11. We write down below the line. 11 is more than 25? No - division cannot be made. Does the dividend still have numbers? No - there is nothing more to share. The calculations are complete.
Answer: the incomplete quotient is 15, the remainder is 11.
And if such a division is proposed, when the two-digit divisor is greater than the first two digits of the multivalued dividend? In this case, the third (fourth, fifth and subsequent) digit of the dividend takes part in the calculations immediately.
Let's give examples per division with three- and four-digit numbers:
75 is a two-digit number. 386 is three-digit. Compare the first two digits on the left with the divisor. 38 over 75? No - the division cannot be done. We take all 3 digits. 386 over 75? Yes - the 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 one step. We write 5 in the incomplete private zone.
Teaching a child to long division is easy. It is necessary to explain the algorithm of this action and consolidate the material covered.
- According to school curriculum, column divisions begin to be explained to children already in the third grade. Students who grasp everything on the fly quickly grasp the topic
- But, if the child gets sick and missed the math lessons, or he did not understand the topic, then the parents must explain the material to the child on their own. It is necessary to convey information to him as much as possible.
- Moms and dads during educational process the child must be patient, showing tact towards his child. In no case should you shout at a child if something does not work out for him, because this way you can discourage him from all the desire to study
Important: For a child to understand the division of numbers, he must thoroughly know the multiplication table. If the kid does not know multiplication well, he will not understand division.
During home extracurricular activities, you can use cheat sheets, but the child must learn the multiplication table before proceeding with the topic "Division".
So how to explain to a child long division:
- Try to explain in small numbers first. Take counting sticks, for example, 8 pieces
- Ask your child how many pairs are in this row of sticks? Correct - 4. So, if you divide 8 by 2, you get 4, and if you divide 8 by 4, you get 2
- Let the child divide another number himself, for example, a more complex one: 24: 4
- When the baby has mastered division prime numbers, then you can proceed to dividing three-digit numbers by single-digit
Division is always a little more difficult for children than multiplication. But diligent additional classes at home will help the kid understand the algorithm of this action and keep up with their peers at school.
Start simple - dividing by a single number:
Important: Calculate in your head so that the division is complete, otherwise the child may get confused.
For example, 256 divided by 4:
- Draw a vertical line on a piece of paper and divide it in half from the right side. On the left, write the first number, and on the right above the line the second
- Ask the kid how many fours fit in a two - not at all
- Then we take 25. For clarity, separate this number from above with a corner. Again ask the child how many fours fit in twenty-five? That's right - six. We write the number "6" in the lower right corner under the line. The child must use the multiplication table for the correct answer.
- Write under 25 the number 24, and underline to write down the answer - 1
- Ask again: how many fours fit in a unit - not at all. Then we demolish the figure "6" to one
- It turned out 16 - how many fours fit in this number? Correct - 4. Write "4" next to "6" in the answer
- Under 16 we write 16, underline and it turns out "0", which means we divided correctly and the answer turned out to be "64"
Written division by a two-digit number
When the child has mastered division by a single number, you can move on. Written division by a two-digit number is a little more difficult, but if the baby understands how this action is performed, then it will not be difficult for him to solve such examples.
Important: Start explaining again with simple steps. The child will learn how to choose the right numbers and it will be easy for him to divide complex numbers.
Do this simple action together: 184:23 - how to explain:
- First divide 184 by 20, it turns out about 8. But we do not write the number 8 in the answer, since this is a trial number
- We check if 8 is suitable or not. We multiply 8 by 23, we get 184 - this is exactly the number that we have in the divisor. The answer would be 8
Important: For the child to understand, try to take 9 instead of eight, let him multiply 9 by 23, it turns out 207 - this is more than in our divisor. The number 9 does not suit us.
So gradually the baby will understand division, and it will be easy for him to divide more complex numbers:
- Divide 768 by 24. Determine the first digit of the quotient - divide 76 not by 24, but by 20, it turns out 3. Write 3 in response under the line to the right
- Under 76 we write 72 and draw a line, write down the difference - it turned out 4. Is this figure divisible by 24? No - we demolish 8, it turns out 48
- Is 48 divisible by 24? That's right - yes. It turns out 2, write this number in response
- It turned out 32. Now we can check whether we have performed the division action correctly. Do long multiplication: 24x32, it turns out 768, then everything is correct
If the child has learned how to perform division by a two-digit number, then it is necessary to move on to the next topic. The algorithm for dividing by a three-digit number is the same as the algorithm for dividing by a two-digit number.
For example:
- Divide 146064 by 716. Take 146 first - ask the child whether this number is divisible by 716 or not. That's right - no, then we take 1460
- How many times does 716 fit in 1460? Correctly - 2, so we write this number in the answer
- We multiply 2 by 716, we get 1432. We write this figure under 1460. It turns out the difference is 28, we write under the line
- We take down 6. Ask the child - is 286 divided by 716? That's right - no, so we write 0 in the answer next to 2. We also demolish the number 4
- We divide 2864 by 716. Take 3 - a little, 5 - a lot, so it turns out 4. Multiply 4 by 716, we get 2864
- Write 2864 under 2864, resulting in a difference of 0. Answer 204
Important: To check the correctness of the division, multiply with the child in a column - 204x716 = 146064. The division is correct.
It's time to explain to the child that division can be not only whole, but also with the remainder. The remainder is always less than or equal to the divisor.
Division with remainder should be explained by simple example: 35: 8 = 4 (remainder 3):
- How many eights fit in 35? Correct - 4. Remaining 3
- Is this figure divisible by 8? That's right - no. It turns out the remainder is 3
After that, the child should learn that division can be continued by adding 0 to the number 3:
- The answer contains the number 4. After it we write a comma, since the addition of zero means that the number will be with a fraction
- It turned out 30. Divide 30 by 8, it turns out 3. We write in the answer, and under 30 we write 24, underline and write 6
- We demolish the number 0 to the number 6. Divide 60 by 8. Take 7 each, it turns out 56. We write under 60 and write down the difference 4
- We add 0 to the number 4 and divide by 8, it turns out 5 - we write in response
- Subtract 40 from 40 to get 0. So the answer is 35: 8 = 4.375
Advice: If the child does not understand something, do not get angry. Let it pass a couple of days and try again to explain the material.
Math lessons at school will also reinforce knowledge. Time will pass and the kid will quickly and easily solve any division examples.
The algorithm for dividing numbers is as follows:
- Make an estimate of the number that will be in the answer
- Find the first incomplete dividend
- Determine the number of digits in the quotient
- Find numbers in each digit of the quotient
- Find the remainder (if any)
According to this algorithm, division is performed both by single-digit numbers and by any multi-digit number(two-digit, three-digit, four-digit, and so on).
When studying with a child, often ask him examples for performing an estimate. He must calculate the answer quickly in his head. For example:
- 1428:42
- 2924:68
- 30296:56
- 136576:64
- 16514:718
To consolidate the result, you can use the following division games:
- "Puzzle". Write five examples on a piece of paper. Only one of them should be with the correct answer.
Condition for the child: Among several examples, only one was solved correctly. Find him in a minute.
Video: Game arithmetic for children addition subtraction division multiplication
Video: Educational cartoon Mathematics Learning by heart multiplication and division tables
In this article, we will analyze division of integers with remainder... Let's start with general principle division of integers with remainder, formulate and prove the theorem on the divisibility of integers with remainder, trace the connections between the dividend, divisor, incomplete quotient and remainder. Next, we will voice the rules by which the division of integers with a remainder is carried out, and consider the application of these rules when solving examples. After that, we will learn how to check the result of dividing integers with a remainder.
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Understanding the division of integers with remainder
We will consider division of integers with remainder as a generalization of division with remainder of natural numbers. This is due to the fact that natural numbers are a constituent part of integers.
Let's start with the terms and designations that are used in the description.
By analogy with division natural numbers with a remainder, we assume that the result of division with a remainder of two integers a and b (b is not equal to zero) are two integers c and d. The numbers a and b are called divisible and divider respectively, the number d - the remainder from dividing a by b, and the integer c is called incomplete private(or simply private if the remainder is zero).
Let us agree to assume that the remainder is a non-negative integer, and its value does not exceed b, that is, (we met such chains of inequalities when we talked about comparing three or more integers).
If the number c is an incomplete quotient, and the number d is the remainder of dividing an integer a by an integer b, then we will briefly write this fact as an equality of the form a: b = c (remainder d).
Note that when dividing an integer a by an integer b, the remainder can be equal to zero... In this case, a is said to be divisible by b without a remainder(or wholly). Thus, dividing integers without a remainder is a special case of dividing integers with a remainder.
It is also worth saying that when dividing zero by some integer, we always deal with division without a remainder, since in this case the quotient will be equal to zero (see the theory section on division of zero by an integer), and the remainder will also be equal to zero.
We have decided on the terminology and designations, now let's figure out the meaning of dividing integers with a remainder.
Dividing a negative integer a by a positive integer b can also make sense. To do this, consider a negative integer as debt. Let's imagine the following situation. The debt, which constitutes the items, must be paid by b people, making the same contribution. The absolute value of the incomplete quotient c in this case will determine the amount of debt of each of these people, and the remainder d will show how many items will remain after the payment of the debt. Let's give an example. Let's say 2 people need 7 apples. If we assume that each of them owes 4 apples, then after paying the debt, they will have 1 apple. This situation corresponds to the equality (−7): 2 = −4 (rest 1).
We will not give any meaning to division with the remainder of an arbitrary integer a by a negative integer, but we will leave it with the right to exist.
Divisibility theorem for integers with remainder
When we talked about dividing natural numbers with remainder, we found out that dividend a, divisor b, incomplete quotient c and remainder d are related by the equality a = b c + d. The integers a, b, c, and d share the same relationship. This relationship is asserted by the following remainder divisibility theorem.
Theorem.
Any integer a can be represented uniquely through an integer and nonzero number b in the form a = b q + r, where q and r are some integers, moreover.
Proof.
First, we prove the possibility of representing a = b q + r.
If integers a and b are such that a is evenly divisible by b, then by definition there exists an integer q such that a = b q. In this case, the equality a = bq + r holds for r = 0.
Now we will assume that b is a positive integer. Let us choose an integer q such that the product b q does not exceed a, and the product b (q + 1) is already greater than a. That is, we take q such that the inequalities b q It remains to prove the possibility of representing a = b q + r for negative b. Since the modulus of the number b in this case is a positive number, then for there is a representation, where q 1 is some integer, and r is an integer satisfying the conditions. Then, taking q = −q 1, we obtain the required representation a = b q + r for negative b. We pass to the proof of uniqueness. Suppose that in addition to the representation a = bq + r, q and r are integers and, there is one more representation a = bq 1 + r 1, where q 1 and r 1 are some integers, and q 1 ≠ q and. After subtracting from the left and right sides of the first equality, respectively, the left and right sides of the second equality, we obtain 0 = b (q − q 1) + r − r 1, which is equivalent to the equality r − r 1 = b (q 1 −q) ... Then an equality of the form From the conditions and we can conclude that. Since q and q 1 are integers and q ≠ q 1, whence we conclude that The equality a = b c + d allows you to find the unknown dividend a if you know the divisor b, the incomplete quotient c, and the remainder d. Let's look at an example. Example. What is the dividend if dividing it by the integer −21 results in an incomplete quotient 5 and a remainder of 12? Solution. We need to calculate the dividend a when we know the divisor b = −21, the incomplete quotient c = 5, and the remainder d = 12. Turning to the equality a = b c + d, we get a = (- 21) 5 + 12. Observing, first we multiply the integers −21 and 5 according to the rule of multiplying integers with different signs, after which we add integers with different signs: (−21) 5 + 12 = −105 + 12 = −93. Answer: −93
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The connections between the dividend, divisor, incomplete quotient and remainder are also expressed by equalities of the form b = (a − d): c, c = (a − d): b and d = a − b · c. These equalities allow you to calculate the divisor, partial quotient, and remainder, respectively. We often have to find the remainder of dividing an integer a by an integer b when the dividend, divisor, and partial quotient are known, using the formula d = a − b · c. To avoid further questions, let's look at an example of calculating the remainder. Example. Find the remainder of dividing the integer −19 by the integer 3 if you know that the incomplete quotient is −7. Solution. To calculate the remainder of the division, we use a formula of the form d = a − b · c. From the condition we have all the necessary data a = −19, b = 3, c = −7. We get d = a − bc = −19−3 ). Answer: As we have noted more than once, positive integers are natural numbers. Therefore, division with remainder of positive integers is carried out according to all division rules with remainder of natural numbers. It is very important to be able to easily perform division with the remainder of natural numbers, since it is this that underlies not only division of positive integers, but also the basis of all division rules with the remainder of arbitrary integers. From our point of view, it is most convenient to perform long division, this method allows you to get both the incomplete quotient (or just the quotient) and the remainder. Consider an example of division with remainder of positive integers. Example. Divide 14 671 by 54 with the remainder. Solution. Let's perform the division of these positive integers by a column: The partial quotient turned out to be 271, and the remainder is 37. Answer: 14 671: 54 = 271 (rest 37). Let us formulate a rule that allows performing division with a remainder of a positive integer by a negative integer. The incomplete quotient of dividing a positive integer a by a negative integer b is the opposite of the incomplete quotient of dividing a by the modulus of b, and the remainder of dividing a by b is equal to the remainder of dividing by. It follows from this rule that the incomplete quotient of dividing a positive integer by a negative integer is a non-positive integer. Let's remake the announced rule into an algorithm for division with the remainder of a positive integer by a negative integer: Here is an example of using the algorithm for dividing a positive integer by a negative integer. Example. Divide the positive integer 17 by the negative integer −5. Solution. Let's use the algorithm of division with the remainder of a positive integer by a negative integer. Dividing The opposite of 3 is −3. Thus, the desired partial quotient of dividing 17 by −5 is −3, and the remainder is 2. Answer: 17: (- 5) = - 3 (rest 2). Example. Divide 45 to -15. Solution. The moduli of the dividend and the divisor are 45 and 15, respectively. The number 45 is divisible by 15 without a remainder, while the quotient is 3. Therefore, the positive integer 45 is divisible by the negative integer −15 without a remainder, the quotient being equal to the opposite number of 3, that is, −3. Indeed, according to the rule of dividing integers with different signs, we have. Answer: 45:(−15)=−3
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Let us give the formulation of the division rule with a remainder of a negative integer by a positive integer. To get an incomplete quotient c from dividing a negative integer a by a positive integer b, you need to take the opposite of the incomplete quotient from dividing the moduli of the original numbers and subtract one from it, then calculate the remainder d by the formula d = a − b c. From this rule of division with a remainder, it follows that the incomplete quotient of dividing a negative integer by a positive integer is a negative integer. From the sounded rule follows the division algorithm with the remainder of a negative integer a by a positive integer b: Let's analyze the solution of the example, in which we will use the written division algorithm with remainder. Example. Find the incomplete quotient and the remainder after dividing the negative integer -17 by the positive integer 5. Solution. The modulus of the dividend −17 is 17, and the modulus of the divisor 5 is 5. Dividing 17 by 5, we get an incomplete quotient 3 and a remainder of 2. The opposite number of 3 is −3. Subtract one from −3: −3−1 = −4. So, the required incomplete quotient is −4. It remains to calculate the remainder. In our example, a = −17, b = 5, c = −4, then d = a − b c = −17−5 (−4) = −17 - (- 20) = - 17 + 20 = 3 ... Thus, the partial quotient of dividing a negative integer -17 by a positive integer 5 is -4, and the remainder is 3. Answer: (−17): 5 = −4 (rest 3). Example. Divide the negative integer -1404 by the positive integer 26. Solution. The modulus of the dividend is 1 404, the modulus of the divisor is 26. Divide 1 404 by 26 with a column: Since the modulus of the dividend was divided by the modulus of the divisor without a remainder, the original integers are divisible without a remainder, and the desired quotient is equal to the number opposite to 54, that is, −54. Answer: (−1 404):26=−54
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Let's formulate the division rule with the remainder of negative integers. To get an incomplete quotient c from dividing a negative integer a by an integer negative number b, you need to calculate the incomplete quotient from dividing the moduli of the original numbers and add one to it, then calculate the remainder d by the formula d = a − b c. From this rule it follows that the incomplete quotient of the division of negative integers is a positive integer. Let's rewrite the stated rule in the form of an algorithm for dividing negative integers: Consider the application of the algorithm for dividing negative integers when solving an example. Example. Find the partial quotient and the remainder of the negative integer -17 divided by the negative integer -5. Solution. Let's use the appropriate modulo division algorithm. The modulus of the dividend is 17, the modulus of the divisor is 5. Division 17 by 5 gives an incomplete quotient of 3 and a remainder of 2. We add one to the incomplete quotient 3: 3 + 1 = 4. Therefore, the required incomplete quotient of dividing −17 by −5 is equal to 4. It remains to calculate the remainder. In this example, a = −17, b = −5, c = 4, then d = a − b c = −17 - (- 5) 4 = −17 - (- 20) = - 17 + 20 = 3 ... So, the incomplete quotient of dividing the negative integer -17 by the negative integer -5 is 4, and the remainder is 3. Answer: (−17): (- 5) = 4 (rest 3). After dividing the integers with remainder, it is useful to check the result. The check is carried out in two stages. At the first stage, it is checked whether the remainder d is a non-negative number, and the condition is also checked. If all the conditions of the first stage of verification are met, then you can proceed to the second stage of verification, otherwise it can be argued that a mistake was made somewhere during division with a remainder. At the second stage, the validity of the equality a = b c + d is checked. If this equality is true, then the division with the remainder was carried out correctly, otherwise, a mistake was made somewhere. Let's consider solutions of examples in which the result of division of integers with remainder is checked. Example. When dividing the number −521 by −12, you got an incomplete quotient 44 and a remainder of 7, check the result. Solution. −2 for b = −3, c = 7, d = 1. We have b c + d = −3 7 + 1 = −21 + 1 = −20... Thus, the equality a = b c + d is incorrect (in our example, a = −19). Therefore, the division with the remainder was carried out incorrectly. Read the topic of the lesson: "Division with remainder." What do you already know on this topic? Can you divide 8 plums evenly 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 that we have performed can be written like this. 8: 2 = 4
Do you think it is possible to divide 8 plums equally on 3 plates (fig. 3)? Rice. 3. Illustration for example We will act like this. First, put one plum in each plate, then the second plum. We will have 2 plums left, but 3 plates. This means that we cannot split further equally. We put 2 plums in each plate, and we have 2 plums left (fig. 4). Rice. 4. Illustration for example Let's continue our observation. Read the numbers. Among these numbers, find those that are divisible by 3. 11, 12, 13, 14, 15, 16, 17, 18, 19
Check yourself. The rest of the numbers (11, 13, 14, 16, 17, 19) are not divisible by 3, or they say "Share with the remainder." Let's find the value of the quotient. 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 performed action can be recorded as follows. 17: 3 = 5 (rest 2) You can also write in a column (Fig. 6) Rice. 6. Illustration for example Consider the drawings. Explain the captions for these figures (Fig. 7). Rice. 7. Illustration for example Consider the first figure (Fig. 8). Rice. 8. Illustration for example We see that 15 ovals were divided by 2. 2 was repeated 7 times, in the remainder - 1 oval. Consider the second figure (Fig. 9). Rice. 9. Illustration for example In this figure, 15 squares were divided into 4. 4 was repeated 3 times, in the remainder - 3 squares. Consider the third figure (Fig. 10). Rice. 10. Illustration for example We can say that 15 ovals were divided into 3. Each of 3 was repeated 5 times equally. In such cases, the remainder is said to be 0. Let's do the division. Divide the seven squares by three. We get two groups, and one square will remain. Let's write down the solution (fig. 11). Rice. 11. Illustration for example Let's do the division. We find out how many times four are contained in the number 10. We see that in the number 10, four are contained 2 times and 2 squares remain. Let's write down the solution (fig. 12). Rice. 12. Illustration for example Let's do the division. Find out how many times two are contained in number 11. We see that number 11 contains two by two 5 times and 1 square remains. Let's write down the solution (fig. 13). Rice. 13. Illustration for example Let's make a conclusion. To divide with remainder means to find out how many times the divisor is contained in the dividend and how many units remain. Division with remainder can also be performed on a numerical ray. On the numerical ray, mark the segments of 3 divisions and see that there were three divisions three times and one division remained (Fig. 14). Rice. 14. Illustration for example Let's write down the solution. 10: 3 = 3 (rest 1) Let's do the division. On the numerical ray, mark the segments of 3 divisions and see that there were three divisions three times and two divisions remained (Fig. 15). Rice. 15. Illustration for example Let's write down the solution. 11: 3 = 3 (rest 2) Let's do the division. On the numerical ray, mark the segments of 3 divisions and see that we got exactly 4 times, the remainder is absent (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 remainder, learned how to perform the named action using a picture and a number ray, and practiced solving examples on the topic of the lesson. Bibliography 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 remainder using the picture. 3. Perform division with the remainder using the number beam. 4. Make an assignment for your classmates on the topic of the lesson.
, and by virtue of the properties of the modulus of a number, the equality 
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... From the obtained inequalities and it follows that an equality of the form impossible under our assumption. Therefore, there is no other representation of the number a, except for a = b q + r.Relationships between dividend, divisor, incomplete quotient and remainder
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