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1 Pigeonhole Principle Pigeonhole Principle: When you organize n things into k categories, one of the categories has at least n/k things in it. Proof: If each category had fewer than n/k things in it then there would fewer than n things. Example: 100 things into 99 boxes. At least one box has 1 1 things in it, so 9 at least one box has or more things in it. Example: 100 things into 49 boxes. At least one box has more than 49 things in it, so at least one box has 3 or more things in it. More generally, when you have enough of something (building blocks or conditions), then you will find some interesting structures. Exercises from Applied Combinatorics, Appendix A n couples. Minimum size subset of these people to contain at least one couple? Maximum size not containing a couple? If it contains k men, then it must exclude k women, so it must have no more than n people in it. So n + 1 people include a couple. Or: n boxes, n + 1 people forces two people into a box.. 0 people at part must include with same number of friends. Reason: number of friends varies from 0 to such numbers. If all different, then someone has 0 friends and someone knows everyone impossible. So only 19 different numbers. of them the same. 3. n person tournament, nobody loses all matches, therefore two players have same number of wins. Reason: Each player has 1 to n 1 wins, and there are n players, so two have same number of wins French, 0 Spanish, 8 German, 15 Russian, 5 Italian books. Number of books chosen to guarantee 1 of same type: In the worst case you have 10 French, 11 Spanish, 8 German, 11 Russian, and 11 Italian books, for 51 books. If you choose 5 then the next book must make a dozen of one kind different pairs of people at a party of 0 implies that somebody has 4 or fewer acquaintances. In other words, 48 edges implies that some vertex degree in a graph of 0 vertices is 4. Reason: if each vertex degree is 5 or greater then the vertex degree some is 100, implying 50 edges. 6. n jokes, subsets of 3, at least 1 of them. Need ( n 3) 1. 6 jokes are minimal. 1

2 7. 7 distinct integers implies with sum or difference divisible by 10. Reason: Just consider the last digit of each number. 7 numbers from the set {0, 1,, 3, 4, 5, 6, 7, 8, 9}. If two the same then a difference is divisible by 10. Now suppose all distinct. Must show that two are complementary. Make categories {0}, {1, 9}, {, 8}, {3, 7}, {4, 6}, {5}. There must be two in a category somewhere since there are 7 numbers and 6 categories. 8. n + 1 numbers in {1,,..., n} must be consecutive. Reason: If not, if the first number is x and successive numbers increase by at least then the last number is x + n > n numbers about a circle, some triple sum is at least 17. Reason: the possible sums are all 6. If they all fall in [6, 16] there are 11 possibilities. Need to cut this down to at most 9. 6 must be formed from {1,, 3}. 7 must be formed from {1,, 4}. 8 must be formed from {1,, 5} or {1, 3, 4}. 9 must be formed from {1,, 6} or {1, 3, 5} or {, 3, 4}. 10 must be formed from {1,, 7} or {1, 3, 6} or {1, 4, 5} or {, 3, 5}. Suppose 6 is the least sum. If 7 also occurs then we must have 413 in the cycle. This rules out 8 or 9. If 8 also occurs then we must have 315 or 134 in the cycle. If 315 then no 7 and no 9. If 134 then no 7, but we can have 9 if we have 6134 in the cycle. The latter rules out 10. On the other hand, suppose 7 is the least sum. Need to rule out one more number. If 7 and 8 then we must have 415 or 143 in the cycle. If 9 also occurs then we must have 6143 in the cycle. This rules out A sequence of 1 non-negative integers adds up to 99. A sum of two consecutive numbers is 17. Reason: suppose every consecutive sum is 16. Then x 1 +x 16,..., x 11 +x 1 16, so the entire sum is 6 16 = 96. Contradiction. 11. Any subset of 8 distinct integers between 1 and 14 contains a pair such that the smaller divides the larger. Reason: We can assume the smallest number is not 1. If it is, throw out 4, 6, 8, 10, 1, 14, leaving only 7 numbers. If it is 3, throw out 6, 9, 1, leaving 8 numbers, among which are 4 and 8. If it is 4, throw out 8, 1, leaving 8 numbers, among which are 5 and 10. If it is 5, throw out 10, leaving 8 numbers, among which are 6 and 1. If it is 6, then 8 numbers are left, including 6 and Let n 3. In any set of n integers there is a pair whose difference is divisible by n 1. Reason: Place each number into congruence class mod n 1. There are n 1 of these, so two fall in same class.

3 13. Every n + 1 integers in [, n] contains a pair with no common divisor. Reason: there must be a pair of consecutive integers, otherwise the first is x and the last is x + n. Consecutive integers have no common divisor unless one of them is Out of 16 positive integers summing to 30 one can produce all the numbers in the range 1 through 9 by a subset sum. More generally, if k positive integers sum to n and k > n then all the numbers in the range [1, n] can be produced. This is best proved by induction on n. If n is then the sum is 1 + 1, which does produce all the numbers in range. Now consider n 3. At least one of the numbers is 1. If all the numbers are 1 then there is nothing to prove. Let x be the smallest number greater than 1. Removing it we have k 1 > n 1 n x, therefore by the induction hypothesis we can create the all the numbers in [1, n x]. Adding x, we can create all the numbers in [1 + x, n]. If n x 1 + x then we can create all numbers in [1, n]. If n x < 1 + x then we have n x x and x n. This can only be the case if the sum in question is 1 +, which generates [1, 3] = [1, n]. 15. More generally, given c computers and p printers, if there is a printer with degree a then there are c a computers vying for p 1 printers, so we need c a p 1 or a c p + 1. Let printer i be hooked up to computers i through i + c p. Then computer i is hooked up to printer i if and only if 0 i j c p. Given computers c 1 < c < < c p, we have the links (c 1, 1),..., (c p, p). Proof: we have i c i and c i + p i c, therefore 0 c i i c p. 16. Treat this as 10 computers and 6 printers. The 6 computers send their requests through the original connections to the 10 old printers, and these send 6 print requests to the 6 new printers. By problem 15, this can be done with 6( ) = 30 connections integers chosen from [1, 60] generate have two disjoint subset sums which are equal. Reason: There are 10 = 104 subset sums, all falling in the range [10, 600], so two of these sums are the same. Use the divided difference of the two sums. 18. Two disks with 10 0s and 10 1s can be aligned to match digits in at least 10 positions. Reason: Superimpose the disks arbitrarily and record the sequence of sums modulo. Now rotate the second disk one position clockwise and record the new sequence of sums modulo. Do this 0 times, 3

4 forming a 0 0 matrix of 0s and 1s. Each column has 10 0s and 10 1s, so there are 00 0s in the matrix. Let the number of 0s in row i be x i. Then x x 0 = 00, so at least one x i 10. This is the number of matches corresponding to the i th pairing of disks weeks, at least one hour a day, no more than 11 hours per week, implies a consecutive sum of 0 somewhere. Reason: Consider the partial sums s 1 through s 35. We want two of them to differ by 0. Consider the 1 partial sums s 15 through s 35. Then two must fall in the same congruence class modulo twenty, hence differ by either 0 or 40. If they differ by 0 then we are done. Suppose they differ by 40. Then 55 s b = 40 + s a 40 + s 15 55, which can only be the case if s a = 15 and a = 15. In other words, the string of numbers begins with 15 consecutive 1s. Let c 16 be least such that s s c 5. Then it can be no larger than 9 since no digit is larger than 5 by the 11 hours per week restriction. Combining s s c with the 0 s 16 s c 1s preceding it forms In an n-vertex graph G, if every vertex degree is at least (n 1)/ then the graph is connected. Proof: By induction on n. True if n = 1. Now consider n > 1. First consider n = k. Then every vertex degree is at least k. Remove one vertex and all its edges. In the remaining subgraph every vertex degree is at least k 1 ((n 1) 1)/, so the subgraph is connected, so the original graph is connected. Now consider n = k + 1. Every vertex degree is at least k. If the graph is not connected, let H be one of its connected components, and let H be the union of its other components. One of these subgraphs has k or fewer vertices and so must have an edge to one of the other subgraphs. Contradiction. Therefore the graph is connected. 1. Every sequence of n +1 distinct integers has a monotonic subsequence of length n + 1. Proof: by induction on n. When n = 1 the sequence has length and is monotonic. Now assume true for n. Consider a sequence of length (n+1) +1 = n +1+(n+1). Suppose there are no monotonic subsequences of length n +. Since there are monotonic subsequences of length n + 1, the initial terms of those that increase have to decrease, so there can be at most n + 1 of these. Remove these initial terms, leaving at least n n terms 4

5 of the sequence. There are still monotonic subsequences of length n + 1, and they all must decrease, so all their initial terms increase, so there can be at most n of these. Remove these initial terms, leaving at least n + 1 terms of the sequence. Now we arrive at a contradiction, because on the one hand we have removed all the initial terms of monotonic sequences of length n + 1, but on the other hand one must still exist by the induction hypothesis. So there must be a monotonic sequence of length n + in the original sequence. More generally, every sequence of rs + 1 distinct integers has an increasing subsequence of length r + 1 or a decreasing subsequence of length s + 1. Proof: by strong induction on rs. If rs = 1 then there are two integers, and either the increase or decrease. Now consider rs > 1. Suppose r > 1. We have rs + 1 = (R + 1)s + 1 = Rs s. Suppose there are no increasing subsequences of length r+1 and no decreasing subsequences of length s+1. There can be at most s increasing subsequences of length r because the initial terms decrease. Remove the initial terms, leaving at least Rs + 1 numbers. Now there are no increasing subsequences of length r = R+1, so by the induction hypothesis there must be a decreasing sequence of length s + 1 contradiction. Now consider r = 1. Then s > 1 and we can use a similar argument.. In every party with at least 6 people, either 3 people know each other or 3 people don t know each other. Reason: We can assume there are exactly 6 people at the party. Let vertex = person and edge = knowledge. Suppose that one cannot find 3 independent vertices in G. Whenever a vertex is not attached to others, those must have an edge between them. So no 3 independent implies that either there is a K 3 or there is a vertex of degree 3. So no 3 independent and no K 3 implies that there is a vertex of degree 3, so if there is a vertex of degree then either there are 3 independent or there are 3 connected. Now one of G and G has a vertex of degree, so one of these two has 3 independent or 3 connected, and if one does they both do. 5

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