Dividing integers with a remainder, rules, examples. Division with remainder
Signs of divisibility of numbers- these are rules that allow you to relatively quickly find out, without dividing, whether this number is divisible by a given number without a remainder.
Some of signs of divisibility quite simple, some more complicated. On this page you will find both signs of divisibility of prime numbers, such as, for example, 2, 3, 5, 7, 11, and signs of divisibility of composite numbers, such as 6 or 12.
I hope this information will be useful to you.
Happy learning!
Test for divisibility by 2
This is one of the simplest signs of divisibility. It sounds like this: if the notation of a natural number ends with an even digit, then it is even (divisible without a remainder by 2), and if the notation of a natural number ends with an odd digit, then this number is odd.
In other words, if the last digit of a number is 2
, 4
, 6
, 8
or 0
- the number is divisible by 2, if not, then it is not divisible
For example, numbers: 23 4
, 8270
, 1276
, 9038
, 502
are divisible by 2 because they are even.
A numbers: 23 5
, 137
, 2303
They are not divisible by 2 because they are odd.
Test for divisibility by 3
This sign of divisibility has completely different rules: if the sum of the digits of a number is divisible by 3, then the number is divisible by 3; If the sum of the digits of a number is not divisible by 3, then the number is not divisible by 3.
This means that in order to understand whether a number is divisible by 3, you just need to add together the numbers that make it up.
It looks like this: 3987 and 141 are divisible by 3, because in the first case 3+9+8+7= 27
(27:3=9 - divisible by 3), and in the second 1+4+1= 6
(6:3=2 - also divisible by 3).
But the numbers: 235 and 566 are not divisible by 3, because 2+3+5= 10
and 5+6+6= 17
(and we know that neither 10 nor 17 are divisible by 3 without a remainder).
Test for divisibility by 4
This sign of divisibility will be more complicated. If the last 2 digits of a number form a number divisible by 4 or it is 00, then the number is divisible by 4, otherwise the given number is not divisible by 4 without a remainder.
For example: 1 00
and 3 64
are divisible by 4 because in the first case the number ends in 00
, and in the second on 64
, which in turn is divisible by 4 without a remainder (64:4=16)
Numbers 3 57
and 8 86
are not divisible by 4 because neither 57
neither 86
are not divisible by 4, which means they do not correspond to this criterion of divisibility.
Divisibility test by 5
And again we have a fairly simple sign of divisibility: if the notation of a natural number ends with the number 0 or 5, then this number is divisible by 5 without a remainder. If the notation of a number ends with another digit, then the number is not divisible by 5 without a remainder.
This means that any numbers ending in digits 0
And 5
, for example 1235 5
and 43 0
, fall under the rule and are divisible by 5.
And, for example, 1549 3
and 56 4
do not end with the number 5 or 0, which means they cannot be divided by 5 without a remainder.
Test for divisibility by 6
We have before us the composite number 6, which is the product of the numbers 2 and 3. Therefore, the sign of divisibility by 6 is also composite: in order for a number to be divisible by 6, it must correspond to two signs of divisibility at the same time: the sign of divisibility by 2 and the sign of divisibility by 3. Please note that such a composite number as 4 has an individual sign of divisibility, because it is the product of the number 2 by itself. But let's return to the test of divisibility by 6.
The numbers 138 and 474 are even and meet the criteria for divisibility by 3 (1+3+8=12, 12:3=4 and 4+7+4=15, 15:3=5), which means they are divisible by 6. But 123 and 447, although they are divisible by 3 (1+2+3=6, 6:3=2 and 4+4+7=15, 15:3=5), but they are odd, which means they do not correspond to the criterion of divisibility by 2, and therefore do not correspond to the criterion of divisibility by 6.
Test for divisibility by 7
This test of divisibility is more complex: a number is divisible by 7 if the result of subtracting twice the last digit from the number of tens of this number is divisible by 7 or equal to 0.
It sounds quite confusing, but in practice it is simple. See for yourself: the number 95
9 is divisible by 7 because 95
-2*9=95-18=77, 77:7=11 (77 is divided by 7 without a remainder). Moreover, if difficulties arise with the number obtained during the transformation (due to its size it is difficult to understand whether it is divisible by 7 or not, then this procedure can be continued as many times as you deem necessary).
For example, 45
5 and 4580
1 have the properties of divisibility by 7. In the first case, everything is quite simple: 45
-2*5=45-10=35, 35:7=5. In the second case we will do this: 4580
-2*1=4580-2=4578. It is difficult for us to understand whether 457
8 by 7, so let's repeat the process: 457
-2*8=457-16=441. And again we will use the divisibility test, since we still have a three-digit number in front of us 44
1. So, 44
-2*1=44-2=42, 42:7=6, i.e. 42 is divisible by 7 without a remainder, which means 45801 is divisible by 7.
Here are the numbers 11
1 and 34
5 is not divisible by 7 because 11
-2*1=11-2=9 (9 is not divisible by 7) and 34
-2*5=34-10=24 (24 is not divisible by 7 without a remainder).
Divisibility test by 8
The test for divisibility by 8 sounds like this: if the last 3 digits form a number divisible by 8, or it is 000, then the given number is divisible by 8.
Numbers 1 000
or 1 088
divisible by 8: the first one ends in 000
, the second 88
:8=11 (divisible by 8 without remainder).
And here are the numbers 1 100
or 4 757
are not divisible by 8 because numbers 100
And 757
are not divisible by 8 without a remainder.
Divisibility test by 9
This sign of divisibility is similar to the sign of divisibility by 3: if the sum of the digits of a number is divisible by 9, then the number is divisible by 9; If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
For example: 3987 and 144 are divisible by 9, because in the first case 3+9+8+7= 27
(27:9=3 - divisible by 9 without remainder), and in the second 1+4+4= 9
(9:9=1 - also divisible by 9).
But the numbers: 235 and 141 are not divisible by 9, because 2+3+5= 10
and 1+4+1= 6
(and we know that neither 10 nor 6 are divisible by 9 without a remainder).
Signs of divisibility by 10, 100, 1000 and other digit units
I combined these signs of divisibility because they can be described in the same way: a number is divided by a digit unit if the number of zeros at the end of the number is greater than or equal to the number of zeros at a given digit unit.
In other words, for example, we have the following numbers: 654 0
, 46400
, 867000
, 6450
. of which all are divisible by 1 0
; 46400
and 867 000
are also divisible by 1 00
; and only one of them is 867 000
divisible by 1 000
.
Any numbers that have less trailing zeroes than the digit unit are not divisible by that digit unit, for example 600 30
and 7 93
not divisible 1 00
.
Divisibility test by 11
In order to find out whether a number is divisible by 11, you need to obtain the difference between the sums of the even and odd digits of this number. If this difference is equal to 0 or is divisible by 11 without a remainder, then the number itself is divisible by 11 without a remainder.
To make it clearer, I suggest looking at examples: 2
35
4 is divisible by 11 because ( 2
+5
)-(3+4)=7-7=0. 29
19
4 is also divisible by 11, since ( 9
+9
)-(2+1+4)=18-7=11.
Here's 1 1
1 or 4
35
4 is not divisible by 11, since in the first case we get (1+1)- 1
=1, and in the second ( 4
+5
)-(3+4)=9-7=2.
Divisibility test by 12
The number 12 is composite. Its sign of divisibility is compliance with the signs of divisibility by 3 and 4 at the same time.
For example, 300 and 636 correspond to both the signs of divisibility by 4 (the last 2 digits are zeros or are divisible by 4) and the signs of divisibility by 3 (the sum of the digits of both the first and third numbers are divisible by 3), but finally, they are divisible by 12 without a remainder.
But 200 or 630 are not divisible by 12, because in the first case the number meets only the criterion of divisibility by 4, and in the second - only the criterion of divisibility by 3. but not both criteria at the same time.
Divisibility test by 13
A sign of divisibility by 13 is that if the number of tens of a number added to the units of this number multiplied by 4 is a multiple of 13 or equal to 0, then the number itself is divisible by 13.
Let's take for example 70
2. So, 70
+4*2=78, 78:13=6 (78 is divisible by 13 without a remainder), which means 70
2 is divisible by 13 without a remainder. Another example is a number 114
4. 114
+4*4=130, 130:13=10. The number 130 is divisible by 13 without a remainder, which means the given number corresponds to the criterion of divisibility by 13.
If we take the numbers 12
5 or 21
2, then we get 12
+4*5=32 and 21
+4*2=29, respectively, and neither 32 nor 29 are divisible by 13 without a remainder, which means the given numbers are not divisible by 13 without a remainder.
Divisibility of numbers
As can be seen from the above, it can be assumed that for any of the natural numbers you can select your own individual sign of divisibility or a “composite” sign if the number is a multiple of several different numbers. But as practice shows, generally the larger the number, the more complex its sign. It is possible that the time spent checking the divisibility criterion may be equal to or greater than the division itself. That's why we usually use the simplest signs of divisibility.
Let's look at a simple example:
15:5=3
In this example we divided the natural number 15 completely by 3, without remainder.
Sometimes a natural number cannot be completely divided. For example, consider the problem:
There were 16 toys in the closet. There were five children in the group. Each child took the same number of toys. How many toys does each child have?
Solution:
Divide the number 16 by 5 using a column and we get:
We know that 16 cannot be divided by 5. The nearest smaller number that is divisible by 5 is 15 with a remainder of 1. We can write the number 15 as 5⋅3. As a result (16 – dividend, 5 – divisor, 3 – incomplete quotient, 1 – remainder). Got formula division with remainder which can be done checking the solution.
a=
b⋅
c+
d
a – divisible,
b - divider,
c – incomplete quotient,
d - remainder.
Answer: each child will take 3 toys and one toy will remain.
Remainder of the division
The remainder must always be less than the divisor.
If during division the remainder is zero, then this means that the dividend is divided completely or without a remainder on the divisor.
If during division the remainder is greater than the divisor, this means that the number found is not the largest. There is a greater number that will divide the dividend and the remainder will be less than the divisor.
Questions on the topic “Division with remainder”:
Can the remainder be greater than the divisor?
Answer: no.
Can the remainder be equal to the divisor?
Answer: no.
How to find the dividend using the incomplete quotient, divisor and remainder?
Answer: We substitute the values of the partial quotient, divisor and remainder into the formula and find the dividend. Formula:
a=b⋅c+d
Example #1:
Perform division with remainder and check: a) 258:7 b) 1873:8
Solution:
a) Divide by column: 
258 – dividend,
7 – divider,
36 – incomplete quotient,
6 – remainder. The remainder is less than the divisor 6<7.
7⋅36+6=252+6=258
b) Divide by column: 
1873 – divisible,
8 – divisor,
234 – incomplete quotient,
1 – remainder. The remainder is less than divisor 1<8.
Let’s substitute it into the formula and check whether we solved the example correctly:
8⋅234+1=1872+1=1873
Example #2:
What remainders are obtained when dividing natural numbers: a) 3 b)8?
Answer:
a) The remainder is less than the divisor, therefore less than 3. In our case, the remainder can be 0, 1 or 2.
b) The remainder is less than the divisor, therefore less than 8. In our case, the remainder can be 0, 1, 2, 3, 4, 5, 6 or 7.
Example #3:
What is the largest remainder that can be obtained when dividing natural numbers: a) 9 b) 15?
Answer:
a) The remainder is less than the divisor, therefore less than 9. But we need to indicate the largest remainder. That is, the number closest to the divisor. This is the number 8.
b) The remainder is less than the divisor, therefore, less than 15. But we need to indicate the largest remainder. That is, the number closest to the divisor. This number is 14.
Example #4:
Find the dividend: a) a:6=3(rest.4) b) c:24=4(rest.11)
Solution:
a) Solve using the formula:
a=b⋅c+d
(a – dividend, b – divisor, c – partial quotient, d – remainder.)
a:6=3(rest.4)
(a – dividend, 6 – divisor, 3 – partial quotient, 4 – remainder.) Let’s substitute the numbers into the formula:
a=6⋅3+4=22
Answer: a=22
b) Solve using the formula:
a=b⋅c+d
(a – dividend, b – divisor, c – partial quotient, d – remainder.)
s:24=4(rest.11)
(c – dividend, 24 – divisor, 4 – partial quotient, 11 – remainder.) Let’s substitute the numbers into the formula:
с=24⋅4+11=107
Answer: c=107
Task:
Wire 4m. need to be cut into 13cm pieces. How many such pieces will there be?
Solution:
First you need to convert meters to centimeters.
4m.=400cm.
We can divide by a column or in our mind we get:
400:13=30(remaining 10)
Let's check:
13⋅30+10=390+10=400
Answer: You will get 30 pieces and 10 cm of wire will remain.
The article examines the concept of dividing integers with a remainder. Let's prove the theorem on the divisibility of integers with a remainder and look at the connections between dividends and divisors, incomplete quotients and remainders. Let's look at the rules when dividing integers with remainders, looking at them in detail using 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 component of integers.
Division with a remainder of an arbitrary number says that the integer a is divided by a number b other than zero. If b = 0, then do not divide with a remainder.
Just like dividing natural numbers with a remainder, integers a and b are divided, with b not zero, by c and d. In this case, a and b are called the dividend and divisor, and d is the remainder of the division, c is an integer or incomplete 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 the integer a by b, which can be briefly stated: a: b = c (remainder d).
The remainder when dividing numbers a by b can be zero, then they say that a is divisible 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, the result is zero. The remainder of the division will also be zero. This can be traced from the theory of dividing zero by an integer.
Now let's look at the meaning of dividing integers with a remainder.
It is known that positive integers are natural numbers, 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 of a that needs to be repaid by b person. To achieve this, everyone needs to contribute equally. To determine the amount of debt for each, you need to pay attention to the value of the private s. The remainder d indicates that the number of items after paying off debts is known.
Let's look at the example of apples. If 2 people owe 7 apples. If we calculate that everyone must return 4 apples, after the full calculation they will have 1 apple left. Let us write this as an equality: (− 7) : 2 = − 4 (from t. 1) .
Dividing any number a by an integer does not make sense, but it is possible as an option.
Theorem on the divisibility of integers with remainder
We have identified that a is the dividend, then b is the divisor, c is the partial quotient, and d is the remainder. They are connected to each other. We will show this connection using the equality a = b · c + d. The connection between them is characterized by the divisibility theorem with remainder.
Theorem
Any integer can only be represented through an integer and 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, and 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, it follows that r = a − b · q. We find that the number a can be represented in the form a = b · q + r.
We now need to consider 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 meets the condition 0 ≤ r< b . Принимаем q = − q 1 , получим, что a = b · q + r для отрицательных b .
Proof of uniqueness
Let's assume that a = b q + r, q and r are integers with the condition 0 ≤ r true< b , имеется еще одна форма записи в виде a = b · q 1 + r 1 , где q 1 And r 1 are some numbers where q 1 ≠ q, 0 ≤ r 1< 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 obtain 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. From here 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 that the number a cannot be represented in any other way except by writing a = b · q + r.
Relationship between dividend, divisor, partial quotient and remainder
Using the equality a = b · c + d, you can find the unknown dividend a when the divisor b with the incomplete quotient c and the remainder d is known.
Example 1
Determine the dividend if, upon division, we get - 21, the partial quotient is 5 and the remainder is 12.
Solution
It is necessary to calculate the dividend a with a known divisor b = − 21, incomplete quotient c = 5 and remainder d = 12. We need to turn to the equality a = b · c + d, from here we get a = (− 21) · 5 + 12. If we follow the order of the actions, we multiply - 21 by 5, after which we get (− 21) · 5 + 12 = − 105 + 12 = − 93.
Answer: - 93 .
The connection between the divisor and the partial quotient and 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 comes down to constantly finding the remainder when dividing an integer of integers 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 when dividing the integer - 19 by the integer 3 with a known incomplete quotient equal to - 7.
Solution
To calculate the remainder of division, we apply a formula of the form d = a − b · c. By condition, all data are available: a = − 19, b = 3, c = − 7. From here we get d = a − b · c = − 19 − 3 · (− 7) = − 19 − (− 21) = − 19 + 21 = 2 (difference − 19 − (− 21). This example is calculated using the subtraction rule a negative integer.
Answer: 2 .
All positive integers are natural numbers. It follows that division is performed according to all the rules of division with a remainder of natural numbers. The speed of division with the remainder of natural numbers is important, since not only the division of positive numbers, but also the rules for dividing arbitrary integers are based on it.
The most convenient method of division is a column, since it is easier and faster to get an incomplete or simply a quotient with a remainder. Let's look at the solution in more detail.
Example 3
Divide 14671 by 54.
Solution
This division must be done in a column:
That is, the partial quotient is equal to 271, and the remainder is 37.
Answer: 14,671: 54 = 271. (rest 37)
The rule for dividing with a remainder 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 the positive integer a by the negative integer b yields a number that is opposite to the incomplete quotient of dividing the moduli of numbers a by b. Then the remainder is equal to 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;
- Let's write down the opposite number to what we got.
Let's look at the example of the algorithm for dividing a positive integer by a negative integer.
Example 4
Divide with remainder 17 by - 5.
Solution
Let's apply the algorithm for dividing with a remainder a positive integer by a negative integer. It is necessary to divide 17 by - 5 modulo. From here we get that the partial quotient is equal to 3, and the remainder is equal to 2.
We get that the required number from dividing 17 by - 5 = - 3 with a remainder equal to 2.
Answer: 17: (− 5) = − 3 (remaining 2).
Example 5
You need to divide 45 by - 15.
Solution
It is necessary to divide the numbers modulo. Divide the number 45 by 15, we get the quotient of 3 without a remainder. This means that the number 45 is divisible by 15 without a remainder. The answer is - 3, since the division was carried out modulo.
45: (- 15) = 45: - 15 = - 45: 15 = - 3
Answer: 45: (− 15) = − 3 .
The formulation of the rule for division with a remainder is as follows.
Definition 2
In order to obtain an incomplete quotient c when dividing a negative integer a by a positive b, you need to apply the opposite of the given 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. To ensure the accuracy of the solution, use the algorithm for dividing a by b with a remainder:
- find the modules of the dividend and divisor;
- divide modulo;
- write down the opposite of the given number and subtract 1;
- use the formula for the remainder d = a − b · c.
Let's look at an example of a solution where this algorithm is used.
Example 6
Find the partial quotient and remainder of division - 17 by 5.
Solution
We divide the given numbers modulo. We find that when dividing, the quotient is 3 and the remainder is 2. Since we got 3, the opposite is 3. You need 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 column and module.
We got the division of the 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 remainder for negative integers, examples
It is necessary to formulate a rule for division with a remainder of negative integers.
Definition 3
To obtain an incomplete quotient c from dividing a negative integer a by a negative integer b, it is necessary to perform modulo calculations, then add 1, then we can perform calculations using the formula d = a − b · c.
It follows that the incomplete quotient of dividing negative integers will be a positive number.
Let us 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 look at this algorithm using 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 this we get that the partial quotient = 3 and the remainder is 2. According to the rule, you need to add the incomplete quotient and 1. We get that 3 + 1 = 4. From here we get that the partial quotient of dividing the given numbers is equal to 4.
To calculate the remainder we will use the formula. By condition we have that a = − 17, b = − 5, c = 4, then, using the formula, we get d = a − b c = − 17 − (− 5) 4 = − 17 − (− 20) = − 17 + 20 = 3 . The required answer, that is, the remainder, is equal to 3, and the partial quotient is equal to 4.
Answer:(− 17) : (− 5) = 4 (remaining 3).
Checking the result of dividing integers with a remainder
After dividing numbers with a remainder, you must perform a check. This check involves 2 stages. First, the remainder d is checked for non-negativity, the condition 0 ≤ d is satisfied< 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
The division is made - 521 by - 12. The quotient is 44, the remainder is 7. Perform check.
Solution
Since the remainder is a positive number, its value is less than the modulus of the divisor. The divisor is - 12, which means its modulus is 12. You can move on to the next check point.
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. Verification passed.
Example 10
Perform division check (− 17): 5 = − 3 (remaining − 2). Is equality true?
Solution
The point of the first stage is that it is necessary to check the division of integers with a remainder. From this it is clear that the action was performed incorrectly, since a remainder equal to - 2 is given. The remainder is not a negative number.
We have that the second condition is met, but not sufficient for this case.
Answer: No.
Example 11
The number - 19 was divided by - 3. The partial quotient is 7 and the remainder is 1. Check whether this calculation was performed correctly.
Solution
Given a remainder equal to 1. He's positive. The value is less than the divider module, which means that the first stage is being completed. 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 = − 3, c = 7, d = 1, which means, substituting the numerical values, we get b · c + d = − 3 · 7 + 1 = − 21 + 1 = − 20. It follows that a = b · c + d the equality does not hold, since the condition gives a = - 19.
From this it follows that the division was made with an error.
Answer: No.
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In this article we will look at division of integers with remainder. Let's start with the general principle of dividing integers with a remainder, formulate and prove the theorem on the divisibility of integers with a remainder, and trace the connections between the dividend, divisor, incomplete quotient and remainder. Next, we will outline the rules by which integers are divided with a remainder, and consider the application of these rules when solving examples. After this, we will learn how to check the result of dividing integers with a remainder.
Page navigation.
General understanding of dividing integers with a remainder
We will consider division of integers with a remainder 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 designations that are used in the description.
By analogy with the division of natural numbers with a remainder, we will 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 accordingly, the number d – the remainder from dividing a by b, and the integer c is called incomplete private(or simply private, if the remainder is zero).
Let us agree to assume that the remainder is a non-negative integer, and its value does not exceed b, that is, (we encountered similar chains of inequalities when we talked about comparing three or more integers).
If the number c is an incomplete quotient, and the number d is the remainder of dividing the integer a by the integer b, then we will briefly write this fact as an equality of the form a:b=c (remaining d).
Note that when dividing an integer a by an integer b, the remainder may 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 are always dealing with division without a remainder, since in this case the quotient will be equal to zero (see the theory section of dividing zero by an integer), and the remainder will also be equal to zero.
We’ve decided on the terminology and notation, now let’s understand the meaning of dividing integers with a remainder.
Dividing a negative integer a by a positive integer b can also be given meaning. To do this, consider a negative integer as debt. Let's imagine this situation. The debt that constitutes the items must be repaid by b people by making an equal contribution. The absolute value of the incomplete quotient c in this case will determine the amount of debt of each of these people, and the remainder d will show how many items will remain after the debt is paid. Let's give 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 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 reserve its right to exist.
Theorem on the divisibility of integers with remainder
When we talked about dividing natural numbers with a remainder, we found out that the dividend a, divisor b, partial quotient c and remainder d are related by the equality a=b·c+d. The integers a, b, c and d have the same relationship. This connection is confirmed as follows divisibility theorem with remainder.
Theorem.
Any integer a can be uniquely represented through an integer and non-zero number b in the form a=b·q+r, where q and r are some integers, and .
Proof.
First, we prove the possibility of representing a=b·q+r.
If integers a and b are such that a is divisible by b, then by definition there is an integer q such that a=b·q. In this case, the equality a=b·q+r at r=0 holds.
Now we will assume that b is a positive integer. Let us choose an integer q so 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 for there is a representation where q 1 is some integer, and r is an integer that satisfies the conditions. Then, taking q=−q 1, we obtain the representation we need a=b·q+r for negative b. Let's move on 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 the left and right sides of the second equality from the left and right sides of the first equality, respectively, we obtain 0=b·(q−q 1)+r−r 1, which is equivalent to the equality r−r 1 =b·(q 1 −q) . Then an equality of the form From the conditions we can conclude that. Since q and q 1 are integers and q≠q 1, then we conclude that The equality a=b·c+d allows you to find the unknown dividend a if the divisor b, partial quotient c and remainder d are known. Let's look at an example. Example. What is the value of the dividend if, when divided by the integer −21, the result is an incomplete quotient of 5 and a remainder of 12? Solution. We need to calculate the dividend a when the divisor b=−21, the partial quotient c=5 and the remainder d=12 are known. Turning to the equality a=b·c+d, we obtain a=(−21)·5+12. Observing, we first multiply the integers −21 and 5 according to the rule for multiplying integers with different signs, after which we perform the addition of integers with different signs: (−21)·5+12=−105+12=−93. Answer: −93
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The connections 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 you to calculate the divisor, partial quotient, and remainder, respectively. We will often have to find the remainder when dividing an integer a by an integer b when the dividend, divisor and partial quotient are known, using the formula d=a−b·c. To avoid any further questions, let’s look at an example of calculating the remainder. Example. Find the remainder when dividing the integer −19 by the integer 3 if you know that the partial quotient is equal to −7. Solution. To calculate the remainder of 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 (we calculated the difference −19−(−21) using the rule of subtracting a negative integer ). Answer: As we have 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 this that underlies the division of not only positive integers, but also the basis of all rules for division with a remainder of arbitrary integers. From our point of view, it is most convenient to perform column division; this method allows you to obtain both an incomplete quotient (or simply a quotient) and a remainder. Let's look at an example of division with a remainder of positive integers. Example. Divide with remainder 14,671 by 54. Solution. Let's divide these positive integers with a column: The partial quotient turned out to be equal to 271, and the remainder is equal to 37. Answer: 14 671:54=271 (rest. 37) . Let us formulate a rule that allows us 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 equal to the remainder of dividing by. From this rule it follows that the partial quotient of dividing a positive integer by a negative integer is a non-positive integer. Let’s transform the stated rule into an algorithm for dividing with a remainder a positive integer by a negative integer: Let's give an example of using the algorithm for dividing a positive integer by a negative integer. Example. Divide with a remainder of the positive integer 17 by the negative integer −5. Solution. Let's use the algorithm for dividing with a remainder a positive integer by a negative integer. By 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 (remaining 2). Example. Divide 45 by −15. Solution. The modules of the dividend and divisor are 45 and 15, respectively. The number 45 is divisible by 15 without a remainder, and the quotient is 3. Therefore, the positive integer 45 is divided by the negative integer −15 without a remainder, and the quotient is equal to the number opposite 3, that is, −3. Indeed, according to the rule for dividing integers with different signs, we have . Answer: 45:(−15)=−3
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Let us give the formulation of the rule for dividing with a remainder a negative integer by a positive integer. To obtain 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 moduli 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 partial quotient of dividing a negative integer by a positive integer is a negative integer. From the stated rule follows an algorithm for dividing with a remainder a negative integer a by a positive integer b: Let's analyze the solution to the example, in which we use the written division algorithm with a remainder. Example. Find the partial quotient and remainder when dividing the negative integer −17 by the positive integer 5. Solution. The modulus of the dividend −17 is equal to 17, and the modulus of the divisor 5 is equal to 5. By dividing 17 by 5, we get the partial quotient 3 and the remainder 2. The opposite of 3 is −3. Subtract one from −3: −3−1=−4. So, the required partial quotient is equal to −4. All that remains is to calculate the remainder. In our example a=−17 , b=5 , c=−4 , then d=a−b·c=−17−5·(−4)= −17−(−20)=−17+20=3 . Thus, the partial quotient of dividing the negative integer −17 by the positive integer 5 is −4, and the remainder is 3. Answer: (−17):5=−4 (remaining 3) . Example. Divide the negative integer −1,404 by the positive integer 26. Solution. The module of the dividend is 1404, the module of the divisor is 26. Divide 1,404 by 26 using a column: Since the module of the dividend is divided by the module of the divisor without a remainder, the original integers are divided without a remainder, and the desired quotient is equal to the number opposite 54, that is, −54. Answer: (−1 404):26=−54
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Let us formulate the rule for division with a remainder of negative integers. To obtain 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 which the remainder d is calculated using the formula d=a−b·c. From this rule it follows that the partial quotient of dividing negative integers is a positive integer. Let's rewrite the stated rule in the form of an algorithm for dividing negative integers: Let's consider the use of the algorithm for dividing negative integers when solving an example. Example. Find the partial quotient and remainder when dividing a negative integer −17 by a negative integer −5. Solution. Let's use the appropriate division algorithm with a remainder. The module of the dividend is 17, the module of the divisor is 5. Division 17 over 5 gives the partial quotient 3 and the remainder 2. To the incomplete quotient 3 we add one: 3+1=4. Therefore, the required partial quotient of dividing −17 by −5 is equal to 4. All that remains is 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 dividing a negative integer −17 by a negative integer −5 is 4, and the remainder is 3. Answer: (−17):(−5)=4 (remaining 3) . After dividing integers with a remainder, 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 checks whether the condition is satisfied. 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 look at solutions to examples in which the result of dividing 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. Solution. −2 for b=−3, c=7, d=1. We have b·c+d=−3·7+1=−21+1=−20. Thus, the equality a=b·c+d is incorrect (in our example a=−19). Therefore, division with a remainder was carried out incorrectly.
, and due to the properties of the modulus of numbers, the equality 
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. From the obtained inequalities and it follows that an equality of the form impossible under our assumption. Therefore, there is no other representation of the number a other than a=b·q+r.Relationships between dividend, divisor, partial quotient and remainder
Division with remainder of positive integers, examples
The rule for dividing with a remainder a positive integer by a negative integer, examples
Division with remainder of a negative integer by a positive integer, examples
Division rule with remainder for negative integers, examples
Checking the result of dividing integers with a remainder
