Read the topic of the lesson: "Division with a remainder." What do you already know about this topic?
Can you divide 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 as follows.
8: 2 = 4
What do you think, is it possible to divide 8 plums equally into 3 plates (Fig. 3)?
Rice. 3. Illustration for example
Let's act like this. First, put one plum in each plate, then the second plum. We will have 2 plums left, but 3 plates. So we can't split it evenly. We put 2 plums in each plate, and we have 2 plums left (Fig. 4).
Rice. 4. Illustration for example
Let's continue monitoring.
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 "divide with the remainder."
Let's find the value of the private.
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 are left.
The action taken can be written as follows.
17: 3 = 5 (rest. 2)
It can also be written in a column (Fig. 6)
Rice. 6. Illustration for example
Review 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 by 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. 3 was repeated 5 times equally. In such cases, the remainder is said to be 0.
Let's do the division.
We divide the 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.
We find out how many times four is contained in the number 10. We see that in the number 10 four is 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.
We 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 make a conclusion. To divide with a remainder means to find out how many times the divisor is contained in the dividend and how many units remain.
Division with a remainder can also be performed on a number line.
On the number line, we mark segments of 3 divisions and we will see that three divisions turned out to be 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 beam, we mark segments of 3 divisions and we will see that three divisions turned out to be 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, we mark segments of 3 divisions and we will 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 picture and a number beam, practiced solving examples on the topic of the lesson.
Bibliography
- M.I. Moro, M.A. Bantova and others. Mathematics: Textbook. Grade 3: in 2 parts, part 1. - M .: "Enlightenment", 2012.
- M.I. Moro, M.A. Bantova and others. Mathematics: Textbook. Grade 3: in 2 parts, part 2. - M .: "Enlightenment", 2012.
- M.I. Moreau. Mathematics lessons: Guidelines for teachers. Grade 3 - M.: Education, 2012.
- Regulatory document. Monitoring and evaluation of learning outcomes. - M.: "Enlightenment", 2011.
- "School of Russia": Programs for elementary school. - M.: "Enlightenment", 2011.
- S.I. Volkov. Mathematics: Testing work. Grade 3 - M.: Education, 2012.
- V.N. Rudnitskaya. Tests. - M.: "Exam", 2012.
- Nsportal.ru ().
- Prosv.ru ().
- Do.gendocs.ru ().
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 the drawing.
3. Perform division with a remainder using the number line.
4. Make a task for your comrades on the topic of the lesson.
In this article, we will analyze integer division with remainder. Let's start with the general principle of dividing integers with a remainder, formulate and prove a theorem on the divisibility of integers with a remainder, and trace the connections between the dividend, divisor, partial quotient, and remainder. Next, we will announce 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.
Page navigation.
General idea of division of integers with a remainder
The division of integers with a remainder we will consider as a generalization of division with a remainder of natural numbers. This is due to the fact that natural numbers are a component of integers.
Let's start with the terms and notation that are used in the description.
By analogy with division of 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) is two integers c and d . The numbers a and b are called divisible and divider respectively, the number d is remainder from dividing a by b, and the integer c is called incomplete private(or simply private if the remainder is zero).
Let's agree that the remainder is a non-negative integer, and its value does not exceed b, that is, (we met similar chains of inequalities when we talked about comparing three or more integers).
If the number c is a partial 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 (residual d) .
Note that when an integer a is divided by an integer b, the remainder can be zero. In this case, we say that a is divisible by b without a trace(or completely). Thus, division of integers without a remainder is a special case of division of 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 section on the theory of division of zero by an integer), and the remainder will also be equal to zero.
We have decided on the terminology and notation, 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 a debt. Let's imagine such a situation. The debt that makes up the items must be repaid 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 paying off the debt. Let's take an example. Let's say 2 people owe 7 apples. If we assume that each of them owes 4 apples, then after paying the debt they will have 1 apple left. This situation corresponds to the equality (−7):2=−4 (remaining 1) .
We will not attach any meaning to division with a remainder of an arbitrary integer a by a negative integer, but we will leave it the right to exist.
Divisibility theorem for integers with remainder
When we talked about dividing natural numbers with a remainder, we found out that the dividend a, the divisor b, the partial quotient c and the remainder d are related by the equality a=b c+d. The integers a , b , c and d share the same relationship. This connection is confirmed by the following divisibility theorem with remainder.
Theorem.
Any integer a can be represented in a unique way through an integer and a non-zero number b in the form a=b q+r , where q and r are some integers, and .
Proof.
Let us first 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=b q+r holds for r=0 .
Now we will assume that b is a positive integer. We choose an integer q in such a way that the product b·q does not exceed the number 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 there is a representation for , where q 1 is some integer, and r is an integer that satisfies the conditions . Then, assuming q=−q 1 , we obtain the required representation a=b q+r for negative b . We turn to the proof of uniqueness. Suppose that in addition to the representation a=b q+r, q and r are integers and , there is another representation a=b q 1 +r 1 , where q 1 and r 1 are some integers, and q 1 ≠ q and . After subtracting from the left and right parts of the first equality, respectively, the left and right parts 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 the equality of the form From the conditions and we can conclude that . Since q and q 1 are integers and q≠q 1 , then , whence we conclude that The equality a=b c+d allows you to find an unknown dividend a if the divisor b, the partial quotient c and the remainder d are known. Consider an example. Example. What is the dividend equal to if its division by the integer −21 results in an incomplete quotient of 5 and a remainder of 12? Decision. We need to calculate the dividend a when we know the divisor b=−21 , the partial 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 carry out the multiplication of integers −21 and 5 according to the rule of multiplication of integers with different signs , after which we perform the addition of integers with different signs : (−21) 5+12=−105+12=−93 . Answer: −93
.
Relationships between the dividend, divisor, partial 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 us to calculate the divisor, partial quotient, and remainder, respectively. We often need 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 . In order to avoid further questions, we will analyze an example of calculating the remainder. Example. Find the remainder of dividing the integer −19 by the integer 3 if the partial quotient is known to be −7. Decision. 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−b c=−19−3 (−7)= −19−(−21)=−19+21=2 (the difference −19−(−21) we calculated by the rule of subtracting a negative integer ). Answer: As we have already noted more than once, positive integers are natural numbers. Therefore, division with a remainder of positive integers is carried out according to all the rules for division with a remainder of natural numbers. It is very important to be able to easily perform division with a remainder of natural numbers, since it is it that underlies the division of not only positive integers, but also the basis of all division rules with a remainder of arbitrary integers. From our point of view, it is most convenient to perform division by a column, this method allows you to get both an incomplete quotient (or just a quotient) and a remainder. Consider an example of division with a remainder of positive integers. Example. Perform a division with a remainder of 14671 by 54 . Decision. Let's perform the division of these positive integers by a column: The incomplete quotient turned out to be 271, and the remainder is 37. Answer: 14 671:54=271 (rest 37) . Let's formulate a rule that allows you to perform division with a remainder of a positive integer by a negative integer. The partial quotient of dividing a positive integer a by a negative integer b is the opposite of the partial quotient of dividing a by the modulus of b, and the remainder of dividing a by b is 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 voiced rule into an algorithm for dividing with a remainder of a positive integer by a negative integer: Let us give an example of using the algorithm for dividing a positive integer by a negative integer. Example. Divide with a remainder of a positive integer 17 by a negative integer −5 . Decision. Let's use the division algorithm with the remainder of a positive integer by a negative integer. Dividing The opposite number of 3 is −3. Thus, the required partial quotient of dividing 17 by −5 is −3, and the remainder is 2. Answer: 17 :(−5)=−3 (rest 2). Example. Divide 45 by -15 . Decision. The modules of the dividend and 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, while the quotient is equal to the number opposite to 3, that is, −3. Indeed, according to the rule of division of integers with different signs, we have . Answer: 45:(−15)=−3
.
Let us formulate the rule of division 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 number opposite to the incomplete quotient from dividing the modules of the original numbers and subtract one from it, after which the remainder d is calculated using 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 voiced 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 use the written division algorithm with a remainder. Example. Find the partial quotient and the remainder of the negative integer −17 divided by the positive integer 5 . Decision. 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 of 3 and a remainder of 2 . The opposite of 3 is −3 . Subtract one from −3: −3−1=−4 . So, the desired 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 negative integer −17 divided by positive integer 5 is −4, and the remainder is 3. Answer: (−17):5=−4 (rest. 3) . Example. Divide the negative integer −1 404 by the positive integer 26 . Decision. The dividend modulus is 1404, the divisor modulus is 26. Divide 1404 by 26 in a column: Since the modulus of the dividend was divided by the modulus of the divisor without a remainder, the original integers are divided without a remainder, and the desired quotient is equal to the number opposite to 54, that is, −54. Answer: (−1 404):26=−54
.
Let us formulate the division rule with the remainder of negative integers. To get an incomplete quotient c from dividing a negative integer a by a negative integer b, you need to calculate the incomplete quotient from dividing the modules of the original numbers and add one to it, after that, calculate the remainder d using 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 voiced 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. Decision. We use the appropriate division algorithm with a remainder. The dividend modulus is 17 , the divisor modulus is 5 . Division 17 times 5 gives the incomplete quotient 3 and the remainder 2. We add one to the incomplete quotient 3: 3+1=4. Therefore, the desired incomplete quotient of dividing −17 by −5 is 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 partial quotient of negative integer −17 divided by negative integer −5 is 4 , and the remainder is 3 . Answer: (−17):(−5)=4 (rest 3) . After the division of integers with a remainder is performed, it is useful to check the result. The verification is carried out in two stages. At the first stage, it is checked whether the remainder d is a non-negative number, and also the condition is 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 an error was made somewhere when dividing 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 a remainder was carried out correctly, otherwise, an error was made somewhere. Let's consider the solutions of examples in which the result of division of integers with a remainder is checked. Example. When dividing the number -521 by -12, the partial quotient was 44 and the remainder was 7 , check the result. Decision. −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, division with a remainder was carried out incorrectly. Division with remainder is the division of one number by another so that the remainder is not zero. It is not always possible to perform division, as there are cases when one number is not divisible by another. For example, the number 11 is not divisible by 3, since there is no such natural number that, when multiplied by 3, would give 11. When the division cannot be performed, it was agreed to divide not all the divisible, but only the largest part of it, which can only be divided into a divisor. In this example, the largest part of the dividend that can be divided by 3 is 9 (as a result we get 3), the remaining smaller part of the dividend - 2 will not be divided by 3. Speaking of dividing 11 by 3, 11 is still called divisible, 3 is a divisor, the result of division is the number 3, they call incomplete private, and the number 2 - remainder of division. The division itself in this case is called division with a remainder. An incomplete quotient is the largest number that, when multiplied by a divisor, gives a product that does not exceed the divisible. The difference between the dividend and this product is called the remainder. The remainder is always less than the divisor, otherwise it could also be divided by the divisor. Division with remainder can be written like this: 11: 3 = 3 (remainder 2) If, when one natural number is divided by another, the remainder is 0, then the first number is said to be evenly divisible by the second. For example, 4 is evenly divisible by 2. The number 5 is not even divisible by 2. The whole word is usually omitted for brevity and they say: such and such a number is divisible by another, for example: 4 is divisible by 2, and 5 is not divisible by 2. You can check the result of division with a remainder in the following way: multiply the incomplete quotient by the divisor (or vice versa) and add the remainder to the resulting product. If the result is a number equal to the dividend, then division with a remainder is done correctly: 11: 3 = 3 (remainder 2) In this article, we will take a close look at division with remainder. Let's start with a general idea about this action, then find out the meaning of dividing natural numbers with a remainder, and introduce the necessary terms. Then we outline the range of problems solved by dividing natural numbers with a remainder. In conclusion, let's dwell on all kinds of connections between the dividend, the divisor, the incomplete quotient and the remainder of the division. Page navigation. Answer: The dividend is 79. It should also be noted that checking the result of dividing natural numbers with a remainder is carried out by checking the validity of the resulting equality a=b·c+d . In its meaning, the remainder d is the number of elements that remains in the original set after the exclusion from its a elements b times c elements each. Therefore, by virtue of the sense of multiplication of natural numbers and the sense of subtraction of natural numbers, the equality d=a−b c. Thus, the remainder d of dividing a natural number a by a natural number b is equal to the difference between the dividend a and the product of the divisor b and the incomplete quotient c. The resulting connection d=a−b·c allows you to find the remainder when the dividend, divisor and incomplete quotient are known. Let's consider an example solution. From the general idea of dividing natural numbers with a remainder, we will move on, and in this article we will deal with the principles by which this action is carried out. Generally division with remainder has much in common with division of natural numbers without a remainder, so we will often refer to the material of this article. First, let's deal with the division of natural numbers with a remainder in a column. Next, we will show how you can find the result of dividing natural numbers with a remainder by sequential subtraction. After that, we will move on to the method of selecting an incomplete quotient, while not forgetting to give examples with a detailed description of the solution. Next, we 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 division of natural numbers with a remainder. Page navigation. One of the most convenient ways to divide natural numbers with a remainder is division by a column. In the article division of natural numbers by a column, we analyzed this method of division in great detail. We will not repeat ourselves here, but simply give a solution to one example. Example. Perform division with a remainder of the natural number 273844 by the natural number 97 . Decision. Let's divide by a column: So the partial quotient of 273844 divided by 97 is 2823 and the remainder is 13. Answer: 273 844:97=2 823 (rest 13) . You can find the incomplete quotient and the remainder of the division of natural numbers by successively subtracting the divisor. The essence of this approach is simple: from the elements of the existing set, sets are sequentially formed with the required number of elements until it is possible, the number of sets obtained gives an incomplete quotient, and the number of remaining elements in the original set is the remainder of the division. Let's take an example. Example. Let's say we need to divide 7 by 3 . Decision. Imagine that we need to put 7 apples into bags of 3 apples. From the initial 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. Of these, we again take 3 pieces, and put them in the second bag. After that, we are left with 4−3=1 apple. It is clear that the process ends here (we cannot form another package with the required number of apples, since the remaining number of apples 1 is less than the number we need 3). As a result, we have two packages with the required number of apples and one apple in the balance. Then, by virtue of the sense of dividing natural numbers with a remainder, it can be argued that we have obtained the following result 7:3=2 (remainder 1) . Answer: 7:3=2 (rest. 1) . Consider the solution of another example, while we present only mathematical calculations. Example. Divide the natural number 145 by 46 by subtracting successively. Decision. 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 subtract again, that is, this is where the process of sequential subtraction ends. As a result, we needed to sequentially subtract the divisor 46 from the dividend 145 3 times, after which we got the remainder 7. Thus 145:46=3 (res. 7) . Answer: 145:46=3 (rest. 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 incomplete quotient is equal to zero, and the remainder is equal to the dividend. That is, if a
It must also be said that it is good to perform division of natural numbers with a remainder in the considered way only when a small number of successive subtractions are required to obtain the result. When dividing given natural numbers a and b with a remainder, the incomplete quotient c can be found. Now we will show what the selection process is based on and how it should work. First, let's decide among which numbers to look for an incomplete quotient. When we talked about the meaning of dividing natural numbers with a remainder, we found out that the incomplete quotient can be either zero or a natural number, that is, one of the numbers 0, 1, 2, 3, ... Thus, the desired incomplete quotient is one of the written numbers, and it remains for us to sort through them to determine which number is the incomplete quotient. Next, we need an equation of the form d=a−b c , specifying , 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. Divisible a and divisor b are known to us from the beginning, 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. Moreover, 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. Perform division with a remainder of the natural number 267 by 21. Decision. Let's choose an incomplete quotient. In our example, a=267 , b=21 . We will successively give c the values 0 , 1 , 2 , 3 , …, calculating the value d=a−b·c at each step 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 of 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 got the number 15 , which is less than 21 , so the process can be considered completed. We picked up an incomplete quotient c=12 , while the remainder d turned out to be 15 . Answer: 267:21=12 (rest. 15) . In this subsection, we will consider an algorithm that allows us to carry out division with a remainder of a natural number a by a natural number b in cases where the method of successive subtraction (and the method of choosing an incomplete quotient) requires too many computational operations. We note right away that if the dividend a is less than the divisor b, then we know both the incomplete 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 incomplete quotient c and the remainder d . The equality a=b c+d defines the relationship between the dividend, divisor, partial quotient and remainder. It follows from the written equality that if we represent 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. It remains only 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 carry out the solution of 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 cyclically repeated over and over again until all the terms are found that add up to the dividend. In the final sixth paragraph, the resulting sum is converted to the form b c + d (if the resulting sum no longer has this form), from which the desired incomplete quotient and the remainder become visible. So, we proceed to the representation of the dividend 899 as the sum of several terms. First, we calculate how much the number of characters in the dividend entry is greater than the number of characters in the divisor entry, and remember this number. In our example, there are 3 digits in the dividend record (899 is a three-digit number), and in the divisor record there are two digits (47 is a two-digit number), therefore, there is one more sign in the dividend record, 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 subtract 1 from the number memorized in the previous paragraph. Let's return to our example. In the record of the divisor 47, we add one digit to the right 0, 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 that, to the number 1 on the right, we attribute the numbers 0 in the amount determined by the number memorized in the previous paragraph. In this case, we get a unit of discharge, with which we will work further. In our example, to the number 1 we assign 1 the number 0, in this case we get the number 10, that is, we will work with the tens digit. 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 divisible. We found out that in our example, the working digit is the tens digit. Therefore, we first multiply the divisor by one unit of the tens place, that is, we multiply 47 by 10, we get 47 10 \u003d 470 . The resulting number 470 is less than the dividend 899, so we proceed to multiply the divisor by two units of the tens digit, 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 in sequential multiplication is the first of the required terms. In the example being analyzed, the desired term is the number 470 (this number is equal to the product 47 100 , we will use this equality later). After that, we find the difference between the dividend and the first term found. If the resulting number is greater than the divisor, then proceed to find the second term. To do this, we repeat all the described steps of the algorithm, but we already take the number obtained here as a dividend. If at this point again a number is obtained that is greater than the divisor, then we proceed to finding the third term, once again repeating the steps of the algorithm, taking the resulting number as a 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 has happened, then we take the number obtained here as the last required term (looking ahead, let's say that it is equal to the remainder), and proceed 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 a dividend and repeat all the steps of the algorithm with it. In the entry of the number 429 there is one sign more than in the entry of the number 47, therefore, remember the number 1. Now, in the record 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 memorized 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 you do not 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 digit of units. Now we successively 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 required term 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 step. Well, here we come to the final stage. All previous actions were aimed at presenting the dividend as the sum of several terms. Now it remains to convert the resulting sum to the form b·c+d . The distributive property of multiplication with respect to addition will help us cope with this task. After that, the desired incomplete quotient and the remainder will become visible. In our example, the dividend 899 is equal to the sum of the 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 has been converted to the form we need 899=47 19+6 , from which it is easy to find the incomplete quotient 19 and the remainder 6 . So, 899:47=19 (res. 6) . Of course, when solving examples, you will not describe the process of division with a remainder in such detail.
, and due to the properties of the modulus of the number - and the equality 
.
. 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, partial quotient, and remainder
Division with a remainder of positive integers, examples
The rule of division with a remainder of a positive integer by a negative integer, examples
Division with a remainder of a negative integer by a positive integer, examples
Division rule with a remainder of negative integers, examples
Checking the result of dividing integers with a remainder
Checking division with a remainder
Finding the remainder if the dividend, divisor and incomplete quotient are known
Division of natural numbers in a column with a remainder
Division of natural numbers with a remainder through successive subtraction
Selection of an incomplete quotient
Algorithm for dividing natural numbers with a remainder, examples, solutions
