For example, if 27 is divided by 6, the quotient is 4 and the remainder is 3.
2. If a number is divided by another number, if it is smaller than the other number, the quotient is 0, and the remainder is itself.
For example, if 1 is divided by 2, the quotient is 0 and the remainder is 1. When 2 is divided by 3, the quotient is 0 and the remainder is 2. In the division of integers, there are only two situations: divisible and non-divisible. When it is not divisible, it will produce a remainder, so the remainder problem is very important in primary school mathematics.
Remainder operation:
A mod b = c means that the remainder obtained by dividing the integer a by the integer b is C.
For example, 7 mod 3 = 1 remainder has the following important properties (A, B and C are natural numbers):
(1) The absolute value of the difference between the remainder and the divisor is less than the absolute value of the divisor (applicable to the real number field);
(2) Dividend = divisor × quotient+remainder;
Divider = (dividend-remainder) ÷ quotient;
Quotient = (dividend-remainder) divider;
Remainder = dividend-divisor × quotient.
(3) If the remainders of A and B divided by C are the same, then the difference between A and B can be divisible by C. For example, if the remainders of 17 and1divided by 3 are 2, then17-1can be divisible by 3.
(4) The sum of A and B divided by the remainder of C (except that A and B divided by C have no remainder) is respectively equal to the sum of the remainder of A and B divided by C (or the remainder of this sum divided by C). For example, 23, the remainder of 16 divided by 5 is 3 and 1 respectively, so the remainder of (23+ 16) divided by 5 is equal to 3+ 1=4. Note: When the sum of remainders is greater than the divisor, the remainders are equal to the sum of remainders and divided by the remainder of c ... For example, the remainders of 19 divided by 5 are 3 and 4 respectively, so the remainders of (23+ 19) divided by 5 are equal to the remainders of (3+4) divided by 5.
(5) The product of A and B divided by C is equal to the product of A and B divided by C (or the product divided by C). For example, 23, the remainder of 16 divided by 5 is 3 and 1 respectively, so the remainder of (23× 16) divided by 5 is equal to 3× 1=3. Note: When the product of the remainder is greater than the divisor, the remainder is equal to the product of the remainder divided by the remainder of c, for example, 23, and the remainder of 19 divided by 5 is 3 and 4 respectively, then the remainder of (23× 19) divided by 5 is equal to that of (3×4) divided by 5.
Properties (4) and (5) can be extended to the case of multiple natural numbers. Example 1 5 120 is divided by a two-digit number and the remainder is 64. Find this two-digit number.
Analysis and solutions:
According to the property (2), divisor × quotient = dividend-remainder.
5 120-64=5056,
5056 should be an integer multiple of the divisor. Decomposition of 5056 into prime factors to obtain
5056=64×79。
According to the property (1), the divisor should be greater than 64, then the divisor is a two-digit number, and the divisor is between 67 and 99.
The divisor of 5056 is only 79, so this two-digit number is 79.
Example 2 The sum of dividend, divisor, quotient and remainder is 2 143, the known quotient is 33 and the remainder is 52. Find dividend and divisor.
Solution: Because dividend = divisor× quotient+remainder = divisor× 33+52,
Dividend =2 143- divisor-quotient-remainder =2 143- divisor -33-52=2058- divisor,
So divisor ×33+52=2058- divisor, so divisor =(2058-52)÷34=59,
Dividend =2058-59= 1999.
A: The dividend is 1999 and the divisor is 59.
Example 3 The sum of numbers A and B is 1088, and the quotient of number A divided by number B is 1 1. Find numbers A and B..
Solution: Because A = B× 1 1+32,
So a+b = b×11+32+b = b×12+32 =1088,
So b = (1088-32) ÷12 = 88,
A = 1088- B = 1000。
A: A's number is 1000 and B's number is 88.
Example 4 has an integer, and the sum of the three remainders obtained by dividing 70, 1 10 and 160 is 50. Find this number.
Analysis and solution: first, find out the approximate range of this number from the subject conditions. Because 50 ÷ 3 = 16...2, at least one of the three remainders is greater than 16, and the inferred divisor is greater than 16. The sum of the three remainders is 50, and the divisor should not be greater than 70, so the divisor is between 17 ~ 70.
Judging from the meaning of the title, (70+1/kloc-0+160)-50 = 290 should be divisible by this number. The prime factorization of 290, 290=2×5×29 and the divisors of 290 between 17 ~ 70 are 29 and 58.
Because110 ÷ 58 =1... 52 > 50, 58 is irrelevant. The integer is 29.
Example 5 Find the remainder of 478×296×35 1 divided by 17.
Analysis and solution: it takes a lot of calculation to calculate the product first and then the remainder. According to property (5), we can first calculate the remainder of each factor divided by 17, and then find the product of the remainder divided by 17.
The remainders of 478,296,351divided by 17 are 2,7 and 1 1, (2× 7×1/) ÷1respectively.
The remainder is 1.
Example 6 Two delegations, A and B, went to visit by bus, and each bus can take 36 people. After the two delegations were filled with several cars, the remaining 1 1 people of Group A and the remaining members of Group B just filled with another car. After the visit, members of group A and group B took a group photo as a souvenir. If each film can take 36 photos, how many photos can the film in the camera take after taking the last photo?
Analysis and solution: After multiple vehicles are full, the number of group A is 1 1, that is, the number of group A is divided by 36+01; The remaining two groups just filled a car, that is to say, the number of group B is 36- 1 1=25 (people), that is, the number of group B (abbreviated as B) divided by 36 equals 25; Each member of group A and each member of group B take a photo in pairs, and take a photo of "A number x B number". Because each film takes 36 photos, the number of photos taken in the last film is equal to the remainder of "A number x B number" divided by 36.
Because A is divided by 36 remainder 1 1, and B is divided by 36 remainder 25, the remainder of "A × B" divided by 36 is equal to1× 25 divided by 36.
( 1 1×25)÷36=7……23,
That is to say, 23 shots were taken in the last film, and 36-23= 13 (shots) can be taken.
It can be seen from Example 6 that turning practical problems into familiar mathematical problems is helpful for us to think and solve problems.
Example 7 When 5397 is divided by a prime number, the remainder is 15. Find this prime number.
Solution: This prime number is divisible.
5397- 15=5382,
And 5382 = 2× 31997×13× 23.
Because the divisor is greater than the remainder 15 and the divisor is a prime number, it can only be 23.
When the dividend is large, a simple way to find the remainder is to gradually remove the integer multiple of the dividend from the dividend to get the remainder.
Example 8 Find the remainder of 645763 divided by 7.
Solution: You can first remove the multiple of 7, 630,000, greater than 15763, then remove 1763, then remove 1400, and then remove 350, greater than 13. The final remainder is 6. This process can be simply recorded as
645763→ 15763→ 1763→363→ 13→6.
If the calculus ability is strong, the above process can be written more simply:
645763→ 15000→ 1000→6.
The following useful conclusions can be drawn by division with remainder:
If two numbers are divided by the same divisor and have the same remainder, then the difference between the two numbers can be divisible by this divisor.
Example 9 has an integer greater than 1. Divide by 967, 1000 and 200 1 to get the same remainder. What is this integer?
Solution: From the above conclusion, the integer should be divisible by the difference of 967, 1000 and 200 1, that is
1000-967=33=3× 1 1,
200 1- 1000= 100 1=7× 1 1× 13,
200 1-967= 1034=2× 1 1×47.
This integer is the common divisor of these three differences 1 1.
Please note that we don't have to find three differences, just two are enough. Because there is always another difference from these two differences.
For example, business trips of 1000-967 and 200 1- 1000,
Too bad.
200 1-967=(200 1- 1000)+( 1000-967)
= 100 1+33
= 1034
From the formula with remainder, we can also draw the following conclusions:
If two numbers A and B are divided by the same divisor to get two remainders, then the sum of two numbers A and B is divided by this divisor, and its remainder is the sum of two remainders divided by this divisor.
For example, if 57 is divided by 13,5, 152 is divided by 13 and 9, then 57+ 152=209 is divided by13, and the remainder is 5+9= 14 divided by/kloc.
The example 10 has a series of consecutive numbers, where the first number is 15 and the second number is 40. Starting with the third number, each number is exactly the sum of the first two numbers. What is the remainder of 1998 divided by 3?
Solution: We can write this series of numbers according to the conditions of the topic, and then see what the law is when each number is divided by 3, but it is too troublesome to do so. According to the conclusion mentioned above, we can take the following measures: starting from the third number, add the remainder obtained by dividing the first two numbers by 3, and then divide by 3 to get the remainder of this number divided by 3, so it is easy to calculate the remainder of the first ten numbers divided by 3, as shown in the following table:
As can be seen from the table, the remainder of the ninth and twelfth numbers divided by 3 is the same as that of the first and second numbers divided by 3. Therefore, the remainder of this series of numbers divided by 3 is once every eight cycles, because
1998= 8×249+ 6,
So the remainder of 1998 divided by 3 should be the same as the remainder of the sixth number divided by 3, which is 2.
Some regular numbers often appear in cycles. Our calculation method is circular system. The calculation time is
1,2,3,4,5,6,7,8,9, 10, 1 1, 12
These twelve numbers form a cycle.
According to the seven-day round, the number of days is
One day, one, two, three, four, five, six. This is also a cycle, which is equivalent to the remainder of some continuous natural numbers divided by 7.
0, 1, 2, 3, 4, 5, 6 cycle, using the cycle system to calculate the time: clock, week, month, season, indicating that people have discovered the cycle phenomenon very early. It is also natural to use numbers to reflect the circular phenomenon.
Cyclic phenomenon, we also call it a cycle with "periodicity" and 12 numbers, that is, a cycle with a period of 12 and a number of 7, that is, a cycle with a period of 7. For example, the cycle of the remainder in 10 is 8. It is interesting to study the cycle of numbers, find periodicity and determine periodicity.
Give two more examples of remainder circulation. Before talking about examples, let's talk about a conclusion drawn by remainder division:
Two numbers A and B are divided by the same divisor to get two remainders. Then the product of two numbers a and b is divided by this divisor, and its remainder is the product of two remainders and the remainder obtained by dividing by this divisor.
For example, 37 divided by 1 1, 4,27 divided by 1 1, 5,37× 27 = 999 divided by1,and the remainder is 4×5=20 divided by/kloc-0.
1997=7×285+2, and we know that the remainder of 1997× 1997 divided by 7 is 2×2=4.
Example:11191997 divided by 7?
Solution: From the above conclusion, we know that the remainder of 19 1997 divided by 7 is the same as that of 2 1997 divided by 7. We only need to consider some multiplication of 2, and divide the remainder by 7.
Write a column of numbers first.
2,2×2=4,2×2×2 =8,
2×2×2×2= 16,…
Then divide them by 7 one by one and make a list to see what the rules are. The list is as follows:
In fact, you can multiply the last number by 2 and then divide it by 7 to get the remainder of the last number divided by 7. (Why? Please think about it. )
As can be seen from the table, the remainder of the fourth number is the same as that of the first number, both of which are 2. According to the calculation of the remainder above, we know that the remainder of the fifth number is the same as the remainder of the second number ... Therefore, the remainder circulates every three numbers. The period of the cycle is 3.
1997=3×665 +2
It is known that the remainder of 2 1997 divided by 7 is the same as that of 2 1997 divided by 7, and this remainder is 4.
Let's look at a slightly more complicated example.
Example 12 70 Numbers are arranged in a row. Except for the two numbers at both ends, three times of each number is exactly equal to the sum of the two numbers at both sides. The leftmost number in this line is as follows:
0, 1,3,8,2 1,55,…
Q: What is the rightmost number (the 70th number) divided by 6?
Solution: First of all, it should be noted that starting from the third number, each number is exactly equal to three times the previous number minus the previous number:
3= 1×3-0,
8=3×3- 1,
2 1=8×3-3,
55=2 1×3-8,
……
However, it is really too much trouble to count one by one and then divide by six one by one. Can you work out the remainder from the previous remainder? Sure！ The same method as calculating the number of rows (why? ), starting from the third number, the calculation method of the remainder is as follows:
Multiply the remainder of the previous number by 3, subtract the remainder of the previous number, and then divide by 6. The remainder is
So the remainder can be worked out one by one, and the list is as follows:
Note that when calculating the remainder of the eighth number, there will be 0×3- 1, which is not allowed within the scope of primary school mathematics, because we are asking for the remainder to be divided by 6, so we can add 0×3 to 6 and then subtract 1.
As can be seen from the table, the remainders of the thirteenth and fourteenth numbers correspond to the remainders of the first and second numbers, so we know that the cycle period of the remainders is 12.
70 = 12×5+ 10
Therefore, the remainder of the seventieth number divided by 6 is the same as the remainder of the tenth number, that is, 4.
In Sun Tzu's calculation more than a thousand years ago, there was such an arithmetic problem:
"I don't know the number of things today. 3322, 5522, 7722. What is the geometry of things? " According to today's words:
Divide a number by 3 and 2, by 5 and 3, by 7 and 2, and find this number.
This kind of problem is also known as "Han Xin points soldiers". Form a kind of problem, that is, the solution congruence formula in elementary number theory. The conditional solution of this kind of problem is called "China's Remainder Theorem", which was put forward by China people. Many primary school mathematics extracurricular reading materials like to talk about this kind of problem, but its general solution is by no means understandable to primary school students. Here, we introduce a popular solution for smaller numbers through two examples.
Example 13 has a number, divided by 3 and 2, divided by 4 and 1. What is this number divided by 12?
Answer: The number divided by 3 and 2 is:
2, 5, 8, 1 1, 14, 17, 20, 23…
The remainder after dividing by 12 is:
2,5,8, 1 1,2,5,8, 1 1,…
Divided by 4, the remaining 1 number is:
1, 5, 9, 13, 17, 2 1, 25, 29,…
The remainder after dividing by 12 is:
1, 5, 9, 1, 5, 9,…
The remainder of a number divided by 12 is unique. Only 5 in the upper two lines is the same * * *, so the remainder of this number divided by 12 is 5.
In the above solution, we list the integers divided by 3 and 2 and those divided by 4 1 one by one, and then consider the remainder divided by 12 one by one to find the same remainder as that divided by 12. This enumeration method is very useful in the case of a small number, and it is also the most acceptable to students.
If we change the problem in Example 23, instead of finding the remainder divided by 12, we find this number. Obviously, there are many numbers that meet the requirements. They are
5+ 12× integer
Integer can take 0, 1, 2, …, endless. In fact, after we find out 5, we notice that 12 is the least common multiple of 3 and 4, plus the integer multiple of 12, which are all numbers that meet the conditions. This is the division of "divide by 3 and 2, divide by 4 and 1". There are three conditions for Sun Tzu to put forward this question. We can combine the two conditions into one, and then combine with the third condition to find the answer.
Example 14 Divide a number by 3 and 2, by 5 and 3, and by 7 and 2 to find the minimum number that meets the conditions.
Solution: First list the numbers divided by 3 and 2:
2, 5, 8, 1 1, 14, 17, 20, 23, 26,…,
Then list the numbers divided by 5 and 3:
3, 8, 13, 18, 23, 28,….
In these two columns, the first common divisor is 8.3, and the least common multiple of 5 is 15. These two conditions are combined into one.
8+ 15× integer,
The number of this string in the list is
8, 23, 38,…,
Then list the number of 2 divided by 7.
2, 9, 16, 23, 30,…,
The minimum number to meet the requirements of the topic is 23.
In fact, we have combined the three conditions in the topic into one: divide by105,23.
Finally, look at an example.
There are three consecutive natural numbers between 15 100 and 200, of which the smallest is divisible by 3, the middle by 5 and the largest by 7. Write these three consecutive natural numbers.
Solution: First find two consecutive natural numbers, the first one is divisible by 3, and the second one is divisible by 5 (also divisible by 3, 1). For example, find 9 and 10, and the next continuous natural number is 1 1.
The least common multiple of 3 and 5 is 15. Considering the integer multiple of 1 1 plus 15, the added number can be divisible by 7. 1 15438+05× 3 = 56 is divisible by 7, so 54,55.
In order to meet the requirement of "between 100 and 200", add 54, 55 and 56 to the least common multiple of 3, 5 and 7 105 respectively.
159, 160, 16 1.