General idea of division of natural numbers with a remainder. Division of integers with a remainder, rules, examples How to perform division with a remainder
Division of multi-digit numbers is easiest to do in a column. Column division is also called corner division.
Before we begin performing division by a column, let us consider in detail the very form of recording division by a column. First, we write down the dividend and put a vertical bar to the right of it:
Behind the vertical line, opposite the dividend, we write the divisor and draw a horizontal line under it:
Under the horizontal line, the quotient resulting from the calculations will be written in stages:
Under the dividend, intermediate calculations will be written:
The full form of division by a column is as follows:
How to divide by a column
Let's say we need to divide 780 by 12, write the action in a column and start dividing:
The division by a column is carried out in stages. The first thing we need to do is define the incomplete dividend. Look at the first digit of the dividend:
this number is 7, since it is less than the divisor, then we cannot start dividing from it, so we need to take one more digit from the dividend, the number 78 is greater than the divisor, so we start dividing from it:
In our case, the number 78 will be incomplete divisible, it is called incomplete because it is just a part of the divisible.
Having determined the incomplete dividend, we can find out how many digits there will be in the quotient, for this we need to calculate how many digits are left in the dividend after the incomplete dividend, in our case there is only one digit - 0, which means that the quotient will consist of 2 digits.
Having found out the number of digits that should turn out in a private one, you can put dots in its place. If, at the end of the division, the number of digits turned out to be more or less than the indicated points, then a mistake was made somewhere:
Let's start dividing. We need to determine how many times 12 is contained in the number 78. To do this, we successively multiply the divisor by natural numbers 1, 2, 3, ... until we get a number as close as possible to the incomplete divisible or equal to it, but not exceeding it. Thus, we get the number 6, write it under the divisor, and subtract 72 from 78 (according to the rules of column subtraction) (12 6 \u003d 72). After we subtracted 72 from 78, we got a remainder of 6:
Please note that the remainder of the division shows us whether we have chosen the right number. If the remainder is equal to or greater than the divisor, then we did not choose the correct number and we need to take a larger number.
To the resulting remainder - 6, we demolish the next digit of the dividend - 0. As a result, we got an incomplete dividend - 60. We determine how many times 12 is contained in the number 60. We get the number 5, write it into the quotient after the number 6, and subtract 60 from 60 ( 12 5 = 60). The remainder is zero:
Since there are no more digits left in the dividend, it means that 780 is divided by 12 completely. As a result of performing division by a column, we found the quotient - it is written under the divisor:
Consider an example where zeros are obtained in the quotient. Let's say we need to divide 9027 by 9.
We determine the incomplete dividend - this is the number 9. We write it into the quotient 1 and subtract 9 from 9. The remainder turned out to be zero. Usually, if in intermediate calculations the remainder is zero, it is not written down:
We demolish the next digit of the dividend - 0. We recall that when dividing zero by any number, there will be zero. We write to private zero (0: 9 = 0) and subtract 0 from 0 in intermediate calculations. Usually, in order not to pile up intermediate calculations, the calculation with zero is not written down:
We demolish the next digit of the dividend - 2. In intermediate calculations, it turned out that the incomplete dividend (2) is less than the divisor (9). In this case, zero is written into the quotient and the next digit of the dividend is taken down:
We determine how many times 9 is contained in the number 27. We get the number 3, write it into a quotient, and subtract 27 from 27. The remainder is zero:
Since there are no more digits left in the dividend, it means that the number 9027 is divided by 9 completely:
Consider an example where the dividend ends in zeros. Let's say we need to divide 3000 by 6.
We determine the incomplete dividend - this is the number 30. We write it into the quotient 5 and subtract 30 from 30. The remainder is zero. As already mentioned, it is not necessary to write down zero in the remainder in intermediate calculations:
We demolish the next digit of the dividend - 0. Since when dividing zero by any number there will be zero, we write it to private zero and subtract 0 from 0 in intermediate calculations:
We demolish the next digit of the dividend - 0. We write one more zero into the quotient and subtract 0 from 0 in intermediate calculations. at the very end of the calculation, it is usually written to show that the division is complete:
Since there are no more digits left in the dividend, it means that 3000 is divided by 6 completely:
Division by a column with a remainder
Let's say we need to divide 1340 by 23.
We determine the incomplete dividend - this is the number 134. We write in the quotient 5 and subtract 115 from 134. The remainder turned out to be 19:
We demolish the next digit of the dividend - 0. Determine how many times 23 is contained in the number 190. We get the number 8, write it into a quotient, and subtract 184 from 190. We get the remainder 6:
Since there are no more digits left in the dividend, the division is over. The result is an incomplete quotient of 58 and a remainder of 6:
1340: 23 = 58 (remainder 6)
It remains to consider an example of division with a remainder, when the dividend is less than the divisor. Suppose we need to divide 3 by 10. We see that 10 is never contained in the number 3, so we write it to the quotient 0 and subtract 0 from 3 (10 0 = 0). We draw a horizontal line and write down the remainder - 3:
3: 10 = 0 (remainder 3)
Column Division Calculator
This calculator will help you perform division by a column. Just enter the dividend and divisor and click the Calculate button.
What does 3rd grade do in math? Division with remainder, examples and tasks - that's what is studied in the lessons. Division with a remainder and the algorithm for such calculations will be discussed in the article.
Peculiarities
Consider the topics included in the program that Grade 3 is studying. Division with a remainder is 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, we divide 21 by 6. It turns out 3, but the remainder remains 3.
In cases where, during the division of natural numbers, the remainder is equal to zero, they say that the division was made by an integer. For example, if 25 is divided by 5, the result is 5. The remainder is zero.
Solution of examples
In order to perform division with a remainder, a specific notation is used.
Let's give examples in mathematics (Grade 3). Division with a remainder can be left out. It is enough to write in a line: 13:4=3 (remainder 1) or 17:5=3 (remainder 2).
Let's analyze everything in more detail. For example, when 17 is divided by three, the integer five is obtained, in addition, the remainder is two. What is the procedure for solving such an example for division with a remainder? First you need to find the maximum number up to 17, which can be divided without a remainder by three. The largest will be 15.
Next, 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 divisible, that is, subtract 15 from 17, we get two. Mandatory action is the reconciliation of the divisor and the remainder. After verification, the response of the action taken is necessarily recorded. 17:3=15 (remainder 2).
If the remainder is greater than the divisor, the action was not performed correctly. It is according to this algorithm that the 3rd class division with a remainder performs. 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 topics that grade 3 faces is division with a remainder. Examples can be complex, especially when additional column calculations are required.
Let's say you need to divide the number 190 by 27 to get the minimum remainder. Let's try to solve the problem using multiplication.
We select a number that, when multiplied, will give a 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, the remainder will be 28. It turned out to be more than the original divisor. Therefore, the number six is not suitable for our example as a multiplier. 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 obtained 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 at school (Grade 3, division with remainder). Examples always include a response record. The whole mathematical expression can be formulated as follows: 190:27=7 (remainder 1). Similar calculations can be made in a column.
This is how class 3 division with a remainder performs. The examples given above will help to understand the algorithm for solving such problems.
Conclusion
In order for primary school students to form the correct computational skills, the teacher, during mathematics classes, must pay attention to explaining the algorithm of the child’s actions when solving tasks for division with a remainder.
According to the new federal state educational standards, special attention is paid to an individual approach to learning. The teacher should select tasks for each child, taking into account his individual abilities. At each stage of teaching the rules of division with a 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 get the desired result.
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. At all 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.
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Division of natural numbers in a column with a remainder
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 .
Solution.
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) .
Division of natural numbers with a remainder through successive subtraction
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 .
Solution.
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.
Solution.
145−46=99 (if necessary, refer to the article subtraction of natural numbers). Since 99 is greater than 46 , we subtract the divisor a second time: 99−46=53 . Since 53>46 , we subtract the divisor a third time: 53−46=7 . Since 7 is less than 46, we will not be able to 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.
Selection of an incomplete quotient
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. Dividend 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 partial 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.
Solution.
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) .
Algorithm for dividing natural numbers with a remainder, examples, solutions
In this section, we 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 does not already have 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, while 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, 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.
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.
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.
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)
