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January 2008

Awesome Sets

by Iurie Boreico, Ivan Borsenco, and Radu Sorici

In this fth AMY Segment 1 we will talk about sets. As you will see sets arethe structures that are very comfortable to use in many mathematical areas.To simplify the understanding of the material we have divided the segment inthree chapters

Chapter 1. Basics of counting with sets

The basic principles: sum, product, function, bijection Inclusion Exclusion Principle

Chapter 2. Sets and bounds: Pigeonhole extended Ordered and partially ordered sets Additive sets

Chapter 3. Sets and useful tricks. Convex sets Counting in two ways

Look for the matrix!

Every chapter contains a theoretical part which is presented together with theexamples. At the end of each chapter we give you a list of problems related tothe material discussed in it. We encourage you to spend time to think aboutthe given problems and send us your solutions, partial or complete. Good luckand success in your challenges!

The authors

1 c 2008, AwesomeMath LLC. All Rights Reserved.

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AMY 2007-2008 Awesome Sets 2

Chapter 1. Basics of counting with sets

The basic principles: sum, product, function, bijection

A set is simply a collection of objects. For example one can consider theset of apples in a box, the set of services offered by your mobile operator, oreven the set of all molecules in the universe. Each set is uniquely determinedby its objects, which are called elements of a set. If an element x belongs toa set S , we write x S . By denition, each element appears in a set exactlyonce, or does not appear at all. For example, {1, 1, 2, 3} does not signify aset, because 1 appears two times, whereas {1, 2, 3} denes a set (the sets inwhich elements can appear with repetitions are called multisets, but we will notconsider them). One can even consider more complicated sets, which consist of objects of different types, for example we have the set {1, {2, 3},Harry }, whichconsists of a number, a wizard, and a set. However in our presentation theexamples will all be homogeneous, containing only elements of the same type.

A basic set is the so-called empty-set , denoted as or { }. It is the set whichhas no elements.Sets can also interact with each other. The basic operations are union and

intersection. The union of some sets, as the name suggests, consists of allelements that belong to at least one of some sets, whereas the intersectioncontains all elements that belong to all the considered sets. Unions are denotedby whereas intersections with . A set A is called a subset of another set B if B contains all the elements of A. We use the notation A B for this situation.

Thus, the empty set is a subset of all the sets but its only subset is itself. If aset contains n elements, it has 2 n subsets.From this basic denitions a lot more can arise. For example one can consider

the set difference A B or A \ B as the set of all elements belonging to A butnot to B . If we consider a framework where all sets are subsets of a general bigset X , then we can dene the complement of a set A, denoted as A, as simplyX A. We have the De Morgan Laws, which are simple exercises for the reader:

Theorem 1.1. Let A and B be two subsets of the same set. Then

A B = A B and A B = A B.

In other words, union and intersection interchange under complementation.Using two sets A, B we can also dene the product A B , which is the setof all pairs (a, b) such that a A, b B . The product can be dened similarlyfor more than two sets.A more complicated but very important concept is also the concept of func-

tion . A function f from set A to set B, also denoted f : A B , is a rule thatassociates to each element x of A and element f (x) of B . The set of all values

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f (x) is called the image of f , and is a subset of B. It is sometimes denoted

as Im (f ) or Im (A) or Im f (A). The function is called injective if to differentelements of A correspond different elements of B , surjective if the image is thewhole set B, and bijective if it is both injective and surjective. We can also say f is an injection, surjection, respectively bijection. If f is a bijection, then to eachelement y of B there corresponds a unique x = g(y) in A such that f (x) = y.By denition, if f : A B is injective, then f is a bijection between A and theimage of f . It is easy to see that f (g(y)) = y, g(f (x)) = x, and the function gis called the inverse of the function f , sometimes written f 1 . The converse isalso true: if there exists a function g: B A such that f (g(y)y, g(f (x)) = xfor any x A, y B , then f is a bijection (and g is too). If there is a bijectionfrom A to B, then we say that A and B are in bijection. It is also easy to seethat of f 1 , f 2 are bijections from A1 to B1 and A2 to B2 , then f that sends(a1 , a 2) to ( f (a1), f (a2)) is a bijection from A1

A2 to B1

B2 .

We will denote the number of elements of a nite set S by |S |. It is oftenalso called cardinality .Exercise: Prove that |A| = n if and only there is a bijection from A to

{1, 2, . . . , n }.Our basic problem is easy to state: given a set S of interest, what is |S |?There are two basic rules that are useful in dealing with this problem. Using

these two rules, we can break down a complicated counting problem into simplesubproblems.

If A B = , then |A B | = |A|+ |B |.The second rule is expressed in terms of the Cartesian product A B ={(a, b)|a A, b B}.

|A B | = |A| |B |The Sum Rule generalizes to n sets: if A1 , A2 , . . . , A n are pairwise disjoint sets,then

|A1 A2 An | = |A1|+ |A2|+ + |An |.The Product Rules generalizes to

|A1 A2 An | = |A1| |A2| |An |.More generally still, if a collection of ntuples ( a1 , a 2 , . . . , a n ) is formed bya sequence of choices such that for any possible choice of ( a1 , a 2 , . . . , a k

1),

the number of available choices for ak is N k , independent of the choices madethus far, then the total number of ntuples than can be formed is the productN 1N 2 N n . The two rules can be easily proved using the previously statedexercise, we leave this to the reader.

Our standard example of an nelement set, using the exercise, is {1, 2, . . . , n },which we will denote by [ n]. By the Product Rule, the number of permutations of

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[n] is n(n1) 21 = n!. By the same argument, there are n(n1) (nk+1)ktuples of elements from [ n]. To count unordered selections (subsets), we di-vide by k! and get the familiar result that the number of kelement subsetsof [n] is n (n 1) (n k + 1) /k ! = nk . The Product Rule is often used inthis way. On purpose, we rst count a larger class of congurations, and thencompensate for the overcount by dividing by the appropriate factor.

Example 1.1. How many ways are there to distribute the elements of [ n]among k unlabeled boxes if a1 of the boxes contain just one element, a2 containtwo elements, and so on, and no box is empty?

Solution. Such an arrangement is called a partition of [ n] into k blocks;the ntuple ( a1 , a 2 , . . . , a n ) species the type of the partition. Note that thenonnegative integers a1 , a 2 , . . . , a n must satisfy a1 + 2a2 + + na n = n anda1 + a2 +

+ an = k. In order to count the number of partitions of [ n] of

the type ( a1 , a 2 , . . . , a n ), imagine rst a different situation in which the boxesare labeled, and that a box containing j elements has j separate compart-ments where individual elements can be placed. In this case, there are n dis-tinguished positions for the elements of [ n] and thus n! possible arrangements.Now we must compensate for the fact that the boxes are not labeled and donot have such compartments. For each j, any permutation of the aj labeledboxes containing j elements gives the same partition; hence we need to divideby (1!) a 1 (2!)a 2 (n!)a n because permutations of elements within a box givesthe same partition. Hence we arrive at the desired formula for the number of partitions of the type ( a1 , a 2 , . . . , a n ), namely

P (a1 , a 2 , . . . , a n ; n) = n!

a1!a2! an !(1!)a 1

(2!)a 2

(n!)a n .

The most natural and attractive way to solve a new counting problem is to showit can be reduced to one we have already solved.

The following idea is very useful - the method of bijective function.

Theorem 1.2. Suppose there exists a bijection f : A B . Then |A| = |B |.An argument showing that |A| = |B | by exhibiting such a function is calleda bijective or combinatorial proof. In practice, the proof that a given mapping

f is a bijection is often carried out by exhibiting the inverse function g = f 1 .Several examples follow.

Example 1.2. Prove that the number of subsets of [2n] having the samenumber of elements as odd elements is 2nn .

Solution. Given X [2n] with k even elements and k odd elements replacethe k odd elements of X by then n k odd elements not in X to obtain f (X )[2n] with k even elements and n k odd elements. Thus f maps the givencollection of subsets of [2n] to the collection of all nelement subsets of [2 n]. Itis easy to see that f is a bijection, with f 1(Y ) given by exchanging the oddelements of Y for the odd elements of [2 n] not in Y .

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Example 1.3. Find the number of ktuples ( S 1 , S 2 , . . . , S k ) satisfyingS 1 S 2 S k [n].

Solution. The desired number is ( k + 1) n . To see this, we construct abijection that associates with each set the sequence ( S 1 , S 2 , . . . , S k ) satisfying

S 1 S 2 S k [n]a correspondence codeword ( a1 , a 2 , . . . , a n ) in which each aj assumes one of k + 1 possible values. To construct the codeword that goes with a given nestedsequence of sets, we proceed as follows. For each j between 1 and n, if j S klet a j be the index of the rst set in the chain S 1 , S 2 , . . . , S k that contains j. If j / S k , set aj = k + 1 . To see that this mapping is a bijection, note that the

chain S 1 , S 2 , . . . , S k is uniquely constructed from its codeword by settingS j = { j |a j i}, i = 1 , 2, . . . , k .

Thus there are just as many chains as there are codewords, namely ( k + 1) n .A 0 1 string of length n is an ntuple ( a1 , a 2 , . . . , a n ) in which each elementis either 0 or 1.

Example 1.4. Prove that the number of 0 1 strings of length n such thatexactly m of the zeroes are followed immediately by ones is n +12m +1 .Solution. Given a 0 1 string ( a1 , a 2 , . . . , a n ) enlarge it to a string of length n + 2 by adding a0 = 1 and an +1 = 0 . Then associate with the enlarged

string a corresponding one ( b1 , b2 , . . . , bn +1 ) by setting bi = 0 if ai = a i

1 andbi = 1 otherwise. It is easy to check that the mapping just dened is a bijectionfrom the set of 0 1 strings ( a1 , a 2 , . . . , a n ) that have exactly m zeros that arefollowed immediately by ones to the set of 0 1 strings ( b1 , b2 , . . . , bn +1 ) thathave 2m + 1 ones. There are n +12m +1 strings in the latter set, so the proof iscomplete.

Example 1.5. How many A B sequences consisting of n A s and n B shave the property that, when read from left to right, the number of A s neverlags behind the number of B s. (Think of an election in which the votes arecounted one at a time; at the end there is a tie, but the candidate A never trailscandidate B as the votes are counted.)

(Ballot Problem)

Solution. We will refer to the fact that A never trails B as the votesare counted as ballot condition. Without the ballot condition, there are 2nnpossible sequences. Of these sequences, certain ones are bad ( A trails B atsome point.) In a bad sequence, there is a rst time where the number of B sexceeds the numbers of A s. At this point, there is one more B than A, andreversing all symbols up to that point gives a sequence with n +1 A s and n1

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B s. This reection (exchanging A s and B s) provides a bijection from the

set of bad sequences onto the set of all sequences with n + 1 A s and n 1 Bs.This shows the number of bad sequences is 2nn +1 , so there are2nn

2nn + 1

= 1n + 1

2nn

acceptable once. The number

C n = 1n + 1

2nn

is called the nth Catalan number. The rst six Catalan numbers are C 0 =1, C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14, C 5 = 42. Remember that Catalan numbersshow up in many important combinatorial problems.

Here is an example that may justify the denition of cardinality for innitesets.

Example 1.6. Assume the we have sets A and B, such that there existsan injective function f from A to B and an injective function g from A to B.Prove that these two sets are in bijection.

Solution. Let A1 be the image of f . Clearly A is in bijection with A1 , butB may comprise some other elements. Say C = B \ B1 . The image of C underg is C 1 . The image of C 1 under f is C 2 , and notice that we have C 2 , C aredisjoint because C 2 A1 . Likewise we construct the sets C 3 , C 4 , . . . with theproperties that C 2i+1 = g(C 2i ), C 2i = f (C 2i1).

Consider sets A = C 1 C 3 . . . , B = C C 2 . . . . The sets A , B are inbijection as the pairs ( C, C 1), (C 2 , C 3), . . . are. The sets A \A , B \B are also inbijection, and the bijection is provided by f . If x in A \ A , then f (x) B \ B ,because if f (x) B , then f (x) C 2k . The latter implies x C 2k1 , so x A ,by injectivity of f . Moreover, f is a surjection: if y B \B , then as y C,there exists a unique x A such that f (x) = y and since y C 2k , x C 2k1 .Hence x A \ A . This is the required bijection.

Remark. If the sets are nite, the problem is easy: if there is an injectionfrom A to B, then the cardinality of A is less than or equal to the cardinalityof B . Also, if there is an injection from B to A, then the cardinality of B is lessthan or equal to the cardinality of A. Therefore the two cardinalities are equaland we can construct a bijection. This argument fails in the innite case, of

course. However, our problem justies the abstract denition of cardinality,applied for all kinds of sets. One says that a set A has greater cardinality thanB if there exists an injection from A to B, and the cardinalities are equal if and only if the sets are in bijection. What our problems proved is the usualordering property: if the cardinality of a set is both greater and less than thecardinality of another, then these two cardinalities are in fact equal.

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The Inclusion-Exclusion Principle

When we have a problem with many intersections or unions of some sets anidea comes handy:

Assume that we have the sets A1 , A2 , . . . , A n . For a subset {i1 , i2 , . . . , i k} of [n], we can form the set M i 1 ,i 2 ,...,i k consisting of all elements that belong to eachof the sets A i 1 , Ai 2 , . . . , A i k and to none of the other. It is often comfortable tocount these sets instead of the original ones. A simplication is also possible,by setting M k the set of elements that belong to exactly k of the sets. Thefollowing very important principle can be proved using this method.

Theorem 2.1. If A1 , A2 , . . . , A n are some sets, then

|A1 A2 . . . An

|=

n

i=1 |Ai

| 1i

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where = 13 , = 12 , =

16 , and = 0.

f (Ai ) = p2i + (2 pi )2 + . . . +

n pi

pi2

= p2i n

pi

3+

n pi

2+

n pi

.

Analogously,

f (Ai Aj ) = pi p2j n

pi pj

3

+ n

pi pj

2

+ n

pi pj, i = j

Using the Inclusion-Exclusion Principle we get

a21 + a22 + . . . a

2 (n ) = f (A) f (

k

i=1

Ai ) =

= n 3 + n 2 + n k

i =1

p2i c n

pi

3

+ n

pi

2

+ n

pi+

+1i

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Here is a rather quaint application of the Inclusion-Exclusion Principle:

Example 2.2. How many 2008-digit numbers are there having no even digiton an odd place? (Starting from the dominant digit)

Solution. There are 1004 odd positions in a 2008-digit number: 1 , 3, . . . , 2007.Let us denote by Ai the set of numbers having an even digit on place 2 i 1.What we seek are the numbers that do not belong to any of the A i . Their totalnumber is 10 2008 | Ai |, as there are 10 2008 2008-digit numbers (we considerthat a number can start with the digit 0). But

| Ai | = |Ai | |Ai Aj |+ . . . .Let us evaluate what is the term |Ai 1 Ai 2 . . . Ai k |. These are all numberswhich have an even digit on each of the k places 2i1

1, 2i2

1, . . . , 2ik

1.

As there are 5 even digits and 10 digits in total, we have 5 possibilities foreach of these place and 10 for each of the remaining 2008 k places. Hence|Ai 1 Ai 2 . . . Ai k | = 5 k 102008 k . Now, there are 1004 sets so i1 , i2 , . . . , i k canbe chosen in 1004k ways. Thus,

|Ai | |Ai Aj |+ . . . =1004

k=1

(1)k11004

k5k 102008 k .

Therefore the answer to the problem is

102008 (1004

k=1

(1)k11004

k5k 102008 k ) =

1004

k=0

(1)k1004

k5k 102008 k .

We recognize here the formula of Newton Binomial Expansion, the sum is just

101004 (1004

k =0

(1)k1004

k5k 101004 k ) = 10 1004 (10 5)1004 = 50 1004 .

Remark. A simpler and much more natural way is as follows: for each of the1004 odd positions, we have only 5 admissible digits (odd), and all 10 digits areallowed for the remaining 1004 positions, so the total number is 5 1004 101004 =501004 . However, we have chosen the other solution to illustrate the method of Inclusion-Exclusion.

Example 2.3 A permutation of a set S is called a derangement if itdoes not have xed points, i.e., if (x) = x for all x S. Find the number of derangements of the set {1, 2, . . . , n }.

Solution. We count instead the permutations that are not derangements.Denote by Ai the set of permutations with (i) = i. Because the elementsin Ai have the value at i already prescribed, it follows that |Ai | = ( n 1)!.

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And for the same reason, |Ai 1 A i 2 A i k | = ( n k)! for any distincti1 , . . . , i k , 1 k n. Applying the inclusion-exclusion principle, we nd that

|A1 A2 An | =n1

(n 1)! n2

(n 2)! + + ( 1)nnn

1!.

The number of derangements is therefore

n!n1

(n 1)! +n2

(n 2)! + + ( 1)nnn

0!.

This number can also be written as

n! 1 11!

+ 12! +

(1)nn!

,

which is approximately equal to n !e , when n goes to innity.In the end, we want to exemplify the method mentioned in the beginning of

the chapter by showing how it helps solve a difficult problem.

Example 2.4. We have n nite sets A1 , A2 , . . . , A n such that the cardinalityof the intersection of any collection of them is even with the exception that thecardinality of the intersection of all them is odd. Find the least possible numberof elements that A1 A2 . . . An can have.

Solution. Set A i = ( A1 A2 . . . An ) \ Ai . For a partition into subsetsI, J of {1, 2, . . . , n } denote [I, J ] = i I A i j J AJ .Let us prove by induction on

|I

| that [ I, J ] has odd cardinality for

|I

| > 0.

By using the Inclusion-Exclusion Principle we deduce that |A1 A2 . . . An |is odd while the unions of other collections of subsets have even cardinality. For|I | = 1 we have [I, J ] = Ai \ ( j = i Aj ). If we set B = j = i Aj , then |B | is even,|Ai | is even |Ai B | is odd. Therefore

|[I , J ]| = |A \ B | = |A| |A B | = |A| (|A|+ |B | |A B |).As |A|, |B | are even, while |A B | is odd, so we deduce [I, J ] is odd. If |I | 2,then we can assume n 1, n I . We set An 1 = An 1 An . It suffices toapply the induction hypothesis for I = I \{n} and for A1 , A2 , . . . , A n 2 , An 1 .This yields the induction step.

As we can partition

{1, 2, . . . , n

} in 2n

1 ways with

|I

| = 0 and the sets

[I, J ] are clearly distinct, we deduce that |A1 A2 . . . An | 2n 1. It canequal 2n 1. The example is inspired by the solution: If we let Ai be the numberof all numbers between 1 and 2 n 1 whose ith binary digit is 1, then we cansee these sets satisfy the conditions and their union has cardinality 2 n 1.

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Ordered and partially ordered sets.

An important branch of combinatorics investigates the order relationshipsthat arise between objects. An example of such relationship may be the usualordering of real numbers, ordering of sets by inclusion or the positive integersby divisibility. One of the fundamental theorems of this branch is DilworthsTheorem.

Consider a set S and let us introduce a partial order on S . For each a andb in S, we may either have a relationship between a and b, a b, or have none,a b, that is a and b are not comparable. The partial order relationship mustsatisfy three conditions; it is

reexive: for all a S we have a a.

anti-symmetric: for all a, b S, if a b, then b a. transitive: for all a, b, c S, a b and b c implies a c.The set equipped with this relationship is called a poset (partial ordered

set). The following denitions are common for posets:A chain is a sequence of pairwise comparable elements a1 a2 . . . an ,a totally ordered subset.An antichain is a sequence of pairwise incomparable elements a1 , a2 , . . . , a n ,

in other words any two elements in the sequence are not related.The partial order length L is the length of the longest chain. The partial

order width W is the length of the longest antichain.Theorem 2.2. The width W of a poset S equals the minimum number of

chains needed to cover S , equivalently S is a disjoint union of W chains.(Dilworths Theorem)

Proof. We prove by induction on |S | that we can cover it by k chains, wherek is the number of elements of the longest antichain. The fact that S cannotbe covered by less than k chains is clear, because otherwise by the PigeonholePrinciple two of the elements of the longest antichain would belong to the samechain, a contradiction. Now we prove that we can cover S by at most k chains.In other words, we prove by induction on |S | that if from any k + 1 numbers wehave two comparable, then S may be covered by k chains.

The base case is clear. Now take a maximal element a (we mean an elementnot smaller than any other element). By the induction hypothesis S

\{a}

canbe partitioned into k chains C 1 , C 2 , . . . , C k . Let xi be the smallest element of C i and yi be the largest element of C i . If no element is comparable with a, thenwe actually see that in S \{a} there are no k pairwise incomparable elements,otherwise adding a to them would produce k+1 pairwise incomparable elementsin S . It follows that we can partition S \{a} into k 1 chains and add a as asingleton to get a cover of S by k chains.

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Assume there are some elements comparable with a, thus smaller than a. Let

zi be the largest element of chain C i smaller than a (if it exists). We can assumez1 , z2 , . . . , z m exist and zm +1 , . . . , z n do not exist. If zi = yi we can add a to C iand we are done. So assume zi < y i . Let us remove from C 1 , C 2 , . . . , C m thesubchains C i = xi . . . z i . There are no k pairwise incomparable elements betweenthe new chains C 1 , C 2 , . . . , C k , otherwise adding a to them would produce k + 1pairwise incomparable elements in S . Thus by the induction hypothesis newchains can be partitioned into at most l k 1 chains D1 , D 2 , . . . , D l . Let q ibe the smallest element of Di . If q i are from distinct original chains C i , thenwe can fulll the condition by the following reasoning: if q i C j , then add C jto the beginning of D i . We are left with at most ( k l) of C i then. If there isno C i left we add a as a singleton and get l + 1 k chains and we are done.Otherwise we have at most ( k l) + l = k chains and we can add a to one of the C i to get again at most k chains.

Thus what we are left to prove is that we can ensure that q i belong to differentchains C j . Let us color elements in k colors such that element has color i if itbelonged to C i . For a chain D i , denote its length by the number of times wechange the color when we pass the chain (i.e. 1122 has length 1, because wechange color one time from 1 to 2). Now assume that q k , q l belong to the sameC i . Assume q k . . . r k and q l . . . r l are the longest consecutive subchains of Dkand D l that consist of elements of C i (elements having the same color). We mayassume rk r l and then we can easily mix q k . . . r k and q l . . . r l into one subchainthat ends in r l (just sort them in increasing order). Thus we can move q k . . . r kand q l . . . r l to Dl . The length of D l is the same whereas the length of D kdecreases or Dk can even disappear. Applying this transformations the sum of lengths of D i decreases and we cannot perform this operation indenitely. So

at one time we will ensure that the smallest elements of D k belong to differentC i , which nishes the proof.

Remark. In a similar way it can be proved that the length L of poset S equals the minimum number of antichains needed to cover S , equivalently S isa disjoint union of L antichains. The important result of Dilworths Theoremwhich we would like to emphasize is |S | L W .

Example 2.1. Let n 1 be an integer and let X be a set of n2 +1 positiveintegers such that in any subset of X with n+1 elements there exist two elementsx = y such that x | y. Prove that there exist a subset {x1 , x2 , . . . , x n +1 } X such that x i |x i+1 for all i = 1 , 2, . . . , n .

(Romanian Team Selection Test, 2005)

Solution. Consider the poset such that a is related to b if and only if adivides b. Using Dilworth Theorem we get that there is an antichain of sizen +1 or a chain of size n +1, otherwise |X | = n2 + 1 n n = n2 , contradiction.There is no antichain of size n + 1, because we are told if we pick n + 1 elementsfrom X there exist two elements x = y such that x | y. Thus there is a chain of size n + 1, and this chain gives us the desired set.

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Example 2.2. There are mn + 1 different numbers in a row. Prove that

one can either select n + 1 numbers which form an increasing sequence if onelooks from left to right, or m + 1 numbers which for a decreasing sequence if one looks from left to right.

Solution. The problem is a direct consequence of Dilworth Theorem: onesays that a, b are comparable if the number to the left is smaller than the numberon the right, then we need to nd either a chain of length n + 1 or an antichainof length m + 1. However, as we know the length of the longest chain timesthe length of the longest antichain is at least mn + 1 the number of elements,which nishes the problem. Another solution, which emulates the easier partof the proof of Dilworth Theorem, that the length of the longest chain equalsthe minimal number of antichains to cover S (which was not proved, left as anexercise to the reader), is at follows:

For each number a, let us set the degree of a be the length of the longestincreasing sequence ending in a. It is clear that the largest degree is exactly thelength of the longest increasing sequence. Now any two elements of the samedegree are not comparable: if a < b, a is to the left of b and the degree of a isk, then the degree of b is at least k + 1 as each increasing sequence ending in amay be enlarged by adding b in the end. Therefore the set of elements of samedegree for an antichain, i.e. a decreasing sequence. It remains to see that if thelength of the longest chain is at most n, then there are at most n degrees soone of the degrees will correspond at least m + 1 numbers, by the pigeonholeprinciple.

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Additive sets.

A whole class of problems in combinatorial number theory is concerned withcertain additive properties of sets. For example, one can investigate all setsthat if contain two elements a, b, then they contain the sum a + b (these setsare called additive sets). Often, the problem is complicated by considering setsin Z p, i.e. sets that contain only residues modulo p (0, 1, . . . , p 1) and theaddition is replaced by the addition of the residues (for example, in Z p we have3 + ( p 1) = p + 2 = 2 , because p + 2 and 2 give the same residue modulo p). This class of problems is quite hard and requires a good mastery of specialtechniques.

First we would like to prove the famous Cauchy-Davenport Theorem.

Theorem 2.3. Let p be a prime number and A and B two nonempty subsetsof Z p . Then

|A + B | min{|A|+ |B | 1, p},where A + B is the set {a + b | a A, b B}.

(Cauchy-Davenport Theorem)

Proof. We prove the statement by induction on |A|, the case when |A| = 1is clear. Also, we may assume that |A| > 1 and |B | < p . Because A has at leasttwo elements, by shifting it we may assume that A contains 0 and x = 0. Also,because B is nonempty and B = Z p , there is an integer n such that nx Bbut ( n + 1) x does not belong to B. By shifting B this time we may supposethat 0 B , but x is not in B. Thus, A B is a proper nonempty subset of Aand we may use induction hypothesis for it and A B . Because A + B contains(A B ) + ( A B ) and |A|+ |B | = |A B |+ |A B |, the conclusion follows.

We can apply k 1 times Cauchy-Davenport Theorem to obtain|A1 + A2 + . . . + Ak | min{|A1|+ |A2|+ . . . + |Ak | (k 1), p}.

Alternative version. Let p be a prime and let 0 a1 < a 2 < < a m < pand 0 b1 < b2 < < bn < p be arbitrary integers. Let k be the number of distinct residues modulo p that a i + bj give when i runs from 1 to m and j runsfrom 1 to n. Prove that

a) if m + n > p then k = p;b) if m + n

p then k

m + n

1.

Solution. Written in a more symbolic way, the problem says that

|A + B | min( |A|+ |B | 1, p),if A, B Z p . Now we prove it by induction on min( |A|, |B |). Assume that|A| |B |. If |B | = p or |A| = 1 the problem is clear (the case |A| = 1 is actuallythe base case). Otherwise we prove that for some k, A + k is not contained in

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B but |(A + k) B | > 0. Indeed, x two elements a1 , a 2 A. If x B impliesthat x + a2 a1 B then we deduce x B, x + a2 a1 B, x + 2( a2 a1) Band so on, concluding that B = Z p . Otherwise there exists x B such thatx + a2 a1 B. Then we can pick k = x a1 . Now as |(A + k)| = |A| and|(A + k) + B | = |A + B | we can suppose that k = 0. The key observationis that |A B | + |A B | = |A| + |B | and A B + A B A + B . Also0 < |A B | < |A|. Therefore we can apply the induction hypothesis to A Band A B to verify the statement.

Remark. This theorem is much harder than its version for Z or R, statedas follows: if nite sets sets A, B are subsets of R, then the set A + B hascardinality at least |A| + |B | 1. This can be proved by showing a simpleexample of |A| + |B | 1 different sums: if A = {a1 < a 2 < . . . < a m } andB = {b1 < b2 < . . . < b n }, then we have m + n 1 different numbers in A + B :a1 + b1 < a 2 + b1 < . . . < a m + b1 < a m + b2 < a m + b2 < . . . < a m + bn . Thisproof fails in the case of sets in Z p because even if a number is smaller than theother, they can still be equal modulo p.

Another important and famous result is the following:

Theorem 2.4. From any 2 n 1 integers we can choose n integers such thattheir arithmetic mean is also an integer.(Erd os-Ginsburg-Ziv Theorem)

Proof. We prove the statement by induction on the number of divisors of n. The induction step is not difficult to prove. If we have 2 ab1 numbers, thenwe can select a numbers whose sum is divisible by a by induction hypothesisfor a. Of the remaining 2 ab

a

1 we can select one more group of a numbers

whose sum is divisible by a, and so on until we get 2 b1 of such groups. Indexthese 2b1 groups as 1, 2, . . . , 2b1. Let the sum of elements of group i be ax i .From 2 b 1 numbers x1 , . . . , x 2b1 by the induction hypothesis we can selectb whose sum is divisible by b. Let them be xi 1 , x i 2 , . . . , x i b . Then the groupsnumbers i1 , i2 , . . . , i b satisfy the desired condition, because we have in total abinitial numbers in them and their sum is divisible by ab.

The remaining part is the base case for n prime. It is enough to consider2 p1 residues modulo p. Write 2 p1 numbers in increasing order

a1 a2 ... a2 p1 ,where 0 a i p1. Let us split the rst 2 p2 numbers into p 1 pairs

(a1 , a p), (a2 , a p+1 ), . . . , (a p1 , a 2 p2).

If the numbers in a pair are equal, say ai = a p1+ i , then we have have pconsecutive equal numbers ai , a i +1 ,...,a p1+ i and summing them we obtain 0modulo p. Otherwise we can apply Cauchy-Davenport Theorem to the setsA1 = {a1 , a p}, A2 = {a2 , a p+1 },...,A p1 = {a p1 , a 2 p2} we get |A1 + A2 +. . . + A p1| p. Therefore every residue modulo p can be represented as a i 1 +

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a i 2 + ... + a i p 1 , where ik {k, p1+ k}. Particularly a2 p1 can be representedin such a way. Thus we have a i 1 + a i 2 + ... + a i p 1 + a2 p1 0 modulo p.

Example 2.3. Let p be a prime number, p = 2, and m1 , m 2 , . . . , m p positiveconsecutive integers and is the permutation of the set {1, 2,...,p}. Prove thatthere exist numbers k, l {1, 2,...,p}, k = l with the property

mk m (k ) m l m ( l) (mod p)Solution. It is clear that numbers m1 , m 2 ,...,m p give different residues

modulo p. Without loss of generality suppose m 1 = 0 , m 2 = 1 , . . . , m p = p 1.Let be a permutation of the set {1, 2, . . . , p}. If m (1) = 0 then there existk {2, 3, . . . , p} such that m (k ) = 0, thus m1 m (1) mk m (k ) 0(mod p)and the problem is solved in this case.It follows that m 1 = m (1) = 0, then using Wilson Theorem we get

m2 m3 m p = ( p1)! 1(mod p)m (2) m (3) m ( p) = ( p 1)! 1(mod p).

Multiplying them

m2m (2) m3m (3) m pm ( p) 1(mod p)Suppose that numbers m2m (2) , m 3m (3) , . . . , m pm ( p) are different modulo pand different from 0, then

m2m (2) m3m (3) m pm ( p) 1(mod p),a contradiction. Thus there exist k, l {2, 3, . . . , p}, k = l such thatmk m (k ) m l m ( l) (mod p),

and we are done.

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Chapter 3. Sets and useful tricks

Convex sets

Mathematical competitions check your knowledge bringing problems fromvarious amount of topics. One of these topics, which is rare, but very useful andrelated with sets, is convex sets. So let us introduce the concept of a convexset which plays an important role in connecting geometry with combinatorics,number theory and other areas of math.

Denition. In Euclidean space, an object is convex if for every pair of pointswithin the object, every point on the straight line segment that joins them is

also within the object. For example, a solid cube is convex, but anything thatis hollow or has a dent in it, is not convex.

Theorem 3.1. If F is a family of more than n bounded closed convex setsin Euclidean n-space Rn , and if every n + 1 members of F have at least onepoint in common, then all the members of F have at least one point in common.

(Hellys Theorem)

We will prove a plane version of the Hellys Theorem. Assume that we haveA1 , A2 , . . . , A n , n 3, convex sets in the plane such that A i Aj Ak = forall i, j,k {1, 2, . . . , n }. Then A1 A2 . . . An = .

Proof. We use mathematical induction. For the base case n = 4 we denoteby B1 A2 A3 A4 , B 2 A1 A3 A4 , B 3 A1 A2 A4 , B 4 A1 A2 A3 ,where B1 , B 2 , B 3 , B 4 are the points at the intersection of the corresponding sets.Consider four possible cases.

1st case. All four points lie on a line. Then let P [B1B2], where [XY ] isa segment with endpoints in X and Y . Clearly, P A1 A2 A3 A4 .2nd case. Without loss of generality B1 , B 2 , B 3 form a triangle and B4 [B2B3]. Then B4 A1 A2 A3 A4 .3rd case. B1 , B 2 , B 3 form a triangle and B4 lies inside the triangle B1B2B3 .

Because B1 , B 2 , B 3 A4 , it follows that B1B2B3 A4 . Thus B4 B1B2B3and therefore B4 A1 A2 A3 A4 .4th case. B1 , B 2 , B 3 , B 4 form a quadrilateral. Let P be the intersection of the

diagonals, without loss of generality P = B1B3 B2B4 . Because P B1B3 andP B2B4 , we have P A2 A4 and P A1 A3 . Thus P A1 A2 A3 A4 .The base case is proved. Assume that the statment is true for all k n. Toprove it for n + 1 denote by B i = Ai An +1 for 1 i n. Using the base caseof the induction we have Ai Aj Ak An +1 = . Thus Bi B j Bk = .Using the induction hypothesis we conclude that A1 A2 . . . An +1 = , andwe are done.

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Example 3.1. Any two of nite number of polygons have a common point.

Prove that there is a line which has a common point with all these polygons.Solution. Take any line g in a plane, and project all polygons onto g. Weget several segments any two of which have a common point. Using HellysTheorem we get that there is a point P belonging to all segments. It is clearthat the perpendicular to g through P intersects all polygons.

We continue to discuss convex sets that exist in R2 . Recall that for twovectors v, w R2 the two-dimensional lattice generated by v, w is the set of points {nv + mw | m, n Z}. For example we can have the unit square latticegenerated by v = (1 , 0) and w = (0 , 1) or the unit triangular lattice generatedby v = (1 , 0) and w = ( 12 ,

32 ). The determinant of the two-dimensional lattice

generated by v, w is |v w|, or the area of the parallelogram dened by vectorsv, w.The next notion that can be intuitively easily understood is a notion of centro-symmetric convex set.

Denition. A convex set K is centro-symmetric, also called centrally sym-metric, if it has a center O that bisects every chord of K through O (i.e. it issymmetric with respect to O).

Theorem 3.2. Given a two-dimensional lattice with determinant , everyconvex centrally-symmetric gure with respect to the origin that does notcontain a lattice point other than the origin. Then has the area at most 4.

(Minkowskis Theorem)

Proof. Let the area of be A. For every vector u from the lattice considerthe translation u of by vector 2u. If for u1 = u2 gures u 1 and u 2intersect, then both of them contain point u1 + u2 which also belongs to thelattice. Indeed, 2 u1 u 1 , 2u2 u 2 , and because they intersect it followsthat u1 + u2 , the midpoint of the segment which connects translated centersalso belongs to the intersection, as is centrally-symmetric. For a positiveinteger n consider kv + lw with 0 k, l < n . From the previous observationwe get these gures are pairwise disjoint. The total area of constructed guresis n2A. On the other hand, they lie within a parallelogram of area at most4n2 + cn, where c is a constant depending on (draw a picture to see this).Taking n , we obtain A 4 .

Remark. The theorem is true for any dimension n, replacing the constant4 by 2n , being the n-dimensional volume of the basic n-dimensional paral-lelepiped of the lattice.

Here is a very nice application of this theorem:

Example 3.2. Let n > 1 be a positive integer. Consider a closed disk Dwith center in the origin and radius n. One draws a disk of radius r = 1n 1

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around every lattice point (i.e. point with integer coordinates) in the disk.

Prove that every line through the origin intersects the disk.Solution. Assume, for the sake of contradiction, that there is a line dthrough O that does not intersect any of the disks. Consider the strip S formedby the two lines d1 , d2 parallel to d, at distance r form d. Then the strip S mustnot contain any of the lattice points of D. If d1 , d2 intersect the circumferenceof D in points A, B,C,D then the rectangle ABCD must not contain any of the lattice points, hence it must have area at most 4 by Minkowski Theorem,because ABCD is clearly convex and centro-symmetric, while the determinantof the lattice of points with integer coordinates if 1. However, as AC = BD =2n,BC = AD = 2r , we conclude that AB = BC = 4n2 4r 2 . Thus the areaof ABCD is

4

n2 r 2r > 4(n 1)

1n

1

= 4 ,

a contradiction.

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Counting in two ways

Many of the results in combinatorics are obtained by counting the numberof elements of a set in two different ways, and comparing the results of thesetwo calculations. Here is a very simple example:

Example 3.3. There are n kids who send each other gifts. It is known thatn 1 of the kids have received as many gifts as they have sent. Prove that then-th kid has has received as many gifts as has sent, too.

Solution. Let us number the kids 1 , 2, . . . , n and let a i , bi be the number of gifts the kid i has sent, respectively received. Let us count the total number of sent gifts. From one side, it is a1 + a2 + . . . + an , because each gift has been sentexactly once, and from the other side it is b1 + b2 + . . . + bn , because each gift hasbeen received exactly once. Therefore a1 + a2 + . . . + an = b1 + b2 + . . . + bn (thisis the conclusion obtained from counting the number of gifts in two differentways!). As we know a1 = b1 , a 2 = b2 , . . . , a n 1 = bn 1 , we conclude thatan = bn , and this means that the n-th kid has received as many gifts as hassent.

Counting in two ways is best illustrated with the help of a matrix: We cansum the entries of a matrix by considering the row sums and summing thesealtogether, or by summing the column sums. This should give the same result.Often for better understanding it is good to use this method, placing some valuesin a table and summing over rows and columns to deduce a useful relation, likein the example below.

Example 3.4. Let A1 , . . . , A k be subsets of {1, 2, . . . , n }, each of them withat least n2 elements, and such that for every i = j we have |Ai Aj | n4 . Provethat k

i=1

Ai kk + 1

n.

Solution. Let B = ki =1 Ai and m = |B |. Denote B = {b1 , . . . , bm } andlet xi be the number of subsets Aj in which bi appears. Clearly we have that

x i 1, because B is the union of all those subsets. Consider an m k table,in which the intersection of row i with column j contains 1 if bi A and 0otherwise. The sum of all the numbers in our table is

m

1x i (summing by rows)

and it equals the sum of |Aj | (summing by columns.) We thus have the equality

S =m

i

x i =k

1|Aj |

nk2

. (1)

In our table, let us calculate the number of pairs of 1 s situated on the samerow. Summing by columns, this is i

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m

1

x i

2=

x2i x i2

. Therefore,

k(k 1)4

n2 i m and we can bound S from below withthe value obtained from (1). Therefore we get the desired

nk (k 1)8

nk4m

nk2 m m(k 1) nk 2m

m(k + 1) nk m kk + 1

n.

Finally, we present a very hard problem from the Chinese Olympiad:

Example 3.5. Let A be a set with |A| = n, and let A1 , A2 , . . . , A n besubsets of A with |Ai | 2, 1 i n. Suppose that for each 2 element subsetof A of A, there is a unique i such that A Ai . Prove that A i Aj = for all1 i < j n.Solution. By the given conditions, we have

n

i=1

|Ai |2

=n2

. (1)

Let A = {x1 , x2 , . . . , x n }, and let di denote the number of subsets Aj , 1 j n,such that x i Aj . Hencen

i =1di =

n

i=1|Ai |. (2)

(double-count the pairs ( x i , Aj ) such that xi Aj over xi and over Aj to getthis identity) On the other hand,n

i=1

|di |2

=1i

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1i

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Look for the matrix!

The next trick that helps to solve some problems with sets is connected withlinear algebra. The main idea of it is to consider an incident matrix A for thegiven subsets, usually of the set {1, 2, . . . , n }. The next step is to explore A At .As you will see from linear algebra, from properties of ranks, determinants andlinear independency the desired conclusion might follow. This is more advancedthan the other parts of the presentation: the reader should be very familiar withlinear algebra.

First of all let us prove a technical result as a lemma that is pretty helpfulto solve these problems.

Lemma. Let a1 , a 2 , . . . , a n , x be real numbers. Prove that

a1 x x xx a2 x xx x a 3 x x x x an

= x(a1x) (an x)1x

+ 1

a1 x + +

1an x

.

In case of division by zero we will consider the following sum as the answer:

(a1 x) (an x) + x(a2 x) (an x) + + x(a1 x) (an 1 x).Solution.

a1 x x xx a2 x xx x a 3 x x x x an

=

a1 x a1 x a1 x a1x a2 x 0 0x 0 a3 x 0 x 0 0 an x

=

= x(a1 x)

a 1x (a 1 x ) 1 1 11 a2 x 0 01 0 a3 x 0 1 0 0 an x

= x(a1 x)(a2 x) (an x)a 1x (a 1 x ) 1 1 11

a 2 x 1 0 01a 3 x 0 1 0 1

a n x 0 0 1

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= x(a1x)(a2x) (an x)a 1x (a 1 x ) +

1a 2 x + + 1a n x 0 0 01a 2 x 1 0 01a 3 x 0 1 0 1

a n x 0 0 1= x(a1 x)(a2 x) (an x) a 1x (a 1 x ) +

1a 2 x + +

1a n x

= x(a1 x)(a2 x) (an x) 1x + 1a 1 x + 1a 2 x + +

1a n x .

We present you a few examples to give you idea how the stated method can

be used.Example 3.6. Let X be a set of n 3 elements, and let A1 , . . . , A m ,be proper subsets of X such that every pair of elements of X is contained in

precisely one set A i . Then m n holds.Solution. For x X let rx , be the number of sets Ai containing x. Notethat 2 r < m by the assumptions. Now if x A i , then r x A i , because theAi sets containing x and an element of Ai , must be distinct. Suppose m < n ,

then m|Ai | < nr x , and thus m(n |Ai |) > n (m r x ) for x Ai , and we nd1 =

x X

1n

=x X A i :x A i

1n(m r x )

>A i x :x A i

1m(n |Ai |)

=A i

1m

= 1 ,

which is absurd.Alternative Solution. Let B be the incidence matrix of ( X ; A1 , . . . , A m , ),

that is, the rows in B are indexed by the elements of X , the columns byA1 , . . . , A m , where

BxA =1 if x A0 if x A

Consider the product BB T . For x = x we have (BBT )xx = 1, since x and xare contained in precisely one set A i , hence

det(BB T ) =

r x 1 1 1 11 rx 2 1

11 1 rx 3 1 1 1 1 rx n

> 0.

It follows that BB T is invertible having rank (BB T ) = n. Thus the rank of then m matrix B is at least n, and we conclude that indeed n < m , since therank cannot exceed the number of columns.

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Example 3.7. Let A1 , A2 , . . . , A n and B1 , B 2 ,...,B n be the subsets of the

set {1, 2, . . . , n }, such that for all i and j , A i and B j have exactly one commonelement and for all nonempty subsets T of {1, 2,...,n }, there exists i such thatthe intersection of A i and T has an odd number of elements. Prove or disprovethat B1 = B2 = . . . = Bn .

Solution. Consider the matrices aij = 1 j A i and bij = 1 i B j and observethat the hypothesis says precisely that AB = R, where R is the matrix hav-ing everywhere 1. On the other hand, the hypothesis says that the system

j aij x j = 0 has no nontrivial solution in the eld Z2 , thus A is invertible overZ2 . But then B = A1R and so all elements in any line of B are equal. Butthis means precisely that B1 = B2 = . . . = Bn .

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Easy Problems

E1. Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram(rst, middle, and last initials) will be in alphabetical order with no lettersrepeated. How many such monograms are possible?

E2. Find the number of all nonempty subsets of the set S n = {1, 2, . . . , n }whichdo not contain two consecutive integers. [Hint: Look for an Italian name!]E3. How many triples of nonnegative integers ( a,b,c) are there such that

0 a b c n?E4. Let S be a set of 7 points such that, in any 3-subset of S , there are at least

two points with distance less than 1. Prove that there exists a 4-subset of S which can be covered by a disk of radius 1.

E5. Seven kids write Christmas postcards to each other. Is it possible thatevery kid sends one postcard and receives two, or sends two postcards andreceives one?

E6. For how many pairs of consecutive integers in the set

{1000, 1001, . . . , 2000}is no carrying required when the two integers are added ?

E7. Let S be a set with six elements. In how many different ways can oneselect two not necessarily distinct subsets of S so that the union the twosubsets is S ? The order of selection does not matter; for example thepair of subsets {a, c}, {b,c,d,e,f } represents the same selection as the pair{b,c,d,e,f }, {a, c}.

E8. How many kelement subset of [ n] contain no two consecutive integers?E9. Let S be a set of n integers. Prove that S contains a subset such that sum

of its elements is divisible by n .

E10. In how many ways can {1, 2, 3, . . . , n } be written as A B C, where A B = B C = A C = ?E11. Consider S = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Prove that for any subset, A, of S ,at least one of the sets A and S \A contains three elements in arithmeticprogression.E12. Fifty segments are given on a line. Prove that some eight of the segments

have a common point, or eight of the segments are pairwise disjoint.

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E13. Let F be a set of subsets of the set {1, 2, . . . , n } such thata) if A is an element of F , then A contains exactly three elements;b) if A and B are two distinct elements in F , A and B share at most onecommon element.Let f (n) denote the maximum number of elements in F . Prove that

f (n) n(n 1)

6 .

E14. Prove that from ten distinct two-digit numbers, one can always choose twodisjoint nonempty subsets, so that their elements have the same sum.

E15. Eleven teachers run a conference. Every hour, one or more teachers give aone-hour presentations, while all of the other teachers observe the presen-tations. (If one chooses to observe a presentation, then he has to observeit for the whole period.) Find the least amount of time during which it ispossible for each teacher to observe all other presentations at least once.

E16. The set {1, 2, . . . , 16} is partitioned into three sets. Prove that it is possibleto nd numbers x, y,z (not necessarily distinct) in one of those sets suchthat x + y = z.

E17. Let A be a set with n elements. Prove that

A i ,A j A|Ai Aj | = n 4n 1 .

E18. What is the greatest number of elements can have a subset A of a set

{1, 2, . . . , 2n} with the property for every two different elements x, y Awe have n x + y.E19. Prove that the number of binary sequences of length n that contain exactly

m pairs of consecutive equal numbers is equal to the number of binarysequences of length n that contain exactly m pairs of consecutive differentnumbers.

E20. Let n be an even integer. Let a1 , a 2 , . . . , a n and b1 , b2 , . . . , bn be permuta-tions of the set {1, 2, . . . , n }. Prove that a1 + b1 , a 2 + b2 , . . . , a n + bn takenmodulo n cannot be a permutation of

{1, 2, . . . , n

}.

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Medium Problems

M1. Let X be a nonempty set having n elements. Find the greatest numberof colors such that no matter we color each subset of X in one color therewill exist two distinct subsets A, B of X such that the sets A, B, A B ,A B have the same color.

M2. For {1, 2, . . . , n } and each of its nonempty subsets a unique alternating sum is dened as follows: Arrange the numbers in the subset in decreasingorder and then, beginning with the largest, alternately add and subtractsuccessive numbers. (For example, the alternating sum for {1, 2, 4, 6, 9}) is9 6 + 4 2 + 1 = 6 and for {5} is is simply 5.) Find the sum of all suchalternating sums for n = 7.

M3. Let d(n) denote the number of divisors of n . Prove that2008

i=1

d(i) =2008

i=1

2008i

.

M4. There are ve regions in the plane, A1 , A2 , A3 , A4 , A5 such that Ai hasarea at most 1, while each of the Ai has area at least 12 . Prove that thereare two distinct indices i, j such that A i Aj has area at least 15 .

M5. Consider the set M = {1, 2, 3, . . . , 2008}. Prove that in any way we choosethe subset X with 15 elements of M there exist two disjoint subsets A andB in X such that the sum of the members of A is equal to the sum of themembers of B .

M6. Let A and B be disjoint sets whose union is the set of positive integers.Prove that for every positive integer n there exist distinct a,b > n suchthat

{a,b,a + b} A or {a,b,a + b} B.M7. There are n points in the plane such that any three of them can be covered

by a disk of radius 1. Prove that all of them can be covered by a disk of radius 1.

M8. Given a set A with n2 elements, n 2, and F a family of subsets of A eachof which has n elements, suppose that any two sets of F have at most oneelement in common. Prove that there are at most n2 + n sets in F .M9. Find the sum

A,B,C |A B C

|, where the sum is taken over all subsets

A,B,C of the set {1, 2, . . . , n }M10. Prove that for every set X = {x1 , x2 , . . . , x n } of real numbers, there exista non-empty subset S of X and an integer m such that

m +s S

s 1n + 1

.

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M11. Find the number of nondecreasing functions f : [n] [n] that satisfyf (k) k for k = 1 , 2, . . . , n .M12. Find all positive integers n such that the set

A = {1, 3, 5, . . . , 2n 1}can be partitioned into 12 subsets, the sum of elements in each subset beingthe same.

M13. We have n sets of even cardinality, such that the intersection of at leasttwo of these sets always has odd cardinality. Prove that their union has atleast n + 1 elements, and show an example that achieves the bound n + 1.

M14. The set {1, 2, 3, . . . , 2n} is divided into two disjoint subsets A and B suchthat |A| = |B | = n. Let A = {a1 , a 2 , . . . , a n } and B = {b1 , b2 , . . . , bn }such that a1 < a 2 < . . . < a n and b1 > b2 > . . . > b n . Prove that|a1 b1|+ + |an bn | = n2 .

M15. A triple of different subsets S i , S j , S k of a set with n elements is called atriangle. Dene its perimeter by

|(S i S j ) (S j S k ) (S k S i )| .Prove that the number of triangles with perimeter n is

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(2n 1 1)(2 n 1).M16. Consider a set of n real numbers. One forms sums of three distinctnumbers from this set. Prove that one can obtain at least 3 n

8 different

sums, and characterize all sets for which exactly 3 n 8 distinct sums canbe obtained.M17. Let n be an integer greater than 1 and let X be a set with n + 1 elements.

Let A1 , A2 , . . . , A 2n +1 be subsets of X such that the union of any n has atleast n elements. Prove that among these 2 n + 1 subsets there exist threesuch that any two of them have a common element.

M18. Let A, B be nite sets of at integers, each containing at least two elements,such that the set A + B has exactly |A|+ |B | 1 elements. Prove that Aand B are arithmetic progressions with the same difference.

M19. Let n be a positive integer. Prove that the set {2, 3, . . . , 22n

1} can bepartitioned into n subsets with the property that none of them containstwo elements such that one is an integer power of the other, but the set

{2, 3, . . . , 22n

} cannot.M20. Let X be a set with n elements and let A1 , A2 , . . . , A m be three element

subsets of X such that |Ai Aj | 1 for i = j . Prove that there exist asubset A of X with at least 2n elements containing none of the A i .

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Hard Problems

H1. Let A be a nonempty set and let f : P (A) P (A) be an increasing functionon the set of subsets of A, meaning thatf (X ) f (Y ) if X Y.

Prove that there exist T , a subset of A, such that f (T ) = T .

H2. Let X = {x1 , x2 , . . . , x n }, n 2 be a set of positive integers. Prove thatthere exist no more than 2n n subsets of X , whose sum of elements is equalto 1.

H3. Let T be the set of all positive integer divisors of 2004 100 . What is thelargest possible number of elements that a subset S of T can have, if noelement of S is an integer multiple of any other element of S ?

H4. The plane is covered with k halfplanes. Prove that we can choose threehalfplanes from those k such that they still will cover the entire space.[Hint: Halfplanes are convex sets!]

H5. Let n be an odd integer greater than 1 and let c1 , c2 , . . . , c n be integers.For each permutation a = ( a1 , a 2 , . . . , a n ) of {1, 2, . . . , n } dene S (a) =n

i=1 ci a i . Prove that there exist permutations a = b of {1, 2, . . . , n } suchthatn! | S (a) S (b).

H6. Suppose we have some intervals on a real line, such that for any k of themthere are two whose intersection is non empty. Prove that there exist k 1points in R such that any of these intervals contains one of these points.[Hint: use Dilworth Theorem]

H7. Let S a be the set of numbers of the form na for some positive integern. Prove that if a,b,c are positive real numbers, then the sets S a , S b, S ccannot be disjoint.

H8. The set {1, 2, . . . , 3n} is partitioned into three sets A,B, and C with eachset containing n numbers. Prove that it is always possible to choose onenumber out of each set so that one of those numbers is the sum of the othertwo.

H9. Let n 2 be a positive integer and let S be a set of 2n+1 elements. Let f bea function from the set of two-element subsets of S to {0, 1, . . . , n 1}. As-sume that for any elements x,y, z of S , one of f ({x, y}), f ({y, z}), f ({z, x})is equal to the sum of the other two. Prove that there exist a, b, c in S suchthat f ({a, b}), f ({b, c}), f ({c, a}) are all equal to 0.

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H10. Let n 3 be an integer and X {1, 2, . . . , n 3}a set of 3n2 elements. Provethat one can nd nine distinct numbers a1 , a 2 , . . . , a 9 in X such that thesystem

a1x + a2y + a3z = 0

a4x + a5y + a6z = 0

a7x + a8y + a9z = 0

has a solution ( x0 , y0 , z0) in nonzero integers.

H11. Let n = 4 be a positive integer. Consider a set S {1, 2, . . . , n } with|S | > n2 . Prove that there exist x,y, z S with x + y = 3z.

H12. Let P 1 , . . . , P n be distinct 2-element subsets of {1, 2, . . . , n } such that if P i P j = 0; then {i, j } = P k for some k. Prove that every one of the ai sappears in exactly 2 of the P j s.H13. Find the number of even permutations of the set {1, 2, . . . , n } that do nothave xed points.H14. Let S be a set with 2008 elements, and let N be an integer with 0 N 22008 . Prove that it is possible to color every subset of S either black or

white so that the following conditions hold:

(a) the union of any two white subsets is white;

(b) the union of any two black subsets is black;

(c) there are exactly N white subsets.

H15. Find the number of subsets of

{1, 2, . . . , 2000

}, the sum of whose elements

is divisible by 5.H16. Let n be an even integer and let A1 , A2 , . . . A n be the subset with even

number of elements from the set {1, 2, . . . , n }. Prove that there exist i and j such that |Ai Aj | is an even number.H17. Let m and n be positive integers and S be a subset with (2 m 1)n + 1elements of the set {1, 2, 3, . . . , 2m n}. Prove that S contains m + 1 distinctnumbers a0 , a 1 , . . . , a m such that ak1 |ak for all k = 1, 2, . . . , m .H18. Prove that the set of numbers {1, 2, 3, . . . , 2008} can be colored with twocolors such that any of its arithmetic progressions with 18 terms contains

both colors.

H19. Let A1 , A2 , . . . , A n +1 be distinct subsets of the set {1, 2, . . . , n }, each of which having exactly three elements. Prove that there are two distinctsubsets among them that have exactly one point in common.

H20. Let p > 2 be a prime number and A = {1, 2, . . . , 2 p}. Find the number of subsets of A, each having p elements and the sum of the elements divisibleby p.

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