Number theory is a branch of mathematics that studies the properties of integers. It has a long history and strong vitality. The description of number theory problems is concise. "Many number theory problems can be summed up from experience and explained to an outsider in just a few sentences, but it is far from easy to prove them." So some people say, "There is no better course in elementary mathematics to discover genius than number theory." Any student who can do any exercises in the number theory textbook today should be encouraged and suggested to work in mathematics in the future. "Therefore, in all kinds of mathematical competitions at home and abroad, the problem of number theory always occupies a considerable proportion.
The problems of number theory in mathematical competitions often involve integer division, division with remainder, odd and even numbers, prime numbers and composite numbers, divisors and multiples, and integer decomposition and splitting. The main conclusions are as follows:
1. Division with remainder: If A and B are two integers and B > 0, there are two integers Q and R, so
a=bq+r(0≤r 14，
Description: The basic knowledge used to solve this problem is the decimal representation of positive integers and the simplest indefinite equation.
Example 3 From the natural number 1, 2, 3, …, 1000, how many numbers can be taken out at most, so that the sum of any three taken numbers can be divisible by 18?
Solution: Let A, B, C and D be any four extracted numbers, then
a+b+c= 18m，a+b+d= 18n，
Where m and n are natural numbers. therefore
C-d= 18 (male-female).
The above formula shows that the difference between any two extracted numbers is a multiple of 18, that is, the remainder obtained by dividing each extracted number by 18 is the same. So, let this remainder be R.
a= 18a 1+r，b= 18b 1+r，c= 18c 1+r，
Where a 1, b 1 and c 1 are integers. therefore
a+b+c = 18(a 1+b 1+c 1)+3r .
Since 18|(a+b+c), 18|3r, which is 6|r, it is inferred that r = 0,6, 12. Because1000 = 55×18+10, we can take 6,24,42, ...
Example 4 Find the natural number n so that it can be divisible by 5 and 49. It has 10 divisors, including 1 and n. ..
Solution: Write the number n as the product of prime factors.
Since n can be divisible by 5, 72=49, a3≥ 1, a4≥2, the other indexes ak are natural numbers or zero. According to the meaning of the question, there are
(a 1+ 1)(a2+ 1)……(an+ 1)= 10 .
Because a3+ 1≥2, a4+ 1≥3, 10=2×5,
a 1+ 1 = a2+ 1 = a5+ 1 =…= an+ 1 = 1，
That is, a 1=a2=a5=…an=0, and n can only have two different prime factors 5 and 7, because a4+ 1 ≥ 3 > 2, it is determined by.
(a3+ 1)(a4+ 1)= 10
It is impossible to know that a3+ 1=5 and a4+ 1=2. So a3+ 1=2, a4+ 1=5, that is, n = 52-1× 75-1= 5× 74 =12005.
Example 5 If n is the least common multiple of 1, 2, 3, …, 1998, 1 999,2000, how many odd numbers of 2 and1does n equal?
Solution: Because 2 10= 1024, 2 1 1 = 2048 > 2000, each natural number not greater than 2000 is expressed as a prime factor multiplication, where the number of 2 is not greater than 10,/kl.
Note: the above five situations are based on the characteristics of the topic, starting with the choice of appropriate integer representation, so that the problem can be solved easily.
Second, the enumeration method
Enumeration (also known as exhaustive method) is to divide the object of discussion into several situations (classification), then discuss each situation one by one, and finally solve the whole problem.
The use of enumeration sometimes requires proper classification, and the principle of classification is neither weight nor leakage. Correct classification helps to expose the essence of the problem and reduce the difficulty of the problem. The most commonly used classification methods in number theory are modular residue classification, parity classification and numerical classification.
Example 6 Find such a three-digit number and divide it by 1 1 to get the remainder equal to the sum of the squares of its three digits.
Analysis and solution: there are only 900 three-digit numbers, which can be solved by enumeration. When enumerating, we can first estimate the range of related quantities, narrow the scope of discussion and reduce the amount of calculation.
Let the hundred, ten and one digit of this three-digit number be x, y and z respectively. Because the remainder obtained by dividing any number by 1 1 is not greater than 10, it is equal to.
x2+y2+z2≤ 10，
So 1≤x≤3, 0≤y≤3, 0≤z≤3. The three numbers you need must be among the following numbers:
100, 10 1, 102, 103, 1 10, 1 1 1, 1 12,
120, 12 1, 122, 130,200,20 1,202,
2 1 1,2 12,220,22 1,300,30 1,3 10。
It is not difficult to verify that only 100 and 10 1 meet the requirements.
Example 7 Write the natural number n on the right of any natural number (for example, write 2 on the right of 35 to get 352). If a new number is divisible by n, then n is called a magic number. Q: How many magic numbers are there in a natural number less than 2000?
Discuss that n is one digit, two digits, three digits and four digits respectively.
N| 100, so n N= 10/0,20,25,50;
N| 1000, so N= 100,125,200,250,500;
(4) When n is four digits, N= 1000, 1250, 2000, 2500, 5000 can also be obtained. 1000 and 1250 are eligible.
To sum up, the number of magic numbers is 14.
Note: (1) We can prove that the k-bit magic number must be the divisor of 10k, and vice versa.
(2) Divide the problem into several situations to discuss, and add a precondition to each situation, thus reducing the difficulty of the problem and making it easy to solve.
Example 8 There are three playing cards, all within 10. After washing these three cards, send them to Xiaoming, Liang Xiao and Xiaoguang respectively. After everyone wrote down his card number, he shuffled, dealt cards and counted again. After repeated several times, the number sequence recorded by each person is 13,15,23. Q: What are the numbers of these three cards?
Solution:13+15+23 = 51,5 1=3× 17.
Because it is impossible to touch 17 > 13 times, it is impossible to touch 17 times, so the sum of the numbers of three playing cards is 17. The possible situations are as follows:
① 1,6, 10 ② 1,7,9 ③ 1,8,8
④2,5, 10 ⑤2,6,9 ⑥2,7,8
⑦3,4, 10 ⑧3,5,9 ⑨3,6,8
⑩3,7,7 ( 1 1)4,4,9 ( 12)4,5,8
( 13)4,6,7 ( 14)5,5,7 ( 15)5,6,6
Only the first situation can meet the requirements of the topic, namely
3+5+5= 13; 3+3+9= 15; 5+9+9=23。
The numbers of these three cards are 3, 5 and 9 respectively.
Example 9 Write 12 continuous natural numbers, all of which are composite numbers.
Analysis 1: In the process of finding prime numbers, we can see that within 100, we can write at most seven consecutive composite numbers: 90, 9 1, 92, 93, 94, 95, 96. Let's continue to apply the screening method and expand the scope of the exam.
The solution 1: 1 13 and 127 * * have 12 consecutive natural numbers, all of which are complex numbers:
1 14, 1 15, 1 16, 1 17, 1 18, 1 19, 120,
12 1, 122, 123, 124, 125, 126。
Analysis 2: If 12 is a continuous natural number, then 1 is a multiple of 2, the second is a multiple of 3, and the third is a multiple of 4 ... and 12 is a multiple of 13, then all the numbers of 12 are composite numbers.
M+2, m+3, …, m+ 13 are 12 consecutive integers, so as long as m is a common multiple of 2, 3, …, 13, these 12 consecutive integers must be composite numbers.
Solution 2: Let m be the least common multiple of the numbers 2, 3, 4, …, 13. M+2, m+3, m+4, …, m+ 13 are multiples of 2, 3, 4 … 13 respectively, so the numbers of 12 are all composite numbers.
Explanation: We can also write.
13! +2, 13! +3,…, 13! + 13
(where n! = 1×2×3×…×n) There are 12 continuous composite numbers here.
Similarly,
(m+ 1)！ +2，(m+ 1)！ +3，…，(m+ 1)！ +m+ 1 is the continuous composite of m.
Third, induction.
When we want to solve a problem, we can first analyze several simple and special situations of this problem, find and summarize the general rules or make some guesses, so as to find a solution to the problem. This way of thinking from special to general is called induction.
Example 10 Arranging prime numbers within 100 into a number string from small to large, and completing the following five tasks in turn is called an operation:
(1) Move the first number on the left to the far right of the number string;
(2) From left to right, two digits form a two-digit number;
(3) Cross out the composite number of these two digits;
(4) If the remaining two prime numbers are the same, keep the one on the left and cross out the others;
(5) The remaining two prime numbers keep numerical order to form a new numerical string.
Q: What is the number string after the operation of 1999?
Solution: number string 711111737 of operation 65438;
Number string of the second operation11133173;
The numeric string of the third operation111731;
Number string of the fourth operation1173;
The numeric string of the fifth operation1731;
The number string of the sixth operation is 7311;
The number string of the seventh operation is 3117;
Number string of the eighth process 1 173.
It is not difficult to see that after four cycles, 1999=4×499+3, so the number string obtained by the 1999th operation is the same as that obtained by the 7th operation, which is 3 1 17.
Example 1 1 has a stack of 100 cards. Lingling picked them up and operated them from top to bottom: throw away the top first card and put the next card at the bottom of the pile. Give up the original third card and put the next card at the bottom. Repeat this until there is only one card left in the hand, so which card is left in the original pile?
Analysis and solution: we can start with simple problems that do not lose the essence of the topic and find the law. The list is as follows:
Let the number of cards in this stack be n, and observe the above table:
(1) When n = 2a (a = 0, 1, 2, 3, ...), the remaining cards are the last cards in the original stack, that is, 2a cards;
(2) When n = 2a+m (m < 2a), the remaining cards are 2m cards in the original stack.
Take N= 100, because 100=26+36, 2×36=72, so the remaining card is the 72nd card in the original stack.
Explanation: This question is essentially the famous josephus problem:
Legend has it that in ancient times, a group of people were captured by barbarians. The enemy ordered them to form a circle, numbered 1, 2, 3, … and then killed 1 and killed No.3. In short, every other person killed one person, and finally there was one person left, and that person was Josephus. If there are11prisoners, what is the number of Josephus?
Example 12 How many weights should be used to weigh 1g, 2g, 3g...40g with scales? What are the weights?
Analysis and solution: Generally speaking, weights can be placed on both sides of the balance, so let's start with the simplest case.
(1) Only one weight of 1 g can be used to weigh 1 g, so a weight of 1 g is needed.
(2) Weighing 2g, there are three schemes:
① Add weight1g;
② Use a weight of 2g;
(3) With a 3 g weight, when weighing, put a 1 g weight in the weighing pan and 3 g weight in the weighing pan. From a mathematical point of view, it is 3- 1=2.
(3) Weigh 3g. The above two schemes (② ③) are used without any weight, so scheme ① is excluded.
(4) Weigh 4g, and use the above scheme ③ without adding any weight, so scheme ② was also rejected. In short, any integer gram weight within (3+ 1) gram can be weighed by 1 gram and 3 grams.
(5) Then think that you can make a leap and use it when it weighs 5 grams.
9-(3+ 1)=5,
That is, put a 9-gram weight in the weighing pan, and put 1 gram and 3-gram weights in the weighing pan. In this way, you can weigh any integer grams in 1+3+9= 13 (grams) in turn.
And to weight 14g, add a weight according to the above law, and its weight is
14+ 13=27 (g),
It can be weighed to any integer gram within 1+3+9+27=40 (grams).
In short, when the weight of the weight is 1, 3, 32 and 33 grams, the weight used is the least and the weight is the largest, which is also the answer to this question.
This conclusion can obviously be generalized. When the counterweight can be placed at both ends of the leveling table, use 1, 3,
This is the weight design scheme with the smallest weight and the largest weight.
Exercise 1
1. It is known that a four-digit decimal minus 1 equals its single digits, and a single digit plus 2 equals a hundred digits. The sum of the four digits arranged in reverse order and the original number is equal to 9878. Try to find this four-digit number
3. Let n be the smallest natural number that satisfies the following conditions: They are multiples of 75, which is exactly 75.
4. What is the largest even number that cannot be written as the sum of two odd numbers?
5. Arrange the 999 numbers 1, 2, 3, 4, …, 999 evenly into a big circle, and start counting from 1: cross out 2, 3 every 1, and cross out 5, 6 every 4 … so that two numbers are crossed out every other number, and so on. Q: How many numbers are left in the end? Why?
6. There are n blocks on the circumference, as shown on the right. One block on point B is adjacent to one block on point A. Xiao Hong first takes 1 block at point B, and then takes 2 blocks every 1 block clockwise, turning 10 times continuously and crossing A nine times. When the 10 time was about to cross the chess piece at A to take other chess pieces, Xiaohong found that there were more than 20 chess pieces left on the circumference. If n is a multiple of 14, how many blocks are there on the circumference?
7. Use five digits (0, 1, 2, 3, 4) to form four digits, and every four digits has no repeated digits (such as 1023, 234 1), and find the sum of all these four digits.
8. Twenty-seven countries participated in an international conference, and each country had two representatives. Verification: It is impossible to arrange 54 delegates to sit around a round table, so there are 9 people between two delegates of any country.
Exercise 1
1. 1987。
(a+d)× 1000+(b+c)× 1 10+(a+d)= 9878 .
Comparing the two sides of the equation and paying attention to the characteristics of the sum of numbers and their carry, we can see that
a+d=8，b+c= 17 .
Given c- 1=d and d+2=b, we can get
a= 1，b=9，c=8，d=7 .
That is, the required four digits are 1987.
2. 1324, 1423, 23 14, 24 13, 34 12, ***5.
3.432。
Solution: In order to ensure that n is a multiple of 75 and as small as possible, because 75=3×5×5, it can be assumed that n has three prime factors 2, 3, 5, that is, n=2α×3β×5γ, where α≥0, β≥ 1, γ≥2, and
(α+ 1)(β+ 1)(γ+ 1)=75。
It is easy to know that when α=β=4 and γ=2, the conditions of the topic are met. now
4.38。
Solution: Odd numbers less than 38 are 9, 15, 2 1, 25, 27, 33.
38 cannot be expressed as the sum of any two of them, but an even number A greater than 38 can be expressed as the sum of two odd numbers:
The last digit of a is 0, then A= 15+5n,
If the last digit of a is 2, then A=27+5n,
If the last digit of a is 4, then A=9+5n,
If the last digit of a is 6, then A=2 1+5n,
The last digit of a is 8, so A=33+5n,
Where n is an odd number greater than 1. So, 38 is what you want.
5.406。
Solution: From a special situation, it can be concluded that if it is 3n (n is a natural number), then draw 1 leave 3n- 1, draw 2 and leave 3n-2 ... draw (n- 1) and leave 3, then draw 1, and the last thing left is the beginning.
36 < 999 < 37, if (999-36=)270 numbers are crossed out from 999 numbers and the remaining (36=) 729 numbers, the above conclusion can be applied.
Because two numbers are crossed out each time, the number 270 must be crossed out 135 times. At this time, the 270th number crossed out is (135×3=)405, and the starting number of the remaining 36 numbers is 406. So the last remaining number is 406.
6.23 pieces.
Solution: let there be more than a blocks on the circumference. Because Xiao Hong took two pieces from the ninth crossing A to the 10 crossing A, there were three pieces on the circumference when crossing A for the ninth time. By analogy, there are 32a pieces on the circumference when crossing the A-piece for the eighth time ... There are 39a pieces on the circumference when crossing the A-piece for the first time 1, and before crossing the A-piece for the first time 1, Xiaohong takes [2(39a- 1)+ 1] pieces.
If N=3 10a=59049a- 1 is a multiple of 14, then n is a common multiple of 2 and 7, so a must be an odd number;
If n = (7× 8435+4) a-1= 7× 8435a+4a- 1 is a multiple of 7, then 4a-1must be a multiple of 7. When a = 2 1, 25, 27, 29,
When n is a multiple of 14, there are 23 blocks on the circumference.
7.259980。
Solution: The addition principle of the sum of several four digits expressed in decimal is:
The sum of several four digits = the sum of thousands of digits × 1000+
Sum of percentiles × 100+
Sum of ten digits × 10+
The sum of single digits.
Take one of 1, 2, 3 and 4 as thousands, and there are 4×3×2=24 (digits) that meet the conditions of the topic. This is because when determining thousands of digits, you can choose hundreds of digits from the remaining four digits; After determining thousands and hundreds, ten digits can be selected from the remaining three digits; Similarly, there are two possibilities for single digits. Therefore, the sum of thousands of four digits that meet the conditions is
( 1+2+3+4)×4×3×2=240。
When one of 1, 2, 3 and 4 is taken as the hundredth bit, because 0 cannot be taken as the thousandth bit, there are three choices for the thousandth bit; There are also three options for ten digits (plus 0); There are two options for single digits. Therefore,
Sum of hundreds = (1+2+3+4) ×18 =180.
Similarly, the sum of ten digits and the sum of single digits are 180.
So the sum of the four digits that meet the conditions is
240× 1000+ 180×( 1+ 10+ 100)= 259980。
8. Number the 54 seats counterclockwise: 1, 2, …, 54. Because it is sitting around the round table, starting from 1, turning counterclockwise to No.55 is equivalent to 1; Go to 56, which is equivalent to Block 2；; In this way, it is obvious that going to m is equivalent to dividing m by the remainder block of 54.
It is assumed that there are arrangements to meet the requirements. Suppose 1 and 1 1 are representatives of the same country. Since any country has only two representatives, 1 1 and 2 1 are not representatives of the same country. The following arrangements are:
2 1 and 3 1 are representatives of the same country;
3 1 and 4 1 are not representatives of the same country;
4 1 and 5 1 are representatives of the same country;
5 1 and 6 1 are not representatives of the same country (6 1, block 7).
So 20k+ 1 and 20k+ 1 1 are representatives of the same country. If 20k+ 1, 20k+1/is greater than 54, the number of seats is divided by 54.
Take k= 13, then 26 1 and 27 1 belong to the same country, the remainder of 26 1 divided by 54 is 45, and the remainder of 27 1 divided by 54 is 1, that is to say/kloc. In this way, 1,1and No.45 Theory of Three Represents belong to the same country, which is impossible. Therefore, the arrangement required by the topic is impossible to achieve.
The Methods and Skills of Number Theory (Ⅱ)
Fourth, reduce to absurdity.
The method of reduction to absurdity is to make the opposite assumption to the conclusion of the proposition, and from this assumption, through correct reasoning, get contradictory results, thus denying the assumption as the starting point of reasoning, thus affirming that the original conclusion is correct.
The process of reducing to absurdity can be simply described as the following three steps:
1. Anti-design: Assume that the conclusion to be proved does not hold, but its opposite holds;
2. Reduction to absurdity: Starting from "counter-hypothesis" and through correct reasoning, we can deduce contradictions-contradictions or self-contradictions of known conditions, axioms, definitions, theorems, counter-hypotheses and obvious facts;
3. Conclusion: Because the reasoning is correct, the contradiction lies in the fallacy of "reverse design". Since the opposite of the conclusion is not true, the conclusion is certain.
The key to applying reduction to absurdity is to lead to contradictions. In number theory, many problems are caused by parity analysis or congruence.
Solution: If there are such three digits, then there are.
100a+10b+c = (10a+b)+(10b+c)+(10a+c). The above formula can be simplified to 80a=b+c, which is obviously impossible. This means that the number you are looking for does not exist.
Note: When proving that there is no problem, reduction to absurdity is often used: first, assume that there is at least one element that meets all the requirements stated in the proposition, and then proceed from this existing element until there is a contradiction.
Example 2: Reverse the arrangement order of a number with 17 digits, and then add the obtained number with the original number. Try to explain that at least one number in the sum is even.
Solution: suppose that none of the sum numbers is even, that is, all are odd. In the addition formula shown in the following formula, the sum d+a of the last column is odd, and the first column is odd, so the sum B+C of the second column is ≤ 9. If the first two digits A and B and the last two digits C and D of the known number are removed, the obtained 13 digit still has the property of "inverting its digits, adding the obtained digits, and the sum digits are all odd". According to this, the two digits before and after are removed at one time, and finally a number is obtained, which is even and contradictory when added to itself. Therefore, there must be even numbers in the sum.
Note: Obviously, the conclusion is also valid for (4k+ 1) digits. But not necessarily for other numbers. Such as 12+2 1, 506+605, etc.
There is a magical coin machine. When inserting 1 cent 1 coin, take out 1 cent 1 coin and 5 cent 1 coin. When inserting 1 nickel, take out 4 1 dime; When inserting 1 1 dime, take out 3 1 dime. Xiaohong starts with 1 min 1 coin and 5 1 coin, and repeatedly puts coins into the machine. Xiao Hong, at a certain moment, can the coins of 1 minute be less than those of 1 minute?
Solution: There are only 1 coins with 1 corner at the beginning and no coins with 1 corner at the beginning, so the total sum of 1 corner at the beginning and 1 corner is 0+ 1= 1, which is an odd number. Every time you use the machine, write the total number of 1 points and 1 angles as Q. Let's examine the parity of Q.
If 1 cent/coin is inserted, then q temporarily decreases 1, but we retrieve 1 coin of 1 cent (and 1 coin of 5 cents), so the total q remains unchanged. If you insert 1 nickel (get 4 1 nickel), then q increases by 4, and its parity remains unchanged; If you insert 1 coin with 1 cent, then if Q is increased by 2, its parity remains unchanged. So every time you use the machine, the parity of Q remains the same, because Q is an odd number at first and will remain odd in the future.
In this way, it cannot be concluded that the number of coins in 1 min is just less than that in 1 min, because if we have p coins in 1 min and (p+10) coins in 1 min, then/kloc-. conflict
Example 4, as shown on the right, fills in 9 prime numbers in a 3×3 grid table. Adding the same natural number to three numbers in the same row or column in a table is called an operation. Q: Can you make all nine numbers in the table become the same number through several operations? Why?
Solution: Because the sum of the nine prime numbers in the table is exactly 100, dividing by 3 is 1. After each operation, the sum increases by a multiple of 3, so the sum of 9 numbers in the table divided by 3 is always 1. If the number of 9 in the table becomes equal, then the sum of 9 numbers should be divisible by 3, which creates a contradiction!
Therefore, no matter how many operations are performed, the numbers in the table will not become nine identical numbers.
Verb (abbreviation of verb) structure method
Construction method is an important mathematical method, which is flexible and diverse. Many problems in number theory can be solved by constructing some integers or integer combinations with special structures and properties.
Example 5 9999 and 99! Can it be expressed as the sum of 99 consecutive odd natural numbers?
Solution: 9999 is enough. Because 9999 is equal to the sum of 99998, it can be directly constructed as follows:
9999=(9998-98)+(9998-96)+…+
=(9998-2)+9998+(9998+2)+…+
=(9998+96)+(9998+98)。
99! I can't. Because 99! It's even, and the sum of 99 odd numbers is odd, so 99! It cannot be expressed as the sum of 99 consecutive odd numbers.
Note: To prove the existence of problems by construction method, we only need to construct mathematical objects that meet the requirements of the topic.
Example 6 From the 999 numbers of 1, 2, 3, …, 999, it is required to cross out as few numbers as possible so that each remainder is not equal to the product of the other two numbers. Which numbers should be crossed out?
Solution: We can cross out the 30 numbers of 2, 3, …, 30, 3 1, because the product of any two numbers except 1 will be greater than 322 = 1024 > 999.
On the other hand, it can be proved that 30 is the least number by constructing a ternary array.
(2,6 1,2×6 1),(3,60,3×60),(4,59,4×59),…,
(30,33,30×33),(3 1,32,3 1×32)。
These numbers written above are different from each other, and the maximum number of these numbers is 3 1×32=992. If the number crossed out is less than 30, there is at least one ternary array left, which does not meet the problem setting conditions. Therefore, 30 is the least number.
Sixth, the matching method
There are various forms of pairing, including rounding of numbers and pairing of elements between sets (which can be used for counting). It is said that Gauss Sum (1+2+…+ 100) initiated pairing when he was 8 years old. Like Gauss, being good at collocation skills can often solve some seemingly troublesome and even intractable problems.
Example 7 Find the sum of all numbers in the number 1, 2, 3, …, 9999998, 999999.
Solution: adding a number 0 before these numbers does not affect the sum of all the numbers. Pair the numbers 100000000, because the sum of the numbers 0 and 9999999, 1 and 999998, …, 49999999 and 5000000 is 9×7=63. There are 5000000 pairs of * * so the sum of all the numbers is 63 × 5000000 = 3 1500000.
Example 8 A shopping mall issued 9999 shopping vouchers to customers, each with a four-digit number ranging from 000 1 to 9999. If the sum of the first two digits of the number is equal to the sum of the last two digits, this shopping voucher is called a "lucky voucher". For example, the number 0734 is a lucky ticket because 0+7=3+4. It shows that the sum of all lucky coupons in the shopping vouchers issued by this mall can be divisible by 10 1.
Solution: Obviously, the number 9999 is a lucky ticket. In addition to this lucky coupon, if a certain number n is a lucky coupon, then the shopping coupon with the number m=9999-n is also a lucky coupon. Since 9999 is an odd number, m ≠ n.
Because m+n=9999, there is no carry when adding, so all lucky coupons can appear in pairs except the lucky coupon with the number of 9999, and the sum of the two numbers in each pair is 9999, that is, the sum of all lucky coupon numbers is a multiple of 9999.
Because 9999=99× 10 1, the sum of all lucky ticket numbers can be divisible by 10 1.
Explain that the molecule m is a multiple of the prime number 89.
Scheme 1: imitate the method of Gaussian summation (1+2+3+…+n) and sum.
① ② The two types are obtained by adding them together.
therefore
2m×88！ =89×k(k is a positive integer).
Because 89 is an odd prime number, 89 cannot be divisible by 88! , thus 89 | m.
Solution 2: Pairing.
Divide the fractions in brackets, and the common denominator is
1×88×2×87×3×86×…×44×45=88! ,
therefore
m×88！ =89×k(k=n×q).
Because 89 is an odd prime number, 89 cannot be divisible by 88! , thus 89 | m.
Seven. Estimation method
Estimation method is to determine the range of a certain number or the whole formula by enlarging or narrowing inequalities, so as to obtain the essential characteristics of related quantities and achieve the purpose of solving problems.
In the problem of number theory, in a limited range, there are at most a limited number of integers. When we transition to an integer, we can test the possible situations one by one and determine the solution to the problem.
Find this number and find five different groups of true points that satisfy the meaning of the question.
Solution: Because every true score is satisfied.
The integer s is the sum of four different true fractions, so 2 < s < 4, it is inferred that S=3. Therefore, you can get the following five different sets of true scores:
Example: There are exactly 1 1 known product 1×2×3×…×n with 106 consecutive zeros at the tail. Find the maximum value of natural number n.
Analysis: If the specific value of n is known, it is easier to find the number of tail zeros of 1×2×…×n, but it is not easy to find the value of n if the number of tail zeros is known in turn. We can estimate the approximate value of n first, and then carefully determine the value of n. ..
Therefore, the number of prime factor 5 in the product 1×2×3×…×400 is 80+ 16+3=99 (pieces). There are more prime factors 2 than 5 in the product, so when n=400, there are 99 zeros at the tail of 1×2×…×n, and seven zeros are needed. Note that 425 contains two prime factors of 5, so
When n=430, 1×2×…×n has 106 zeros at its tail;
When n=435, the tail of 1×2×…×n