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hw07sol

hw07sol - Math 55 Discrete Mathematics Fall 2008 Homework 7...

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Math 55: Discrete Mathematics, Fall 2008 Homework 7 Solutions 5.2: 12. 26, since there are 25 possibilities for the pair ( a mod 5 , b mod 5). 14. (a) Consider the list of 14 integers x 1 , . . . , x 7 , 11 - x 1 , . . . , 11 - x 7 . All 14 are between 1 and 10, so there must be at least four equal pairs, since we cannot have three integers equal on this list, and we could have at most a list of 13 with 0 to 3 equal pairs. An equal pair must be of the form x i = 11 - x j , that is, x i + x j = 11. The same x i and x j gives a second equal pair x j = 11 - x i , but since we have four equal pairs, we must have two solutions x i + x j = 11 (b) No. A counterexample is { 1 , . . . , 6 } in which the only pair that sums to 11 is 5 + 6. * 36. As noted in the hint, we cannot have a person who knows no one else and a person who knows everyone else. So if k ( x ) is the number of other people that person x knows, then all the values k ( x ) are either in the set { 0 , 1 , . . . , n - 2 } (if nobody knows everyone) or in { 1 , 2 , . . . , n - 1 } (if nobody knows no one else). Either way, there are n values in a set of n - 1 possibilities, so two must be equal.

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hw07sol - Math 55 Discrete Mathematics Fall 2008 Homework 7...

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