组合数学【第5版】(英文版)第3章答案

Math475Text:Brualdi,Introductory Combinatorics5th Ed. Prof:Paul Terwilliger

Selected solutions for Chapter3

1.For1≤k≤22we show that there exists a succession of consecutive days during which the grandmaster plays exactly k games.For1≤i≤77let b i denote the number of games

played on day i.Consider the numbers{b1+b2+···+b i+k}76

i=0∪{b1+b2+···+b j}77

j=1

.

There are154numbers in the list,all among1,2,...,153.Therefore the numbers{b1+b2+

···+b i+k}76

i=0∪{b1+b2+···+b j}77

j=1

.are not distinct.Therefore there exist integers i,j

(0≤i

2.Let S denote a set of100integers chosen from1,2,...,200such that i does not divide j for all distinct i,j∈S.We show that i∈S for1≤i≤15.Certainly1∈S since 1divides every integer.By construction the odd parts of the elements in S are mutually distinct and at most199.There are100numbers in the list1,3,5,...,199.Therefore each of 1,3,5,...,199is the odd part of an element of S.We have3×5×13=195∈S.Therefore none of3,5,13,15are in S.We have33×7=189∈S.Therefore neither of7,9is in S.We have11×17=187∈S.Therefore11∈S.We have shown that none of1,3,5,7,9,11,13,15 is in S.We show neither of6,14is in S.Recall33×7=189∈S.Therefore32×7=63∈S. Therefore2×32×7=126∈S.Therefore2×3=6∈S and2×7=14∈S.We show10∈S. Recall3×5×13=195∈S.Therefore5×13=65∈S.Therefore2×5×13=130∈S. Therefore2×5=10∈S.We now show that none of2,4,8,12are in S.Below we list the integers of the form2r3s that are at most200:

1,2,4,8,16,32,64,128,

3,6,12,24,48,96,192,

9,18,36,72,144,

27,54,108,

81,162.

In the above array each element divides everything that lies to the southeast.Also,each row contains exactly one element of S.For1≤i≤5let r i denote the element of row i that is contained in S,and let c i denote the number of the column that contains r i.We must have c i

3.See the course notes.

4,5,6.Given integers n≥1and k≥2suppose that n+1distinct elements are chosen from{1,2,...,kn}.We show that there exist two that differ by less than k.Partition

{1,2,...,nk}=∪n

i=1S i where S i={ki,ki−1,ki−2,...,ki−k+1}.Among our n+1

chosen elements,there exist two in the same S i.These two differ by less than k.

1

7.Partition the set{0,1,...,99}=∪50

i=0S i where S0={0},S i={i,100−i}for1≤i≤49,

S50={50}.For each of the given52integers,divide by100and consider the remainder. The remainder is contained in S i for a unique i.By the pigeonhole principle,there exist two of the52integers for which these remainders lie in the same S i.For these two integers the sum or difference is divisible by100.

8.For positive integers m,n we consider the rational number m/n.For0≤i≤n divide the integer10i m by n,and call the remainder r i.By construction0≤r i≤n−1.By the pigeonhole principle there exist integers i,j(0≤i

10iθ=b+

r 10 −1

=b+

r

10

+

r

102

+

r

103

+···

Since the integer r is in the range0≤r≤10 −2this yields a repeating decimal expansion forθ.

9.Consider the set of10people.The number of subsets is210=1024.For each subset consider the sum of the ages of its members.This sum is among0,1,...,600.By the pigeonhole principle the1024sums are not distinct.The result follows.Now suppose we consider at set of9people.Then the number of subsets is29=512<600.Therefore we cannot invoke the pigeonhole principle.

10.For1≤i≤49let b i denote the number of hours the child watches TV on day i.

Consider the numbers{b1+b2+···+b i+20}48

i=0∪{b1+b2+···+b j}49

j=1

.There are

98numbers in the list,all among1,2,...,96.By the pigeonhole principle the numbers

{b1+b2+···+b i+20}48

i=0∪{b1+b2+···+b j}49

j=1

.are not distinct.Therefore there exist

integers i,j(0≤i

11.For1≤i≤37let b i denote the number of hours the student studies on day i.Consider

the numbers{b1+b2+···+b i+13}36

i=0∪{b1+b2+···+b j}37

j=1

.There are74numbers in the list,

all among1,2,...,72.By the pigeonhole principle the numbers{b1+b2+···+b i+13}36

i=0∪

{b1+b2+···+b j}37

j=1are not distinct.Therefore there exist integers i,j(0≤i

such that b i+1+···+b j=13.During the days i+1,...,j the student will have studied exactly13hours.

12.Take m=4and n=6.Pick a among0,1,2,3and b among0,1,2,3,4,5such that a+b is odd.Suppose that there exists a positive integer x that yields a remainder of a(resp.b) when divided by4(resp.by6).Then there exist integers r,s such that x=4r+a and x=6s+bining these equations we obtain2x−4r−6s=a+b.In this equation the

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