DOWNLOAD PDF Mathematical Circles (Russian Experience) Dmitri Fomin Sergey Genkin Ilia Itenberg Translated from the Russian by Mark Saul Universities. "This is a sample of rich Russian mathematical culture written by professional mathematicians with great experience in working with high school students. Mathematical circles: (Russian experience) / Dmitri Fomin, Sergey Genkin, Ilia Itenberg ; translated from the Russian by Mark Saul Genkin, S. A. (Sergeĭ.

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I am looking for something like Mathematical Circles: A Russian Experience. 4, Views · Are there any solutions provided for the book Mathematical Circles: (Russian Where are mathematics book circles available in PDF format?. Mathematical Circles (Russian Experience): Dmitri Fomin, Sergey Genkin & called mathematical circles. The mathematical circles of the former Soviet Union. TABLE OF CONTENTS OF Mathematical Circles (Russian Experience): the book is organized in two parts; each part provides materials it bl f l th i l suitable for a.

There are six letters in the Hermetian language. If two numbers have equal remainders when divided by 9, then their difference is divisible by 9. In how many. The first of them should be devoted to the calculation of remainders. For example, let us prove , the second proposition. Can a knight start at square al of a chessboard, and go to square h8, visiting each of the remaining squares exactly once on the way?

We call a natural number "odd-looking" if all of its digits are odd. In such problems the students may encounter difficulty distinguishing the two numbers m and n. A word in this language is an arbitrary sequence of no more than four letters. How many ways are there to do tllis? We toss a coin three times. How many different colorings of the table are there? How many four-digit odd-looking numbers are there?

No matter where it stands. If n is a natural number. Exercise 2. Thus we are left with 49 squares where the black rook can be placed. The following exercises may be useful Exercise 1. How many ways are there to put one white and one black king on a chessboard so that they do not attack each other? But before this we must digress a little bit. Simplify the expressions a 1O! We have 60 squares left. How many three-digit numbers can be written using the digits I. For convenience of calculations and for consistency.

The white king can be placed on any of the 64 squares. Prove that if p is a prime number. Exercise 3. The white rook can be placed on any of the 64 squares. There are six possible choices of a color for the bottom strip.

Before working with permutations one must know the definition of factorial and learn how to deal with this function. Methodolo2ical remark. Let us reason just the same way we did in solving Problem The first digit can be any of the three given. Such arrangements are called permutations.

After that we have only five colors to use for the middle strip. Now let us go back to the permutations. How many ways are there to put one white and one black rook on a chessboard so that they do not attack each other? The second can be occupied by any of the three remaining balls. This set of problems about words demonstrates one very interesting and impor- tant idea-the idea of multiple counting.

Thinking of the three letters A in this word as different letters AI. We have three letters S and two letters E in this word. Under this assumption we have 5! Here are four more problems using this method. For example. Therefore the answer is 8! In the following five problems you must calculate the number of different words that can be obtained by rearranging the letters of a particular word. Since the letters Ai can be rearranged in their places in 3! Thus we have 3. Any finite sequence of English letters will be called "a word" whether or not it can be found in a dictionary.

This word contains two letters T. The first place in the row can be occupied by any of the given balls. Let us temporarily think of these letters T as two different letters Tl and T2. Temporarily thinking of all of them as different letters. For convenience of notation we introduce the following convention. That is. Since all the letters in this word are different.

How many ways are there to lay four balls. How many diagonals are there in a convex n-gon? It is allowed to rotate a necklace but not to turn it over. Any of the n vertices can be chosen as the first endpoint of a diagonal. How many necklaces can be made using 13 different beads? See Figure We can choose any of the 20 towns in the country say.

Counting the diagonals this way. Let us first assume that it is prohibited to rotate the necklace. How many air routes are there? There are 20 towns in a certain country. Every route connects two towns. A similar problem is discussed in the chapter "Graphs-I" where we count the number of edges of a graph. Assume now that it is allowed to turn a necklace over. A "necklace" is a circular string with several beads on it. How many different necklaces can be made using 13 different beads?

Then it is clear that we have 13! Since these are exactly the numbers with all their digits odd. How many ways are there to buy an envelope and a stamp? We also recommend reviewing the material already covered in previous sessions.

In conclusion we would like to note that it is reasonable to devote a separate session to any idea which ties together the problems of each set in this chapter and.

There are five types of envelopes and four types of stamps in a post office. A word is any sequence of six letters. How many words are there in the Hermetian language? The main idea in this solution was to use the method of complements. Problems for independent solution Problem How many six-digit numbers have at least one even digit?

There are six letters in the Hermetian language. In addition. For this reason we present here a list of problems for independent solution and for homework. Instead of counting the numbers with at least one even digit.

Since there are six-digit numbers in all. How many games will be played if there are 18 participants? How many ways are there to place a set of chess pieces on the first row of a chessboard? The set consists -of a king.

How many "words" can be written using exactly five letters A and no more than three letters B and 10 other letters? How many ways are there to do this? The rules of a chess tournament say that each contestant must play every other contestant exactly once. Seven nouns. Each of two novice collectors has 20 stamps and 10 postcards. How many ways are there to place a two bishops. We can form a sentence by choosing one word of each type. How many ways. How many ways are there to arrange them in pairs for a dance?

There are three rooms in a dormitory: How many six-digit numbers have all their digits of equal parity Call odd or all even?

Every morn- ing. Among all possible outcomes. Mother has two apples. How many ways are there to arrange some or all of them in a stack? The stack may consist of a single book. There are N boy. How many ways are there to carry out one fair exchange between these two collectors?

How many ways are there to split 14 people into seven pairs? Problem 4B! How many nine-digit numbers have an even sum of their digits?

How many ways are there to house seVen students in these rooms? We call an exchange fair if they exchange a stamp for a stamp or a postcard for a postcard. How many ways are there to put eight rooks on a chessboard so that they do Rot attack each other? How many ten-digit numbers have at least two equal digits? Problem 45! In how many. We toss a die three times. There are five books on a shelf. How many ways are there to choose four cards of different suits and different values from a deck of 52 cards?

Even very routine problems like the factoring of integers can be turned into a contest by asking "Who can factor this huge number first?

It can be represented. The number 1 is neither prime nor composite. Try to introduce elements of play in your sessions. Since divisibility also enters into the school curriculum. This theme is not so recreational as some others. Prime and composite numbers Among natural numbers we can distinguish prime and composite numbers. This is the complete "decomposition" of our number its representation as a product of prime numbers. But each of the numbers 42 and 10 is composite.

How can this be done? Let us consider the number It is certainly composite. We just keep factoring the numbers we have into pairs of smaller numbers Prime numbers are like "bricks". A number is composite if it is equal to the product of two smaller natural numbers.

Most of the contents of this section are connected with the Fundamental Theorem of Arithmetic. Is 29 3 divisible by 9? But what if we try to factor the number in some other way?

Is 29 3 divisible by 8? It is called the FUnda- mental Theorem of Arithmetic: Is it true that if a natural number is divisible by 4 and by 3.

Students should unaerstand that the properties of divisibility are almost com- pletely determined by the representation of a nat. It may surprise you that we will always end up with the same representation products which differ only in the order of their factors are considered identical-we usually arrange the factors in increasing order. This may seem evident. The following exercises will help. Is it true that A must be divisible by 67 Answer. The number 5A is divisible by 3.

A might be 2. The number A is even. Since this number is also divisible by 3. Is it true that A must be divisible by 37 Answer. Thus we can only be sure that A is even. The reason is that the number 3. Is it true that if a natural number is divisible by 4 and by 6. Is it possible that the number 2A is divisible by 37 Answer. The number 15A is divisible by 6. The number A is not divisible by 3. Is it true that 3A must be divisible by 67 Answer.

The reason is that if a number is divisible by 4. The Greatest Common Divisor G. Encourage students to form their own conjectures regarding the problems and theorems you discuss. Given two different prime numbers p and q. Problems using relatively prime numbers can be found at the end of the section. The Least Common Multiple L. Methods and ideas introduced in this section will be used for the solution of problems in other sections of this chapter as well as in other chapters of the present book.

This is the common part "intersection" of the decompositions of the numbers. We ask you to think of the material of this section as just a sketch of a scenario for an actual session.

Prove that the product of any three consecutive natural numbers is divisible by 6. Prove that the product of any five consecutive natural numbers is a divisible by In some places you will probably give a series of very similar problems or exercises one after another.

As a teacher. Students should discuss and solve a few examples. There is at least one even number. We give here a list of some such problems. Any number divisible by 2 and by 3 is divisible by 6. For some number n. Add 3 to both sides of the equation. Problem II! Prove that any two natural numbers a and b satisfy the equation gcd o.

Three hundred O's. Then he changed all the digits to letters different digits were changed to different letters. You have a cent coin in your pocket but you do not have any 3-cent coins. What if you had a cent coin? This is a representation of the operation of division of 20 by 3 with a remainder. How many zeros are there at the end of the decimal representation of the number ! Find the smallest natural number n such that n!

Tom multiplied two two-digit numbers on the blackboard. Remainders Assume that you are in a country where coins of certain values are in circulation.

Given a prime number p. Can a number written with one hundred 0'5. After that it gives you coins for this remainder. Prove that if a number has an odd number of divisors. How does our change machine work? It gives out 3-cent coins until the remain- der is less than 3. You should draw the students' attention to the idea of the solution to the last problem.

This could be done. This number is divisible by 3. It is clear that the remainder is zero if and only if the original number the value of the coin you put into the machine is divisible by 3. We call the number r the remainder when the number N is divided by m.

Now we give a more accurate definition: Suppose we replace each of the numbers with its remainder when divided by 3. This is the number 4. Now we can discuss the following problem: What is the change after he or she receives the 3-cent coins? This is easy. This machine would represent the oper- ation of division by m with a. What is remarkable is that we do not have to calculate the sum of all the products.

The solution. The number n can give any of the following remainders when divided by 3: This case-by-case analysis completes the required proof.

Your students must learn how to app. In the Lemma On Remainders the number 3. The reason is that the following statement is always true: Lemma On Remainders. Generalizations of the Lemma On Remainders will be used throughout this section. We claim that the remainder of the original expression that is.

If n has remainder 2. A formal proof of this fact is not very difficult. We do pot think that a discussion of the proof of the Lemma On Remainders is absolutely necessary in the sessions. Methodological remark.

Thus we consider three cases. If n has remainder 0. We recommend solving a number of problems similar to Problem The natural numbers x. This method deserves to be pointed out to the students.

The key moment in the last solution was the idea of a case-by-case analysis. If two numbers have equal remainders when divided by 9. Prove that d is divisible by 6. They should understand that such an analysis indeed gives us a complete and rigorous proof. Prove that n 3. The sum of the squares of three natural numbers is divisible by 9. Three prime numbers p. Prove that we can choose two of these numbers such that their difference is divisible by 9. We hope that the following problems will help achieve this objective.

To calculate the last digit of a power of 9 it is sufficient to multiply by 9 the last digit of the previous power of 9.

Given natural numbers a.

Prove that if we decrease by 7 the sum of the squares of any three natural numbers. Find the last digit of the number Solution. To begin let us note that the last digit of the number is the same as the last digit of the number 9 We write down the last digits of the first few powers of 9: Case-by-case analysis can also be used in many fields other than arithmetic.

It would be excellent for students to learn to determine whether a case-by-case analysis can help in solving a problem. Pl-oblem Prove that the given number is a multiple of both 3 and 8. Prove that the sum of the squares of five consecutive natural numbers cannot be a perfect square. The "art of guessing" requires certain skills and. As exercises to maintain the skills mentioned we suggest the mmposition of multiplication tables for remainders when divided by "the most frequently used" numbers Find remainders when divided by 3.

Find the remainder when the number is divided by 7. Let us write down the last digits of the first few powers of two: We can see that 25 ends with 2. This remainder is Show that the remainder when the given number is divided by 7 is zero.

Since the length of the cycle is 4. Prove that they form another cycle. You can. What is the last digit of ? Find the remainder of 2 when divided by 3. Write down the remainders when several powers of 2 are divided by 3. A few more problems This section contains a series of divisibility problems which are not united by any common statement or method of solution.

We will discuss in more detail the following two problems: Problem 52! Prove that the number Prove that if n. When dealing with problems about cubes of integers like Problems it is often useful to analyze the remainders modulo 7 or modulo 9.

In either case there are only three possible remainders: The material explained in this section may be used to create at least two sessions. The first of them should be devoted to the calculation of remainders. The second can be spent in discussing the idea of case-by-case analysis in solutions of various problems. Given natural numbers x.. Seven natural numbers are such that the sum of any six of them is divisible by 5.

Quite often.. Prove that each of these numbers is divisible by 5. Find the smallest natural number which has a remainder of 1 when divided by 2.

The point is that when divided by 3 or 4. Let us try again to be simple and make this number divisible by 3. We show how it works for the two numbers and Euclid's algorithm In the first section we discussed the concept of the Greatest Common Divisor of two natural numbers.

Assume that there are only n prime numbers. Such a number is easy to find. We will try to explain how one can arrive at a solution. This contradiction completes the proof. The problems of this section should not be given for solution at one session. Note that Euclid's algorithm can be shortened as follows: This is equivalent to saying that n.

It is called Euclid's algorithm. For large numbers. Prove that there are infinitely many prime numbers. This method is based on the following simple reasoning: They can be given in the course of an entire year of classes. We can demonstrate this "improved". Let us try to keep things simple and make this number divisible by 2. Pn' Then the number PIP2. Pn' Therefore. In a sense. Find the G. As you can see. Find gcd 2 However simple it may seem. Euclid's algorithm is a very im- portant arithmetic fact which can be used.

Find gcd For more details. Solution to Problem 1. One million pine trees grow in a forest. Then there would be no more than N pigeons altogether.

Introduction Students who have never heard of the Pigeon Hole Principle may think that it is a joke: If there were no more than one bead of each color among these. This is obvious. Notice the vagueness of the proposition "some pigeon hole must contain " '"two or more "This is. A bag contains beads of two colors: Show that two pine trees in the forest must have the same number of pine needles. We can draw three beads from the bag. It is known that no pine tree has more than pine needles on it.

The following problem also seems to have nothing to do with pigeons and pigeon holes: What is the smallest number of beads which must be drawn from the bag. Suppose no more than one pigeon were in each hole. Twenty-five crates of apples are delivered to a store. Given twelve integers. More general pigeons If you have read the problems above carefully..

It is exactly this kind of problem that can often be solved using the Pigeon Hole Principle. The apples are of three different sorts. They should first solve a few simple exercises. Sometimes they will not even "! On the other hand. Solution to Problem 2. We put each "pigeon" pine tree in the pigeon hole numbered with the number of pine needles on the tree.

It may be necessary to explain the difference between an intuitive understanding and an actual proof.

Since there are many more "pigeons" than pigeon holes. This can be followed by a series of problems in conscious imitation of the arguments just given by the teacher Problems We have one million "pigeons"-the pine trees-and.

In discussing these first few problems. Here the beads play the role of pigeons. From that point on. Notice that the statements of these problems include the same vagueness as the Pigeon Hole Principle itself. The city of Leningrad has five million inhabitants. Show that among these crates there are at least nine containing the same sort of apple.

Show that two of these must have the same number of hairs on their heads. Students have trouble dealing with this vagueness. We leave the proof of the General Principle as an exercise. Show that both neighbors of at least one student are boys. A snail crawls along a plane with constant velocity, turning through a right angle every 15 minutes. Show that the snail can return to its starting point only after a whole number of hours.

Three grasshoppers play leapfrog along a line. At each tum, one grasshopper leaps over another, but not over two others. Can the grasshoppers return to their initial positions after leaps? See Figure 7. Of coins, 50 are counterfeit, and differ from the genu!

Peter has a scale in the form of a balance. He chooses one coin, and wants to find out in one weighing whether it is counterfeit. Can he do this? Is it possible to arrange the numbers from 1 through 9 in a sequence so that there are oddly many numbers between 1 and 2, between 2 and 3,. This page is intentionally left blank.

How many words does the Hermetian language contain? How many "lucky" six-digit numbers are there? How many? These and many other similar questions will be discussed in this chapter.

We will start with a few simple problems. Problem 1. There are five different teacups and three different tea saucers in the "Tea Party" store. How many ways are there to buy a cup and a saucer? First, let us choose a cup. Then, to complete the set, we can choose any of three saucers. Thus we have 3 different sets containing the chosen cup.

There are also four different teaspoons in the "Tea Party" store. How many ways are there to buy a set consisting of a cup, a saucer, and a spoon? Let us start with any of the 15 sets from the previous problem. There are four different ways to complete it by choosing a spoon. In just the same way we can solve the following problem.

There are three towns A, B, and C, in Wonderland. Six r. In how many ways can one drive from A to C? A new town called D and several new roads were built in Wonderland see Figure 9.

How many ways are there to drive from A to C now? Consider two cases: In each case it is quite easy to calculate the number of routes-if we drive through B then we have 24 ways to drive from A to C; otherwise we have 6 ways. To obtain the answer we must add up these two numbers. Thus we have 30 possible routes. Dividing the problem into several cases is a very useful idea. It also helps in solving Problem 5. There are five different teacups, three saucers, and four teaspoons in the "Tea Party" store.

How many ways are there to buy two items with different names? Three cases are possible: IS, 20, and 12 ways respectively. Adding, we have the answer: For teachers. The main goal which the teacher must pursue during a discussion of these problems is making the students understand when we must add the numbers of ways and when we must mUltiply them. Of course, many problems should be presented some can be found at the end of this chapter Problems , in [49], or created by the teacher.

Some possible subjects are shopping, traffic maps, arrangement of objects, etc. We call a natural number "odd-looking" if all of its digits are odd. How many four-digit odd-looking numbers are there? It is obvious that there are 5 one-digit odd-looking numbers. We can add another odd digit to the right of any odd-looking one-digit number in five ways. In the last problem the answer has the form mn Usually, an answer of this type results from problems where we can place an element of some given m-element set in each of n given places.

In such problems the students may encounter difficulty distinguishing the two numbers m and n, therefore confusing the base and the exponent. Here are four more similar problems. We toss a coin three times. How many different sequences of heads and tails can we obtain? Each box in a 2 x 2 table can be colored black or ,white. How many different colorings of the table are there? How many ways are there to fill in a Special Sport Lotto card?

In this lotto you must predict the results of 13 hockey games, indicating either a victory for one of two teams, or a draw. The Hermetian alphabet consists of only three letters: A, B, and C. A word in this language is an arbitrary sequence of no more than four letters. Calculate separately the numbers of one-letter, two-letter, three-letter, and four-letter words.

Let us continue with another set of problems. A captain and a deputy captain must be elected in a soccer team with 11 players. How many ways are there to do tllis?

Any of 11 players can be elected as captain. After that, any of the 10 remaining players can be chosen for deputy. This problem differs from the previous ones in that the choice of captain influences the set of candidates for deputy position, since the captain cannot be his or her own deputy. Thus, the choices of captain and deputy are not independent as the choices of a cup and a saucer were in Problem 1, for example. Below we have four more problems on the same theme.

How many ways are there to sew one three-colored flag with three horizontal strips of equal height if we have pieces of fabric of six colors?

We can distinguish the top of the flag from the bottom. There are six possible choices of a color for the bottom strip. After that we have only five colors to use for the middle strip, and then only four colors for the top strip. Therefore, we have 6. How many ways are there to put one white and one black rook on a chessboard so that they do not attack each other?

The white rook can be placed on any of the 64 squares. No matter where it stands, it attacks exactly 15 squares including the square it stands on.

Thus we are left with 49 squares where the black rook can be placed. How many ways are there to put one white and one black king on a chessboard so that they do not attack each other? The white king can be placed on any of the 64 squares. However, the number of squares it attacks depends on its position. Therefore, we have three cases: We have 60 squares left, and we can place the black king on any of them. Such arrangements are called permutations, and they play a significant role in combinatorics and in algebra.

But before this we must digress a little bit. If n is a natural number, then n! Therefore, 2! For convenience of calculations and for consistency, O! Methodolo2ical remark.

Before working with permutations one must know the definition of factorial and learn how to deal with this function. The following exercises may be useful Exercise 1. Simplify the expressions a 1O! Exercise 2. Exercise 3. Prove that if p is a prime number, then p - I! Now let us go back to the permutations. How many three-digit numbers can be written using the digits I, 2, and 3 without repetitions in some order?

Let us reason just the same way we did in solving Problem The first digit can be any of the three given, the second can be any of the two remaining 2. Thus we have 3. How many ways are there to lay four balls, colored red, black, blue, and green, in a row? The first place in the row can be occupied by any of the given balls.

The second can be occupied by any of the three remaining balls, et cetera. Finally, we have the answer simifar to that of Problem For convenience of notation we introduce the following convention. Any finite sequence of English letters will be called "a word" whether or not it can be found in a dictionary. For example, we can form six words using the letters A, B, and C each exactly once: In the following five problems you must calculate the number of different words that can be obtained by rearranging the letters of a particular word.

Since all the letters in this word are different, the answer is 6! This word contains two letters T, and all the other letters are different. Let us temporarily think of these letters T as two different letters Tl and T2. Under this assumption we have 5! However, any two words which can be obtained from each other just by transposing the letters Tl and T2 are, in fact, identical. Thus, our words split into pairs of identical words. However, any words which can be obtained from each other just by transposing the letters A, are identical.

Since the letters Ai can be rearranged in their places in 3!

Therefore the answer is 8! We have three letters S and two letters E in this word. Temporarily thinking of all of them as different letters, we have 9!

Then, recalling that the letters S are identical, we come to the final answer: This set of problems about words demonstrates one very interesting and important idea-the idea of multiple counting. That is, instead of counting the number of objects we are interested in, it may be easier to count some other objects whose number is some known multiple of the number of objects. Here are four more problems using this method. There are 20 towns in a certain country, and every pair of them is connected by an air route.

How many air routes are there? Every route connects two towns. We can choose any of the 20 towns in the country say, town A as the beginning of a route, and we have 19 remaining towns to choose the end of a route say, town B from. However, this calculation counted every route AB twice: A similar problem is discussed in the chapter "Graphs-I" where we count the number of edges of a graph. How many diagonals are there in a convex n-gon? Any of the n vertices can be chosen as the first endpoint of a diagonal, and we have n - 3 vertices to choose from for the second end any vertex, except the chosen one and its two neighbors.

Counting the diagonals this way, we have counted every diagonal exactly twice. See Figure A "necklace" is a circular string with several beads on it. It is allowed to rotate a necklace but not to turn it over. How many different necklaces can be made using 13 different beads? Let us first assume that it is prohibited to rotate the necklace. Then it is clear that we have 13! However, any arrangement of beads must be considered identical to those 12 that can be obtained from it by rotation. Assume now that it is allowed to turn a necklace over.

How many necklaces can be made using 13 different beads? How many six-digit numbers have at least one even digit? Instead of counting the numbers with at least one even digit, let us find the number of six-digit numbers that do not possess this property. The main idea in this solution was to use the method of complements; that is, counting or, considering the "unrequested" objects instead of those "reguested" Here is another problem which can be solved using this method.

There are six letters in the Hermetian language. A word is any sequence of six letters, some pair of which are the same. How many words are there in the Hermetian language? We also recommend reviewing the material already covered in previous sessions. For this reason we present here a list of problems for independent solution and for homework.

In addition, you can take problems from [49] or create them yourself. Problems for independent solution Problem There are five types of envelopes and four types of stamps in a post office.

How many ways are there to buy an envelope and a stamp? Seven nouns, five verbs, and two adjectives are written on a blackboard. We can form a sentence by choosing one word of each type, and we do not care about how much sense the sentence makes.

How many ways are there to do this? Each of two novice collectors has 20 stamps and 10 postcards. We call an exchange fair if they exchange a stamp for a stamp or a postcard for a postcard. How many ways are there to carry out one fair exchange between these two collectors?

Problem , How many six-digit numbers have all their digits of equal parity Call odd or all even? In how many. How many ways are there to choose four cards of different suits and different values from a deck of 52 cards? There are five books on a shelf. How many ways are there to arrange some or all of them in a stack? The stack may consist of a single book. How many ways are there to put eight rooks on a chessboard so that they do Rot attack each other?

There are N boy. How many ways are there to arrange them in pairs for a dance? The rules of a chess tournament say that each contestant must play every other contestant exactly once. How many games will be played if there are 18 participants?

How many ways are there to place a two bishops; b two knights; c two queens on a chessboard so that they do not attack each other? Mother has two apples, three pears, and four oranges. Every morning, for nine days, she gives one fruit to her son for breakfast. There are three rooms in a dormitory: How many ways are there to house seVen students in these rooms?

How many ways are there to place a set of chess pieces on the first row of a chessboard? The set consists -of a king, a queen, two identical rooks, two identical knights, and two identical bishops. How many "words" can be written using exactly five letters A and no more than three letters B and 10 other letters?

How many ten-digit numbers have at least two equal digits? Problem 45! We toss a die three times. Among all possible outcomes, how many have at least one occurrence of six? How many ways are there to split 14 people into seven pairs? Problem 4B! How many nine-digit numbers have an even sum of their digits? This theme is not so recreational as some others, yet it contains large amounts of important theoretical material.

Try to introduce elements of play in your sessions. Even very routine problems like the factoring of integers can be turned into a contest by asking "Who can factor this huge number first? Since divisibility also enters into the school curriculum, you can use the knowledge acquired by students there. Prime and composite numbers Among natural numbers we can distinguish prime and composite numbers. A number is composite if it is equal to the product of two smaller natural numbers.

Otherwise, and if the number is not equal to 1, it is called prime. The number 1 is neither prime nor composite. Prime numbers are like "bricks", which you can use to construct all natural numbers. How can this be done? Let us consider the number It is certainly composite. It can be represented, for instance, as 42 But each of the numbers 42 and 10 is composite, too.

This is the complete "decomposition" of our number its representation as a product of prime numbers.

But what if we try to factor the number in some other way? It may surprise you that we will always end up with the same representation products which differ only in the order of their factors are considered identical-we usually arrange the factors in increasing order. This may seem evident, but it is not easy to prove. It is called the FUndamental Theorem of Arithmetic: Most of the contents of this section are connected with the Fundamental Theorem of Arithmetic.

Students should unaerstand that the properties of divisibility are almost completely determined by the representation of a nat. The following exercises will help. Yes, since 2 is one of the factors in the decomposition of the given number. No, since the decomposition of this number does not contain the prime number 5.

Is 29 3 divisible by 8? Is 29 3 divisible by 9? Indeed, the decomposition of a natural number which is divisible by 4 must contain at least two 2's.

Since this number is also divisible by 3, there is also at least one 3. For example, the number 12 can serve as a counterexample. The reason is that if a number is divisible by 4, then its decomposition contains at least two 2's; if the same number is divisible by 6, then it means that its decomposition contains 2 and 3.

Therefore, we can be sure that the decomposition has two 2's but not necessarily three! The number A is not divisible by 3. Is it possible that the number 2A is divisible by 37 Answer. No, since 3 does not belong to the decomposition of A, and, therefore, does not belong to the decomposition of 2A. The number A is even. Is it true that 3A must be divisible by 67 Answer. The number 5A is divisible by 3.

Is it true that A must be divisible by 37 Answer. Yes, since the decomposition of 5A contains 3, while the decomposition of 5 does not. The number 15A is divisible by 6. Is it true that A must be divisible by 67 Answer. For example, A might be 2. The reason is that the number 3, which is one of the prime factors of the number 6, also belongs to the decomposition of the number Thus we can only be sure that A is even.

For example, two different prime numbers are, of course, relatively prime. Also, the number 1 is relatively prime to any other natural number.

Students should discuss and solve a few examples. Problems using relatively prime numbers can be found at the end of the section. The Greatest Common Divisor G.

The Least Common Multiple L. This is the common part "intersection" of the decompositions of the numbers. This, as you can see, is the "union" of the numbers' decompositions. We ask you to think of the material of this section as just a sketch of a scenario for an actual session.

As a teacher, you will want to create a more elaborate version. In some places you will probably give a series of very similar problems or exercises one after another, yet try to keep things varied. Encourage students to form their own conjectures regarding the problems and theorems you discuss. However, this topic will be learned best if, in further sessions, there are problems using the ideas explained above. We give here a list of some such problems.

Methods and ideas introduced in this section will be used for the solution of problems in other sections of this chapter as well as in other chapters of the present book. Given two different prime numbers p and q, find the number of different divisors of the number a pq; b p2q; c p2q2; d pnqm Problem 2. Prove that the product of any three consecutive natural numbers is divisible by 6.

There is at least one even number, and at least one number divisible by 3, among any three consecutive numbers. Any number divisible by 2 and by 3 is divisible by 6, so the result follows directly from the hint. Prove that the product of any five consecutive natural numbers is a divisible by 30; b divisible by Given a prime number p, find the number of natural numbers which are a less than p and relatively prime to it; b less than p2 and relatively prime to it.

Find the smallest natural number n such that n! How many zeros are there at the end of the decimal representation of the number ! For some number n, can the number n' have exactly five zeros at the end of its decimal representation? Prove that if a number has an odd number of divisors, then it is a perfect square.

Tom multiplied two two-digit numbers on the blackboard. Then he changed all the digits to letters different digits were changed to different letters, and equa.

Can a number written with one hundred 0'5, one hundred 1's, and one hundred 2's be a perfect square? This number is divisible by 3, but not by 9. This, then, must be true of the number in the problem, regardless of the order in which its digits appear.

You should draw the students' attention to the idea of the For teachers. This could be done, for example, by asking what if the number described had two hundred O's, 1's, and 2's? Three hundred O's, 1's, and 2's? Problem II! Add 3 to both sides of the equation, then factor the left-hand side. Remainders Assume that you are in a country where coins of certain values are in circulation, and you want to buy a stick of gum for 3 cents from a vending machine.

You have a cent coin in your pocket but you do not have any 3-cent coins, which you need to buy the gum. Fortunately, you see a change machine which can give you any number of 3-cent coins. Obviously, you get five 3-cent coins for your I5-cent coin. What if you had a cent coin? Then, of course, you get six 3-cent coins plus two cents change.

This is a representation of the operation of division of 20 by 3 with a remainder. How does our change machine work? It gives out 3-cent coins until the remainder is less than 3. It is clear that the remainder is zero if and only if the original number the value of the coin you put into the machine is divisible by 3. Analogously, we can imagine a machine which gives m-cent coins and change which varies from 0 through m - 1 cents. This machine would represent the operation of division by m with a.

Now we give a more accurate definition: We claim that the remainder of the original expression that is, of the number x is also 1. The reason is that the following statement is always true: Lemma On Remainders. Methodological remark. A formal proof of this fact is not very difficult, though for beginners it may seem full of technicalities. For example, let us prove , the second proposition.

In the Lemma On Remainders the number 3, of course, can be changed to any other natural number: Generalizations of the Lemma On Remainders will be used throughout this section. Your students must learn how to app,ly these ideas when calculating remainders. We recommend solving a number of problems similar to Problem 15, drawing the students' attention to the use of these statements.

We do pot think that a discussion of the proof of the Lemma On Remainders is absolutely necessary in the sessions. The solution.

The number n can give any of the following remainders when divided by 3: Thus we consider three cases. This case-by-case analysis completes the required proof. The key moment in the last solution was the idea of a case-by-case analysis, used to examine all the possible remainders modulo some natural number.

This method deserves to be pointed out to the students. They should understand that such an analysis indeed gives us a complete and rigorous proof. Case-by-case analysis can also be used in many fields other than arithmetic. It would be excellent for students to learn to determine whether a case-by-case analysis can help in solving a problem.

We hope that the following problems will help achieve this objective. Prove that n 5 Problem Prove that n 3 - n is divisible by 24 for any odd n. Prove that the given number is a multiple of both 3 and 8. Given natural numbers a and b such that a2 prove that the same sum of squares is also divisible by 44l. Pl-oblem Three prime numbers p, q, and r, all greater than 3, form an arithmetic progression: Prove that d is divisible by 6.

Prove that if we decrease by 7 the sum of the squares of any three natural numbers, then the result cannot be divisible by 8. The sum of the squares of three natural numbers is divisible by 9. Prove that we can choose two of these numbers such that their difference is divisible by 9. If two numbers have equal remainders when divided by 9, then their difference is divisible by 9.

Find the last digit of the number Solution. To begin let us note that the last digit of the number is the same as the last digit of the number 9 We write down the last digits of the first few powers of 9: Therefore, the last digit of is also 9.

Let us write down the last digits of the first few powers of two: Jl'O"rer is determined by the last digit of the previous power of 2, we have a cycle: Since the length of the cycle is 4, the last digit of the number 2 can be iJund using the remainder of the number 50 when divided by 4.

This remainder is 2. What is the last digit of ? Find the remainder of 2 when divided by 3. Write down the remainders when several powers of 2 are divided by 3. Prove that they form another cycle. Find the remainder when the number is divided by 7. Show that the remainder when the given number is divided by 7 is zero. Find the last digit of the number The "art of guessing" requires certain skills and, while there are some standard tricks, can be quite difficult.

You can. Find remainders when divided by 3. Given the pair of prime numbers p and 8p2 Problem Open Preview See a Problem? Details if other: Thanks for telling us about the problem. Return to Book Page. Preview — Mathematical Circles by Dmitri Fomin.

Mathematical Circles: This text was produced by a remarkable cultural circumstance in the former Soviet Union which fostered the creation of groups of students, teachers, and mathematicians called mathematical circles.

The work is predicated on the idea that studying mathematics can generate the same enthusiasm as playing a team sport - without necessarily being competitive.

Get A Copy. Paperback , pages. Published July 1st by American Mathematical Society. More Details Original Title. Other Editions 1.

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