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**The Pigeonhole Principle: Selected Exercises**

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**The Pigeon-Hole Principle**

Let k be a positive integer. If k + 1 or more objects are placed in k boxes, Then at least 1 box contains 2 or more objects. Copyright © Peter Cappello

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**Copyright © Peter Cappello 2011**

Exercise 10 Let ( xi, yi ), i = 1, 2, 3, 4, 5, be a set of 5 distinct points with integer coordinates in the xy plane. Show that the midpoint of the line joining at least 1 pair of these points has integer coordinates. (x1, y1) ( (x1+ x2 )/2 , (y1+ y2 )/2 ) (x2, y2) Copyright © Peter Cappello 2011

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Exercise 10 Solution (xj + xi)/2 is integer if xj & xi are both odd or both even. (yj + yi)/2 is integer if yj & yi are both odd or both even. Put each ordered pair into 2 x 2 = 4 categories: x odd, y odd x odd, y even x even, y odd x even, y even With 5 distinct ordered pairs, at least 1 category has ≥ 2 points. A line connecting 2 points in the same category has a midpoint with integer coordinates. Copyright © Peter Cappello

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**Copyright © Peter Cappello**

Exercise 30 Show: If there are 100,000,000 wage earners in the US who earn < $1,000,000, there are 2 with the same income, to the penny. (Assume each wage earner’s income > 0.) Copyright © Peter Cappello

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Exercise 30 Solution Show that if there are 100,000,000 wage earners in the US who earn < $1,000,000, there are 2 with the same income, to the penny. Solution: Assume that each income is > 0. Denominate the incomes in pennies. The smallest possible income is 1¢. The largest possible income is $999, = 99,999,999¢. Since there are more wage earners (100,000,000) than distinct income values (99,999,999), at least 2 must earn the same income. Copyright © Peter Cappello

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Exercise 40 There are 51 houses on a street. Each house has an integer address between 1000 and 1099, inclusive. Show that at least 2 houses have consecutive addresses. Copyright © Peter Cappello

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Exercise 40 Solution There are 51 houses on a street. Each house has an integer address between 1000 and 1099, inclusive. Show that at least 2 houses have consecutive addresses. Solution: Partition the address space into 50 intervals: [1000, 1001], [1002, 1003], …, [1098, 1099]. Associate each of the 51 addresses with an interval. There are more addresses than intervals. At least 1 interval is associated with 2 distinct addresses. Copyright © Peter Cappello

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End Copyright © Peter Cappello

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**Copyright © Peter Cappello**

20 Find an increasing subsequence of maximal length and a decreasing subsequence of maximal length in the sequence 22, 5, 7, 2, 23, 10, 15, 21, 3, 17. Solution: Brute force. 5, 7, 10, 15, 21 is an increasing subsequence of length 5. 22, 10, 3 is a decreasing subsequence of length 3 (not unique). Copyright © Peter Cappello

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