LESSON 2: THE INCLUSION-EXCLUSION PRINCIPLE

LESSON 2: THE INCLUSION-EXCLUSION PRINCIPLE The inclusion-exclusion principle (also known as the sieve principle) is an extended version of the rule of the sum. It states that, for two (finite) sets, A and B, the number of elements in the union of the two sets is the sum of the elements in each set, respectively, minus the number of elements that are in both: A B = A + B A B. Similarly, for three sets A, B and C, we have: A B C = A + B + C A B A C B C + A B C. For the general case of the principle, let A 1, A 2,..., A n be finite sets. Then n A i = i=1 n A i i=1 1 i<j n A i A j + 1 i<j<k n + ( 1) n 1 A 1 A n. A i A j A k + The name comes from the idea that the principle is based on over-generous inclusion, followed by compensating exclusion. Examples: 1. How many bit strings of length 8 either start with 1 or end with the two bits 00? Solution: Let A be the set of the strings starting with the digit 1, and B the set of the strings ending with 00. Then A B = A + B A B = 2 7 +2 6 2 5 = 160. 2. How many positive integers not exceeding 2001 are multiple of 3 or 4 but not 5? (AMC 12, 2001) (A) 768 (B) 801 (C) 934 (D) 1067 (E) 1167 Solution: Let A be the set of the multiples of 3 not exceeding 2001, B the set of the multiples of 4 not exceeding 2001, and C the set of the numbers not divisible by 5 and not exceeding 2001. Then the question is to find (A B) C = (A C) (B C) = A C + B C A B C. We also have that if X Y, then Y \ X = Y X. This follows from the rule of the sum applied to the disjoints sets X, Y \ X. Indeed, Y = X (Y \ X), therefore Y = X + Y \ X 1

and the desired formula follows. It is easy to see that A = {3, 6,..., 2001} has A = 667, B = {4, 8,..., 2000} has B = 500, and C = {1, 2,..., 2001} \ {5, 10,..., 2000} has C = 2001 400 = 1601. A C will consist of the multiples of 3 that are not multiples of 5, that is of the multiples of 3 that are not multiples of 15. We will count the multiples of 15 not exceeding 2001 and will subtract their number from A. The multiples of 15 not exceeding 2001 are 15, 30,..., 1995 (i.e. 1995 : 15 = 133 numbers), therefore A C = 667 133 = 534. Similarly, to compute B C, we need to subtract from B the number of the numbers that are multiples of 20 and do not exceed 2001. These numbers are 20, 40,..., 2000 (100 numbers), therefore B C = B 100 = 400. To compute A B C we need to subtract from A B = 166, the number of the numbers not exceeding 2001 that are multiples of 60. There are 33 such numbers, so A B C = 166 33 = 133. Then (A B) C = 534 + 400 133 = 801. 3. Each of the 2001 students at a high school studies either Spanish or French, and some study both. The number of those studying Spanish is between 80 percent and 85 percent of the school population, and the number of those studying French is between 30 percent and 40 percent. Let m be the smallest number of students that could study both languages, and let M be the largest number of students that could study both languages. Find M m. (AIME 2001) Solution: Let S and F be the set of the students of the school that studying Spanish and French, respectively. Then S F = 2001. Moreover, 80 100 S 2001 85 100 and 30 100 F 2001 40. From here we find that S {1601, 1602,..., 1700} and 100 F {601, 602,..., 800}. Then S F = S + F S F is maximum when S and F are both maximum, therefore M = 1700+800 2001 = 499. Similarly, S F is minimum when S and F are minimum. In this case m = 1601+601 2001 = 201. Then M m = 499 201 = 298. (AIME 2001). 4. In the small hamlet of Abaze, two base systems are in common use. Also, everyone speaks the truth. One resident said: 26 people use my base, base 10, and only 22 people speak base 14. Another said, Of the 25 residents, 13 are bilingual and 1 can not use either base. How many residents are there? (Use base 10, please!) [AHSME Dropped problem] (A) 15 (B) 25 (C) 27 (D) 35 (E) 36 Solution: Denote by a the index of the base of the first resident. Then let us 2

translate into base 10 what the first resident said: 2a + 6 people use my base, base a, and only 2a + 2 people speak base a + 4. (Note that the fact that he said he was using base 10 is irrelevant: in each base, the expression of the index of the base is 10 ( one-zero ).) Now, the first resident said, in his base, that there were 26 people speaking his base. The second resident said there were 25 residents all together in Abaze, therefore it is clear that the second resident uses another base than the first resident. We know that the other base is a + 4. Let us translate into base 10 what the second resident said: Of the 2(a + 4) + 5 residents, (a + 4) + 3 are bilingual and 1 can t use any base. We will use inclusionexclusion. Denote by A and B the set of the residents of Abaze using base a and base a + 4, respectively. Then the residents that do use one base or the other has A B = 2(a + 4) + 5 1 = 2a + 12 elements (remember, one resident does not use any base). A = 2a + 6, B = 2a + 2, A B = (a + 4) + 3 = a + 7. Writing inclusion-exclusion: A B = A + B A B, we obtain the following equation: 2a+12 = (2a+6)+(2a+2) (a+7). We obtain a = 11, so there are 2(a+4)+5 = 35 residents in this hamlet. 5. Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is m, where m and n are n relatively prime positive integers. Find m + n. Solution: Since each of the 9 unit squares of the 3-by-3 square can take one of two colors, in total there are 2 9 different ways of coloring the 3-by-3 square. Let us count now the colorings that form a 2-by-2 red square. This red square can be formed in the upper left, upper right, bottom left or bottom right corner of the 3-by-3 square. Let A, B, C, D be, respectively, the sets of the four types of colorings listed above that produce a 2-by-2 red square. Since the other 5 squares of the 3-by-3 square can have any color, we have A = B = C = D = 2 5. These sets are however not disjoint, so we must apply inclusion-exclusion: A B C D = A + B + C + D A B A C A D B C B D C D + A B C + A B D + A C D + B C D A B C D. As A B represents the set of the colorings having the first two rows red (and the 3 squares of the lowest row free to take any of the two colors), we have A B = 2 3. Similarly, A C = 2 3, B D = 2 3 and C D = 2 3. We have a different situation for A D and B C: if two opposing 2-by-2 squares are red, this leaves uncertain only the colors of the two remaining 1-by-1 corners, therefore A D = B C = 2 2. The intersections of 3 of the sets leaves undecided only the color of one corner square, therefore each of these intersections has 2 elements. Finally, the intersection of all the four sets contains only the square completely colored in red, that is, one element. We obtain: 3

A B C D = 4 2 5 (4 2 3 + 2 2 2 ) + 4 2 1 = 95. So from the 512 possible colorings, 512 95 = 417 do not contain any 2-by-2 red square, so the probability of not having a 2-by-2 red square is 417 512. As 417 and 512 are relatively prime, we find m = 417, n = 512, and m + n = 929. 6. Let A 1, A 2, A 3,..., A 12 be the vertices of a regular dodecagon. How many distinct squares in the plane of the dodecagon have at least two vertices in the set {A 1, A 2, A 3,..., A 12 }? Solution: Consider every possible pair of vertices of the dodecagon. The choice of such pairs can be made in 66 ways. Indeed, choose the first point (12 ways), then choose the second point (11 ways). But any unordered pair {A, B} of vertices has been chosen twice (first A, then B, or first B, then A), so we must divide 12 11 by 2. Therefore we have 66 pairs of vertices. Two vertices, A, B of the dodecagon can be vertices of 3 possible squares: one in which A and B are diagonally opposite vertices of the square, and two ways in which AB is a side of the square, the square being constructed on one or the other side of the line AB. So it seems we have 66 3 = 198 squares that have at least two vertices in the set of the vertices of the dodecagon. But this overcounts, as it is possible for a square to contain more then two vertices of the dodecagon, that square being then counted several times, once for each pair of vertices of the dodecagon. Now, if 3 vertices, A, B, C, of the dodecagon are vertices of a square, then they are consecutive vertices of the square and therefore ABC is a rightangle, and AB = BC. This means that AC is a diameter in the circumcircle of the dodecagon, and B is the midpoint of one of the arcs AC. Then the forth vertex of the square is the midpoint of the other arc AC, hence all the 4 vertices of the square must be vertices of the dodecagon. Moreover, the 4 vertices can be: {A 1, A 4, A 7, A 10 }, or {A 2, A 5, A 8, A 11 }, or {A 3, A 6, A 9, A 12 }. Each of these squares was counted 6 times, once for each side and diagonal. In order to avoid overcounting, we need to count these 3 squares only once, so we must subtract from the total we have found (198), 5 times the number of the overcounted squares: we get 198 5 3 = 183 squares. Practice problems 1. Every student in the senior class is taking history or science and 85 of them are taking both. If there are 106 seniors taking history and 109 seniors taking science, how many students are in the senior class? Solution: Let A be the set of the students of the senior class that take history, and 4

B the set of those taking science. The number of the students in the senior class is A B = A + B A B = 106 + 109 85 = 130. 2. Find, with justification, the number of positive integers less than or equal to 1000 that are divisible by 3, 5, or 11. Solution: Let A, B, C be the set of the numbers less than or equal to 1000 that are divisible by 3, 5 and 11, respectively. Then the question translates to computing A B C. We shall apply inclusion-exclusion. The number of elements for each set is easy to compute: A = {3, 6,..., 999} has 333 elements, B = {5, 10,..., 1000} has 200 elements, and C = {11, 22,..., 990} has 90 elements, A B = {15, 30,...990} has 66 elements, A C = {33, 66,..., 990} has 30 elements, B C = {55, 110,..., 990} has 18 elements, and A B C = {165, 330, 495, 660, 825, 990} has 6 elements. Applying inclusion exclusion, A B C = 333+200+90 66 30 18+6 = 515. 3. At an international conference of 120 people, 75 speak English, 60 speak Spanish, and 45 speak French (and everyone present speaks at least one of the three languages). a) What is the maximum possible number of these people who can speak only one language? In this case, how many speak only English, how many speak only Spanish, and how many speak only French, and how many speak all three? b) What is the maximum number of people who speak only French? In this case, what can be said about the number of people who speak only Spanish and the number who speak only English? c) What is the maximum number of people who speak only English? In this case, what can be said about the number of people who speak only Spanish and the number of people who speak only French? Solution: a) In the Venn diagram above, A, B, C represent the numbers of people that speak only English, only Spanish, and only French, respectively, X is the number of people that speak only English and Spanish, Y is the number of people that speak only English and French, Z is the number of people speaking Spanish 5

and French, but not English. Finally, M is the number of people speaking all three languages. Then we have: A+B +C +X +Y +Z +M = 120, A+X +M +Y = 75, B + Z + M + X = 60, and C + Z + M + Y = 45. By adding these last three relations and denoting A + B + C = α, X + Y + Z = β, we get α + 2β + 3M = 180. As α + β + M = 120, we get β + 2M = 60, hence M 30. We prove that M = 30 is indeed the maximum value. If M = 30, then β = 0 and X = Y = Z = 0, then A = 45, B = 30, C = 15. This configuration satisfies all the requirements, hence M = 30 is the desired maximum. b) We want the maximum value of C. As C + Z + M + Y = 45, we get C 45. If C = 45, then Y = Z = M = 0, hence A = 75 X, B = 60 X. Plugging into A + B + C + X + Y + Z + M = 120 yields X = 60, then A = 15, B = 0. This configuration satisfies all the requirements, hence C = 45 is the desired maximum. c) We want the maximum value of A. As A + X + M + Y = 75, we get A 75. If A = 75, then X = Y = M = 0, hence B = 60 Z, C = 45 Z. Plugging into A + B + C + X + Y + Z + M = 120 yield Z = 60, then B = 0, but C = 15! So clearly the desired maximum is not 75. In order to find it, notice that by adding A+X +M +Y = 75, B +Z +M +X = 60, and subtracting A + B + C + X + Y + Z + M = 120 we get M + X C = 15, hence M + X 15. Then, from A + X + M + Y = 75 follows that A 60. If A = 60, we must have: M + X = 15, Y = 0, C = 0, M + Z = 45 = B + Z. Plugging X = 15 M, Z = 45 M, B = M into A + B + C + X + Y + Z + M = 120 yields 0 = 0. Therefore we can yake M = 0, getting X = 15, Z = 45, B = 0. This configuration satisfies all the requirements, hence A = 60 is the desired maximum. 4. In a student campus having 420 residents, everybody speaks at least one of the three languages - French, English and Spanish. The number of people who speak French is 200, the number of people who speak English is 300 and the number of people who speak Spanish is 400. a) What is the minimum possible number of students who speak all the three languages? b) If the number of students who speak all the three languages is 100, what is the number of students who speak exactly one language? c) What is the maximum possible number of students who speak all the three languages? In this case, how many speak only English, how many speak only Spanish, and how many speak only French? 6

Solution: a) In the Venn diagram above, A, B, C represent the numbers of students that only speak French, English and Spanish, respectively, X is the number of students that speak French and English but not Spanish, Y is the number of students that speak French and Spanish but not English, Z is the number of students speaking English and Spanish, but not French. Finally, M is the number of students speaking all three languages. Then we have: A + B + C + X + Y + Z + M = 420, A + X + M + Y = 200, B + Z + M + X = 300, and C + Z + M + Y = 400. By adding these last three relations and denoting A + B + C = α, X + Y + Z = β, we get α + 2β + 3M = 900. As α + β + M = 420, we get β + 2M = 480, hence M α = 60. As α 0, we get M 60. We prove that M = 60 is indeed the minimum value. If M = 60, then A = B = C = 0 and X + Y + Z = β = 360 hence Z = A + M + (X + Y + Z) 200 = 220. Similarly, we get Y = 120 and X = 20. This configuration satisfies all the requirements. b) If M = 100, then from M α = 60 we get α = A + B + C = 40 represents the number of people who speak exactly one language. c) From β + 2M = 480 we get that M 240. On the other hand, from A + X + M + Y = 200 we get M 200. In case M = 200 we obtain A = X = Y = 0, hence B + Z = 100, C + Z = 200. Substituting B = 100 Z, C = 200 Z into A + B + C + X + Y + Z + M = 420 yields Z = 80, hence B = 20, C = 120. 5. The Alpine Club consisting of n members organizes four mountain expeditions for its members. Let E1, E2, E3, E4 be the four teams participating in these expeditions. How many ways are there to form those teams, given that E1 E2, E2 E3, E3 E4? Austrian-Polish Mathematics Competition, 1995 Solution: By the principle of inclusion and exclusion, we get the desired total by counting the number of formations without any constraints, then for each constraint subtracting the number of formations that fail to meet that contraint, then adding for each pair of constraints the number of formations that fail to meet both constraints, then subtracting the number that fail all three constraints. 7

Since each person can choose whether or not to be on each team, there are 16 n formations in total. Suppose one constraint fails, say E 1 E 2 =. Then each person only has 12 choices, since four choices involve joining both E 1 and E 2. If E 1 E 2 and E 3 E 4 are both empty, each person has only 9 choices, while if either of the other pairs fail, each person has 10 choices. Finally, if all three constraints fail, each person has 8 choices. Hence the number of acceptable formations is 16 n 3 12 n + 9 n + 2 10 n 8 n. 8

EVALUATE, THEN GIVE AN EXAMPLE Another useful method of answering questions concerning a minimal or a maximal value of something is evaluate, then give an example. For a maximum, evaluate means show that the maximum can not exceed M. The second part of the proof requires to give an example where this value M is attained. The difficulty often consists in guessing the value of M. Quite often we find ourselves confronted with the following dilemma: the best M for which we have found an example is not the same as the best M for which we have an estimation: there is a gap between the two numbers. Should we try to improve our estimation, maybe using different arguments, or should we keep trying to find better examples? Unfortunately, there is no magical solution to this dilemma... Examples: 1. What is the maximum number of distinct positive integers whose sum is 2011? Solution: First we prove that this maximum can not exceed 62 ( evaluate ). Suppose it was possible to have more then 62 distinct positive integers adding up 63 64 to 2011. Then their sum would be at least 1 + 2 +... + 63 = = 2016 > 2011, 2 contradiction. Thus we have proven that there can be at most 62 numbers. The second part of the proof consists in providing an example in order to prove that there are 62 distinct positive integers whose sum is 2011. Consider the numbers 1, 2,..., 61 and 120. The sum of these 62 distinct positive 61 62 integers is + 120 = 2011. 2 (the example given above is by no means the only available one; any other such example is equally good) 2. What is the maximum number of rooks that can be positioned on a chessboard such that no two rooks attack each other? The same question for queens, bishops, kings and knights. Solution: (a) rooks We can not have two rooks on the same row, therefore we can have at most 8 rooks (the evaluation part). Placing 8 rooks on a diagonal of the square provides an example of how we can place 8 rooks on the board such that no rooks attack each other. We have proven that the desired maximum is 8. 9

that can be placed in a 7 7 grid (without overlapping) such that each figure covers exactly 4 10 (b) queens We can not have two queens on the same row, therefore we can have at most 8 queens (the evaluation part). Finding a layout with 8 queens such that no two of them attack each other is a little bit more difficult to construct. Denote the rows of the chessboard by 1,2,...,8 and the columns by letters from a to h. Then one possible example with 8 queens on a chessboard with no two queens attacking each other is to place them in: a2, b4, c6, d8, e3, f1, g7, and h5. Essentially there are 12 such examples. See http://en.wikipedia.org/wiki/8 queens (c) bishops evaluation: Consider the diagonal joining the bottom-left corner and the upperright corner of the square and all the shorter diagonal lines parallel to this one. There are 15 such diagonals, two of which passing only through one single unit square (the unit squares in the upper-left and bottom-right corners). These 15 lines cover all the unit squares, hence we can not place more than 15 bishops on the board. But the two diagonals that pass only through the corner squares are opposite corners of the board, so we can not place a bishop on both these squares. This means that we can not have more then 14 non-attacking bishops on the board. An example with 14 bishops: place a bishop on each square of the lowest row, then a bishop on each of the squares of the top row with the exception of the two corners. (d) kings Divide the board in 16 2 2 squares. Two kings can not be placed in the same 2 2 square, therefore we can have at most 16 kings. On the other hand, placing a king in the upper-left corner of each of the 16 2 2 squares constitutes an example of how one can place 16 non-attacking kings on a chessboard. In conclusion, the desired maximum is 16. (e) knights For the example part, note that each knight changes the color of its square when it moves. Placing 32 knight on the 32 white squares is an example of 32 non-attacking knights. In order to prove that one can not put more, notice that in any 2 4 rectangle you can place at most 4 non-attacking knights. (One can pair up the 8 fields of such a rectangle in pairs such that if one places knights in the two fields that form a pair, the knights will attack each other.) As one can cover the chessboard with 8 such rectangles, one can not place more than 32 non-attacking knights on the chessboard. 3. Determine the greatest number of figures congruent to

unit squares. Solution: Number the rows and the columns from 1 to 7, then color the squares situated on even rows and even columns. Nine squares will be colored. Each tile placed on the board, whatever its position, will cover exactly one colored square, therefore we can have at most 9 tiles on the board. On the other hand, it is possible to place 9 tiles on the board: you can place 3 on the first two rows, another 3 on the next two rows, and 3 more on the following two rows (the last row, and some isolated other squares remain uncovered). The desired maximum is 9. Remark: The more frequently used coloring schemes (chessboard, stripes) will fail to provide a justification of the fact that more than 9 is impossible. 4. A beetle sits on each square of a 9 9 board. At a signal each beetle crawls diagonally onto a neighboring square. Then it may happen that several beetles will sit on the same square and none on others. Find the minimal possible number of free squares. Solution: Color the columns alternately black and white ( stripes ). We get 45 black and 36 white squares. Every beetle changes the color of its square each time it crawls. Hence at least 9 black squares remain empty. An example with exactly 9 squares remaining unoccupied is easy to construct. (Hint: Make 2 36 of the beetles swap places according to the pattern shown in the figure after problem 8.) 11

grid without overlapping. that can be placed in a 6 6 Solution: The answer is 4. An obvious example: place two neighboring crosses so that they cover entirely the second row, then two other crosses that cover row number 5. Now, for the evaluation part, note that when placed on the grid, any cross occupies 3 squares of a certain row. This row can not be the first or the last row. Moreover, if we place two crosses on the same row, there is no place on the previous and on the next row to place a cross with 3 squares there. So we can place at most 2 crosses on lines 2+3, and at most 2 crosses on rows 4+5. Therefore the maximum number of crosses can not exceed 4. Alternatively, for the evaluation part, notice that the corners can not be covered, and, from the other 4 squares situated on the side, at most 2 can be covered. Therefore we have at least 12 uncovered squares, hence at most 24 covered squares. It follows that we can use at most 4 crosses. 6. Consider a row of 2n squares colored alternately black and white. A legal move consists of choosing a contiguous set of squares (one or more squares but they must be next to each other, no gaps are allowed) and inverting their colors. What is the minimum number of moves necessary to make the row entirely one color? Solution: We shall prove that the minimum number of moves is n. It is clear that in n moves one can make all the squares have a single color (just change the color of a single black square at a time). In order to prove it is not possible to obtain a single colored row in less than n moves, notice that if one changes, by a move, the color of the square no. k, then by the same move he also changes the color of square no. k 1, unless the selection of the squares to operate on stops right there, between squares k 1 and k. But, initially, squares k 1 and k have different colors, and need to have, at the end, the same color, therefore there must be a move that stops right between k 1 and k. (This move might operate on either square k 1, or square k, but not on both.) There are 2n squares, therefore 2n 1 separators between two neighboring squares. Each move goes from one separating place to another (ore uses one or both ends of the row), therefore each move uses at most 2 separating places. Therefore, we need at least n moves to use up all the 2n 1 separating places. 7. Consider an 8 8 board. What is the minimum number of a) 1 1 squares, 12 5. Determine the maximal number of figures

b) 1 2 rectangles, that need to be painted black such that any 2 2 square contains at least one painted 1 1 square? Solution: a) Obviously 16. Divide the board in 16 disjoint 2 2 squares. One needs to paint black at least one unit square from each of these 2 2 squares; on the other hand, painting black all the squares on even rows and even columns is a valid example of 16 black unit squares that intersect any 2 2 square. b) Consider the 3 non-touching 2 2 squares on the top 2 rows, 3 non-touching 2 2 squares on the 2 rows in the middle, and 3 non-touching 2 2 squares on the bottom 2 rows. Each of these 9 squares has to contain at least one black square. On the other hand, the 2 1 rectangles intersecting these 9 squares are distinct, hence we need to paint black at least 9 2 1 rectangles. An example with 9 such rectangles is given below. 8. Consider an 8 8 chessboard. In some of the unit squares a diagonal is drawn such that any two of these diagonals do not have any common points. What is the maximum number of diagonals that can be drawn? Solution: Consider the 9 9 grid made of the vertices of the unit squares. Color these vertices by stripes, alternately with red and black, starting with a red stripe. Any diagonal joins vertices of different color. As there are 36 black vertices and 45 red ones, there can be at most 36 diagonals. An example with 36 non-intersecting diagonals: number the columns from 1 trough 9, and rows by letters from A trough I. Join vertices A1, A2, A3, A4, A5, A6, A7, A8, B1, C1, D1, E1, F1, G1, H1 with the vertex situated diagonally to the right and upwards. This describes an L-shape of 15 diagonals. We continue with these L-shapes. The next one will consist of 11 diagonals, then follows an L-shape with 7 diagonals and finally one with 3 diagonals. In total, 36 diagonals. 13

9. Consider a 5 5 square divided into 25 unit squares. What is the maximum number of unit squares that can be painted black such that each 2 2 square contains at most two black unit squares? [Moscow Olympiad 1996-7] Solution: We have an easy example with 15 blacks: simply color the squares in rows no. 1, 3 and 5. We prove that if one colors more than 15 squares in black, then there will always be a 2 2 square containing more then two black unit squares. Suppose there are at least 16 black squares. Grouping the 16 squares on the first 4 rows and 4 columns, into four 2 2 squares, we have that at most 8 of these 16 squares are black (at most 2 in each of the four squares). Then at least 8 of the remaining 9 squares have to be black. Considering two cases, when the square on the fifth row and fifth column is black, and white respectively, we arrive quickly to a contradiction in both cases. 10. If we write the numbers 1, 2,..., n in some order, we will get an n-chain. For example, one possible 11-chain is 3764581121910. What is the smallest n with n > 1 such that there exists an n-chain that is a palindrome? (A number is a palindrome if it may be read the same way in either direction.) [Chech Olympiad, 2012] Solution: An example of a palindromic 19-chain is 9 18 7 16 5 14 3 12 1 10 11 2 13 4 15 6 17 8 19. For 1 < n. We will show that 19 is the smallest possible value of n. First note that only one digit can appear in a palindromic chain for an odd number of times (namely the middle one). Clearly, for 1 < n 9 this condition cannot be satisfied. Similarly, for 10 n < 19 both digits 0 and 9 appear exactly once, thus such an n-chain cannot be a palindrome. 14

Practice problems: 1. Determine the largest positive integer n for which there exist n positive integers greater than 2 such that the sum of any three of them is a prime number. Solution: Among the n numbers we can neither have three that give the same remainder when divided by 3, nor three that give three different remainders when divided by three, or else their sum would be a multiple of 3, larger than 3, hence not a prime. Therefore the maximum value of n is at most 4 (we can have at most two types of numbers and at most two of each type). An example for n = 4: take 3, 5, 11, 15. 2. Each of the unit squares of an n n square is colored with red, yellow or green. Find the smallest value of n such that, for every possible coloring, there exist a row and a column with at least three unit squares of the same color (the same color on both the row and on the column). Solution: The desired minimum is 7. For n = 7, as 49 = 3 16 + 1, we have at least 17 unit squares of the same color (pigeonhole principle for the 3 colors), say red. As 17 = 7 2 + 3, there exist a row and a column with at least 3 red unit squares (pigeonhole principle for the 7 rows and for the 7 columns). Hence, for n = 7, for any coloring, there exist a row and a column with at least three unit squares of the same color. For n = 6, it is possible, for some colorings, not to have both rows and columns with three squares of the same color on them. Simply color the square (i, j), i, j = 1, 6 with red if i + j 0 (mod 3), yellow if i + j 1 (mod 3), and green is i + j 2 (mod 3). On each row and column there are exactly two squares of the same color, hence n = 6 is not suitable. Is this solution complete? Isn t something missing there? This solution deliberately contains a typical mistake: In order to prove that the minimum suitable value is M it is not enough to prove that M is suitable and M 1 is not. It is possible that M 2 or some other smaller value is suitable again. To complete the proof of the problem above, notice that for n < 6 we can extract from the example given for n = 6 an n n sub-square which will obviously provide a convenient example for the fact that any n < 6 is not suitable. 15