Many numbers cannot be divided completely; when dividing, there is often a remainder other than zero. In this article, we will discuss how to divide natural numbers with the remainder and consider their application in detail with examples.
Let's start with division of natural numbers with a remainder in a column, then we will consider division using successive subtraction. Finally, we finish with an analysis of the method of selecting an incomplete quotient. We present an algorithm for division with a remainder for the most general case and show how to check the result of division of natural numbers with a remainder.
This is one of the most convenient ways to divide. It is described in detail in a separate article devoted to division of natural numbers by a column. Here we will not give the whole theory anew, but we will concentrate on the case of division with a remainder.
We will give an example solution, since it is easiest to understand the essence of the method in practice.
Example 1. How to divide natural numbers with a remainder?
Divide the natural number 273844 by the natural number 97 .
We divide by a column and write:
Result: partial quotient is 2823 , and the remainder is 13 .
Division of numbers with a remainder through successive subtraction
To find the incomplete quotient and remainder, you can resort to successive subtraction of the divisor from the dividend. This method is not always appropriate, but in some cases it is very convenient to use. Let's look at the example again.
Example 2. Division with a remainder through successive subtraction.
Let's say we have 7 apples. We need to put these 7 apples into bags of 3 apples. In other words, 7 divided by 3 .
We take 3 pieces from the initial number of apples and put them in one package. We will have 7 - 3 = 4 apples left. Now, from the remaining apples, we again take away 3 pieces and put them in another bag. There are 4 - 3 = 1 apple left.
1 apple is the remainder of the division, since at this stage we can no longer form another package with three apples and the division is, in fact, completed. Division result:
7 ÷ 3 = 2 (remainder 1)
This means that the number 3, as it were, fits in the number 7 twice, and the unit is a remainder less than 3.
Let's consider one more example. This time, we will give only mathematical calculations, without resorting to analogies.
Example 3. Division with a remainder through successive subtraction.
Let's calculate: 145 ÷ 46 .
The number 99 is greater than 46 , so we continue the sequential subtraction of the divisor:
We repeat this operation one more time:
As a result, we needed to successively subtract the divisor from the dividend 3 times before we got the remainder - the result of subtraction, which is less than the divisor. In our case, the remainder is the number 7.
145 ÷ 46 = 3 (remainder 7) .
The successive subtraction method is not suitable when the dividend is less than the divisor. In this case, you can immediately write down the answer: the incomplete quotient is zero, and the remainder is equal to the most divisible.
If a< b , то a ÷ b = 0 (остаток a) .
For example:
12 ÷ 36 = 0 (remainder 12) 47 ÷ 88 = 0 (remainder 47)
Also regarding the successive subtraction method, it should be noted that it is convenient only in cases where the entire division operation is reduced to a small number of subtractions. If the dividend is many times larger than the divisor, using this method will be impractical and involve a lot of cumbersome calculations.
Incomplete quotient selection method
When dividing natural numbers with a remainder, you can calculate the result by choosing an incomplete quotient. We will show how the selection process can be conducted, and what it is based on.
First, we determine among which numbers we need to look for an incomplete quotient. From the very definition of the division process, it is clear that the incomplete quotient is equal to zero, or is one of the natural numbers 1, 2, 3, etc.
Secondly, we will establish a relationship between the divisor, dividend, partial quotient and remainder. Consider the equation d = a - b c . Here d is the remainder of the division, a is the dividend, b is the divisor, c is the partial quotient.
Thirdly, let's not forget that the remainder is always less than the divisor.
Now let's take a look at the selection process. The dividend a and the divisor b are known to us from the very beginning. As an incomplete quotient, we will successively take numbers from the series 0, 1, 2, 3, etc. Applying the formula d = a - b c and calculating the resulting value with a divisor, we end the process when the remainder d is less than the divisor b . The number taken for c at this step will be an incomplete quotient.
Let's look at the application of this method with an example.
Example 4. Division with a remainder by selection
Divide 267 by 21.
a = 267 b = 21 . Let's choose an incomplete quotient.
Let's use the formula d = a - b · c and iterate over c , giving it the values 0 , 1 , 2 , 3 , and so on.
If c \u003d 0, we have: d \u003d a - b c \u003d 267 - 21 0 \u003d 267. The number 267 is greater than 21 , so we continue the substitution.
With c \u003d 1 we have: d \u003d a - b c \u003d 267 - 21 1 \u003d 246. Because 246 > 21, repeat the process again.
With c \u003d 2 we have: d \u003d a - b c \u003d 267 - 21 2 \u003d 267 - 42 \u003d 225; 225 > 21 .
With c \u003d 3 we have: d \u003d a - b c \u003d 267 - 21 3 \u003d 267 - 63 \u003d 204; 204 > 21 .
With c \u003d 12 we have: d \u003d a - b c \u003d 267 - 21 12 \u003d 267 - 252 \u003d 15; 15< 21 .
Algorithm for dividing natural numbers with a remainder
When the partial quotient and successive subtraction methods discussed above require too cumbersome calculations, the following method is used for division with a remainder. Consider an algorithm for dividing a natural number a by a number b with a remainder.
Recall that if a< b, неполное частное равно нулю, а остаток равен делимомому a . Мы будем рассматривать случай, когда a >b.
We formulate three questions and answer them:
- What is known there?
- What do we need to find?
- How are we going to do it?
Initially, the dividend and the divisor are known: a and b.
You need to find the incomplete quotient c and the remainder d.
Here is a formula that defines the relationship between the dividend, divisor, incomplete quotient and remainder. a = b c + d. It is this ratio that we will take as the basis of the algorithm for dividing natural numbers with a remainder. The dividend a must be represented as a sum a = b c + d, then we will find the required values.
The division algorithm, thanks to which we represent a as a sum a = b c + d, is very similar to the algorithm for dividing natural numbers without a remainder. Below are the steps of the algorithm using the example of dividing the number 899 by 47.
1. First of all, we look at the dividend and the divisor. We find out and remember how many digits the number in the dividend entry is greater than the number in the divisor. In our specific example The dividend has three digits, and the divisor has two.
Let's remember this number.
2. On the right in the divisor entry, add the number of zeros determined by the difference between the number of characters in the dividend and the divisor. In our case, you need to add one zero. If the written number is greater than the divisible, then you need to subtract one from the number memorized in the first paragraph.
In our example, we add zero to the right of 47. Since 470< 899 , запомненное в предыдущем пункте число не нужно уменьшать на единицу. Таким образом, число 1 так и остается у нас в памяти.
3. On the right to the number 1 we attribute the number of zeros, equal to the number defined in the previous paragraph. In our example, assigning one zero to one, we get the number 10. As a result of this action, we received a working unit of the discharge, with which we will work further.
4. We will sequentially multiply the divisor by 1, 2, 3. . etc. units of the working digit until we get a number that is greater than or equal to the divisible.
The working digit in our example is tens. After multiplying the divisor by one unit of the working bit, we get 470.
470 < 899 , поэтому умножаем на еще одну единицу рабочего разряда. Получаем: 47 · 20 = 940 ; 940 > 899 .
The number that we got at the penultimate step (470 = 47 10) is the first of the required terms.
5. Find the difference between the dividend and the first term found. If the resulting number is greater than the divisor, then we proceed to finding the second term.
We repeat steps 1 - 5, however, we take the number obtained here as a dividend. If again we get a number greater than the divisor, again repeat steps 1 - 5 in a circle, but with a new number as a dividend. We continue until the number obtained here is less than the divisor. Let's move on to the final stage. Looking ahead, let's say that the last number received will be equal to the remainder.
Let's look at an example. 899 - 470 = 429, 429 > 47. We repeat steps 1 - 5 of the algorithm with the number 429 taken as a dividend.
1. In the entry of the number 429, there is one sign more than in the entry of the number 47. We remember the difference - the number 1.
2. In the record of the dividend on the right, we add one zero. We get the number 470. Since 470 > 429, subtract 1 from the number 1 memorized in the previous paragraph and get 1 - 1 = 0. We remember 0 .
3. Since in the previous paragraph we got the number 0 and remembered it, we do not need to add any zero to the one on the right. Thus, the working digit is units
4. Sequentially multiply the divisor 47 by 1 , 2 , 3 . . etc. We will not give detailed calculations, but pay attention to the final result: 47 9 = 423< 429 , 47 · 10 = 470 >429 . Thus, the second required term is 47 9 = 423.
5. The difference between 429 and 423 is equal to the number 6 . Since 6< 47 , это третье, и последнее искомое слагаемое. Перейдем к завершающему этапу алгоритма деления столбиком.
6. The purpose of the previous steps was to represent the dividend as the sum of several terms. For our example, we got 899 = 470 + 423 + 6 . Recall that 470 = 47 10 , 423 = 47 9 . Let's rewrite the equation:
899 = 47 10 + 47 9 + 6
Apply the distributive property of multiplication.
899 = 47 10 + 47 9 + 6 = 47 (10 + 9) + 6
899 = 47 19 + 6 .
Thus, we presented the dividend in the form of the previously given formula a = b c + d.
Required unknowns: incomplete quotient c \u003d 19, remainder d \u003d 6.
Of course, when deciding practical examples there is no need to describe all the actions in such detail. Let's show it:
Example 5. Division of natural numbers with a remainder
Divide the numbers 42252 and 68.
Let's use an algorithm. The first five steps give the first term - the number 40800 = 68 600 .
We repeat the first five steps of the algorithm again with the number 1452 = 42252 - 40800 and get the second term 1360 = 68 20
The third time we go through the steps of the aglorhythm, but with the new number 92 = 1452 - 1360. The third term is equal to 68 = 68 1 . The remainder is 24 = 92 - 68.
As a result, we get:
42252 = 40800 + 1360 + 68 + 24 = 68 600 + 68 20 + 68 1 + 24 = = 68 (600 + 20 + 1) + 24 = 68 621 + 24
The incomplete quotient is 621, the remainder is 24.
Division of natural numbers with a remainder. Checking the result
Division of natural numbers with a remainder, especially when large numbers, a rather laborious and cumbersome process. Anyone can make a mistake in calculations. That is why, checking the result of the division will help you understand if you did everything right. Checking the result of division of natural numbers with a remainder is performed in two stages.
At the first stage, we check if the remainder is greater than the divisor. If not, then all is well. Otherwise, we can conclude that something went wrong.
Important!
The remainder is always less than the divisor!
At the second stage, the validity of the equality a = b · c + d is checked. If the equality after substituting the values turns out to be true, then the division was performed without errors.
Example 6. Checking the result of dividing natural numbers with a remainder.
Let's check if it is true that 506 ÷ 28 = 17 (remainder 30) .
Compare the remainder and divisor: 30 > 28 .
So the division is wrong.
Example 7. Checking the result of division of natural numbers with a remainder.
The student divided 121 by 13 and as a result received an incomplete quotient 9 with a remainder of 5. Did he do the right thing?
To find out, we first compare the remainder and the divisor: 5< 13 .
The first checkpoint has been passed, let's move on to the second.
Let's write down the formula a = b c + d. a = 121 ; b = 13; c = 9 d = 5 .
Substitute the values and compare the results
13 9 + 5 = 117 + 5 = 122; 121 ≠ 122
It means that an error has crept into the student's calculations somewhere.
Example 8. Checking the result of division of natural numbers with a remainder.
The student performed laboratory work in physics. During the execution, he needed to divide 5998 by 111 . As a result, he got the number 54 with a remainder of 4. Is everything calculated correctly?
Let's check! The remainder 4 is less than the divisor 111 , so we proceed to the second stage of the check.
We use the formula a \u003d b c + d, where a \u003d 5998; b = 111 ; c = 54 ; d = 4 .
After substitution, we have:
5998 = 111 54 + 4 = 5994 + 4 = 5998 .
The equality is correct, which means that the division is correct.
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
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 called largest number, which, when multiplied by a divisor, gives a product that does not exceed the dividend. 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.
Checking division with a remainder
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)
The article analyzes the concept of division of integers with a remainder. We will prove the theorem on the divisibility of integers with a remainder and look at the connections between divisibles and divisors, incomplete quotients and remainders. Consider the rules when division of integers with remainders is performed, having examined in detail with examples. At the end of the solution, we will perform a check.
General understanding of division of integers with remainders
Division of integers with a remainder is considered as a generalized division with a remainder of natural numbers. This is done because natural numbers are a constituent of integers.
Division with a remainder of an arbitrary number says that the integer a is divisible by the number b , which is different from zero. If b = 0 then no division with remainder is performed.
As well as the division of natural numbers with a remainder, the division of integers a and b is performed, with b different from zero, by c and d. In this case, a and b are called dividend and divisor, and d is the remainder of the division, c is an integer or partial quotient.
If we assume that the remainder is a non-negative integer, then its value is not greater than the modulus of the number b. Let's write it this way: 0 ≤ d ≤ b . This chain of inequalities is used when comparing 3 or more numbers.
If c is an incomplete quotient, then d is the remainder of dividing an integer a by b, you can briefly fix: a: b \u003d c (remain d).
The remainder when dividing the numbers a by b is possible zero, then they say that a is divided by b completely, that is, without a remainder. Division without a remainder is considered a special case of division.
If we divide zero by some number, we get zero as a result. The remainder of the division will also be zero. This can be seen from the theory of division of zero by an integer.
Now consider the meaning of division of integers with a remainder.
It is known that positive integers are natural, then when dividing with a remainder, the same meaning will be obtained as when dividing natural numbers with a remainder.
Dividing a negative integer a by a positive integer b makes sense. Let's look at an example. Imagine a situation where we have a debt of items in the amount a that needs to be repaid by b people. To do this, everyone needs to contribute equally. To determine the amount of debt for each, it is necessary to pay attention to the value of private c. The remainder d indicates that the number of items after paying off debts is known.
Let's take an example with apples. If 2 people need 7 apples. If we calculate that everyone must return 4 apples, after the full calculation they will have 1 apple left. Let's write this as an equality: (− 7) : 2 = − 4 (о с t. 1) .
Dividing any number a by an integer does not make sense, but it is possible as an option.
Divisibility theorem for integers with remainder
We found that a is the dividend, then b is the divisor, c is the partial quotient, and d is the remainder. They are interconnected. We will show this relationship using the equality a = b · c + d . The relationship between them is characterized by the divisibility theorem with remainder.
Theorem
Any integer can only be represented in terms of an integer and a non-zero number b in this way: a = b · q + r , where q and r are some integers. Here we have 0 ≤ r ≤ b .
Let us prove the possibility of the existence of a = b · q + r .
Proof
If there are two numbers a and b, and a is divisible by b without a remainder, then it follows from the definition that there is a number q, that the equality a = b · q will be true. Then the equality can be considered true: a = b q + r for r = 0.
Then it is necessary to take q such that given by the inequality b · q< a < b · (q + 1) было верным. Необходимо вычесть b · q из всех частей выражения. Тогда придем к неравенству такого вида: 0 < a − b · q < b .
We have that the value of the expression a − b · q is greater than zero and not greater than the value of the number b, hence it follows that r = a − b · q . We get that the number a can be represented as a = b · q + r.
Now we need to consider the possibility of representing a = b · q + r for negative values of b .
The modulus of the number turns out to be positive, then we get a = b q 1 + r, where the value q 1 is some integer, r is an integer that fits the condition 0 ≤ r< b . Принимаем q = − q 1 , получим, что a = b · q + r для отрицательных b .
Proof of uniqueness
Assume that a = b q + r , q and r are integers with the condition 0 ≤ r< b , имеется еще одна форма записи в виде a = b · q 1 + r 1 , где q 1 And r1 are some numbers where q 1 ≠ q, 0 ≤ r1< b .
When the inequality is subtracted from the left and right sides, then we get 0 = b · (q − q 1) + r − r 1 , which is equivalent to r - r 1 = b · q 1 - q . Since the module is used, we get the equality r - r 1 = b · q 1 - q.
The given condition says that 0 ≤ r< b и 0 ≤ r 1 < b запишется в виде r - r 1 < b . Имеем, что q And q 1- whole, and q ≠ q 1, then q 1 - q ≥ 1 . Hence we have that b · q 1 - q ≥ b . The resulting inequalities r - r 1< b и b · q 1 - q ≥ b указывают на то, что такое равенство в виде r - r 1 = b · q 1 - q невозможно в данном случае.
It follows from this that the number a cannot be represented in any other way, except by such a notation a = b · q + r.
Relationship between dividend, divisor, partial quotient and remainder
Using the equality a \u003d b c + d, you can find the unknown dividend a when the divisor b is known with an incomplete quotient c and the remainder d.
Example 1
Determine the dividend if, when dividing, we get - 21, an incomplete quotient 5 and a remainder 12.
Solution
It is necessary to calculate the dividend a with a known divisor b = − 21, an incomplete quotient c = 5, and a remainder d = 12. We need to refer to the equality a = b c + d, from here we get a = (− 21) 5 + 12. Subject to the order of operations, we multiply - 21 by 5, after which we get (- 21) 5 + 12 = - 105 + 12 = - 93.
Answer: - 93 .
The relationship between the divisor and the partial quotient and the remainder can be expressed using the equalities: b = (a − d) : c , c = (a − d) : b and d = a − b · c . With their help, we can calculate the divisor, partial quotient and remainder. This boils down to constantly finding the remainder of dividing an integer a by b with a known dividend, divisor, and partial quotient. The formula d = a − b · c is applied. Let's consider the solution in detail.
Example 2
Find the remainder of dividing an integer - 19 by an integer 3 with a known incomplete quotient equal to - 7 .
Solution
To calculate the remainder of a division, we apply a formula of the form d = a − b c . By condition, all data a = − 19 , b = 3 , c = − 7 are available. From here we get d \u003d a - b c \u003d - 19 - 3 (- 7) \u003d - 19 - (- 21) \u003d - 19 + 21 \u003d 2 (difference - 19 - (- 21)... This example is calculated by the subtraction rule whole negative number.
Answer: 2 .
All positive integers are natural. It follows that the division is performed according to all the rules of division with a remainder of natural numbers. The speed of division with a remainder of natural numbers is important, since not only the division of positive ones is based on it, but also the rules for dividing arbitrary integers.
The most convenient division method is a column, since it is easier and faster to get an incomplete or just a quotient with a remainder. Let's consider the solution in more detail.
Example 3
Divide 14671 by 54 .
Solution
This division must be done in a column:
That is, the incomplete quotient is equal to 271, and the remainder is 37.
Answer: 14671: 54 = 271. (rest. 37)
The rule of division with a remainder of a positive integer by a negative integer, examples
To perform division with a remainder of a positive number by a negative integer, it is necessary to formulate a rule.
Definition 1
The incomplete quotient of dividing a positive integer a by a negative integer b gives a number that is opposite to the incomplete quotient of dividing the modules of numbers a by b. Then the remainder is the remainder when a is divided by b.
Hence we have that the incomplete quotient of dividing a positive integer by a negative integer is considered a non-positive integer.
We get the algorithm:
- divide the modulus of the dividend by the modulus of the divisor, then we get an incomplete quotient and
- remainder;
- write down the opposite number.
Consider the example of the algorithm for dividing a positive integer by a negative integer.
Example 4
Perform division with a remainder of 17 by - 5 .
Solution
Let's apply the division algorithm with the remainder of a positive integer by a negative integer. It is necessary to divide 17 by - 5 modulo. From here we get that the incomplete quotient is 3, and the remainder is 2.
We get that the desired number from dividing 17 by - 5 \u003d - 3 with a remainder equal to 2.
Answer: 17: (− 5) = − 3 (remaining 2).
Example 5
Divide 45 by - 15 .
Solution
It is necessary to divide the numbers modulo. We divide the number 45 by 15, we get the quotient 3 without a remainder. So the number 45 is divisible by 15 without a remainder. In the answer we get - 3, since the division was carried out modulo.
45: (- 15) = 45: - 15 = - 45: 15 = - 3
Answer: 45: (− 15) = − 3 .
The formulation of the division rule with a remainder is as follows.
Definition 2
In order to get an incomplete quotient c when dividing a negative integer a by a positive b, you need to apply the opposite of this number and subtract 1 from it, then the remainder d will be calculated by the formula: d = a − b · c.
Based on the rule, we can conclude that when dividing, we get a non-negative integer. For the accuracy of the solution, the algorithm for dividing a by b with a remainder is used:
- find the modules of the dividend and divisor;
- divide modulo;
- write the opposite of the given number and subtract 1 ;
- use the formula for the remainder d = a − b c .
Consider an example of a solution where this algorithm is applied.
Example 6
Find the incomplete quotient and the remainder of the division - 17 by 5.
Solution
We divide the given numbers modulo. We get that when dividing, the quotient is 3, and the remainder is 2. Since we got 3 , the opposite is 3 . It is necessary to subtract 1 .
− 3 − 1 = − 4 .
The desired value is equal to - 4 .
To calculate the remainder, you need a = − 17 , b = 5 , c = − 4 , then d = a − b c = − 17 − 5 (− 4) = − 17 − (− 20) = − 17 + 20 = 3 .
This means that the incomplete quotient of division is the number - 4 with a remainder equal to 3.
Answer:(− 17) : 5 = − 4 (remaining 3).
Example 7
Divide the negative integer - 1404 by the positive 26 .
Solution
It is necessary to divide by a column and by modulus.
We got the division of modules of numbers without a remainder. This means that the division is performed without a remainder, and the desired quotient = - 54.
Answer: (− 1 404) : 26 = − 54 .
Division rule with a remainder of negative integers, examples
It is necessary to formulate a division rule with a remainder of integers negative numbers.
Definition 3
To obtain an incomplete quotient from dividing a negative integer a by a negative integer b, it is necessary to perform modulo calculations, after which add 1, then we can calculate using the formula d = a − b · c.
It follows from this that the incomplete quotient of the division of negative integers will be a positive number.
We formulate this rule in the form of an algorithm:
- find the modules of the dividend and divisor;
- divide the modulus of the dividend by the modulus of the divisor to obtain an incomplete quotient with
- remainder;
- adding 1 to the incomplete quotient;
- calculation of the remainder, based on the formula d = a − b c .
Let's consider this algorithm with an example.
Example 8
Find the partial quotient and remainder when dividing - 17 by - 5 .
Solution
For the correctness of the solution, we apply the algorithm for division with a remainder. First, divide the numbers modulo. From here we get that the incomplete quotient \u003d 3, and the remainder is 2. According to the rule, it is necessary to add the incomplete quotient and 1. We get that 3 + 1 = 4 . From this we obtain that the incomplete quotient of division given numbers equals 4 .
To calculate the remainder, we will apply the formula. By condition, we have that a \u003d - 17, b \u003d - 5, c \u003d 4, then, using the formula, we get d \u003d a - b c \u003d - 17 - (-5) 4 \u003d - 17 - (- 20) = − 17 + 20 = 3 . The desired answer, that is, the remainder, is 3, and the incomplete quotient is 4.
Answer:(− 17) : (− 5) = 4 (remaining 3).
Checking the result of dividing integers with a remainder
After performing the division of numbers with a remainder, it is necessary to perform a check. This check involves 2 stages. First, the remainder d is checked for non-negativity, the condition 0 ≤ d< b . При их выполнении разрешено выполнять 2 этап. Если 1 этап не выполнился, значит вычисления произведены с ошибками. Второй этап состоит из того, что равенство a = b · c + d должно быть верным. Иначе в вычисления имеется ошибка.
Let's look at examples.
Example 9
Produced division - 521 by - 12. The quotient is 44, the remainder is 7. Run a check.
Solution
Since the remainder is a positive number, its value is less than the modulus of the divisor. The divisor is -12, so its modulus is 12. You can move on to the next checkpoint.
By condition, we have that a = - 521 , b = - 12 , c = 44 , d = 7 . From here we calculate b c + d , where b c + d = − 12 44 + 7 = − 528 + 7 = − 521 . It follows that the equality is true. Check passed.
Example 10
Check division (− 17) : 5 = − 3 (remaining − 2). Is equality true?
Solution
The meaning of the first stage is that it is necessary to check the division of integers with a remainder. This shows that the action was performed incorrectly, since the remainder is given, equal to - 2. The remainder is not a negative number.
We have that the second condition is satisfied, but insufficient for this case.
Answer: No.
Example 11
The number - 19 divided by - 3 . The partial quotient is 7 and the remainder is 1. Check if this calculation is correct.
Solution
Given a remainder of 1. He is positive. The value is less than the divider module, which means that the first stage is performed. Let's move on to the second stage.
Let's calculate the value of the expression b · c + d . By condition, we have that b \u003d - 3, c \u003d 7, d \u003d 1, therefore, substituting the numerical values, we get b c + d \u003d - 3 7 + 1 \u003d - 21 + 1 \u003d - 20. It follows that a = b · c + d equality is not satisfied, since the condition is given a = - 19 .
This implies that the division was made with an error.
Answer: No.
If you notice a mistake in the text, please highlight it and press Ctrl+Enter
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. Math Lessons: Guidelines for the teacher. 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: Verification 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.
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.
