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Mathematical Gems III AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 9 Mathematical Gems III Ross Honsberger MATHEMATICAL GEMS III By ROSS HONSBERGER THE DOLCIANI MATHEMATICAL EXPOSITIONS Published by THE MATHEMATICAL ASSOCIATION OF AMERICA COl1znzittee Oil Publicatiolls ALAN TUCKER, Chairman SUbC011111zittee 011 Dolciani Mathenzatical Expositions JOSEPH MALKEVITCH, Chairman D. MCCARTHY D. SMALL 10.1090/dol/009 The Do/ciani Mathell1atical Expositions NUMBER NINE MATHEMATICAL GEMS III By ROSS HONSBERGER University o.f Waterloo Published lind Distributed by THE MATHEMATICAL ASSOCIATION OF AMERICA @1985 by The Mathematical Association ofAmerica (Incorporated) Library of Congress Catalog Card Number 061842 Complete Set ISBN 0-88385-300-0 Vol. 9 ISBN 0-88385-318-3 Printed in the United States ofAmerica Current printing (last digit): 10 9 8 765 4 3 2 The DOLCIANI MATHEMATICAL EXPOSITIONS series of the Math­ ematical Association of America was established through a generous gift to the Association from Mary P. Dolciani, Professor ofMathematics at Hunter College of the City University of New York. In making the gift, Professor Dolciani, herself an exceptionally talented and successful expositor of mathematics, had the purpose of furthering the ideal of excellence in mathematical exposition. The Association, for its part, was delighted to accept the gracious gesture initiating the revolving fund for this series from one who has served the As­ sociation with distinction, both as a member of the Committee on Publications and as a member of the Board of Governors. It was with genuine pleasure that the Board chose to name the series in her honor. The books in the series are selected for their lucid expository style and stimulating mathematical content. Typically, they contain an ample supply of exercises, many with accompanying solutions. They are intended to be suffi­ ciently elementary for the undergraduate and even the mathematically inclined high-school student to understand and enjoy, but also to be interesting and sOluetilnes challenging to the more advanced mathematician. DOLCIANI MATHEMATICAL EXPOSITIONS 1. Mathematical Gems, Ross Honsberger 2. Mathematical Gems II, Ross Honsberger 3. Mathematical Morsels, Ross Honsberger 4. Mathematical Plums, Ross Honsberger (ed.) 5. Great Moments in Mathematics (Before 1650), Howard Eves 6. Maxima and Minima without Calculus, Ivan Niven 7. Great Moments in Mathematics (After 1650), Howard Eves 8. Map Coloring, Polyhedra, and the Four-Color Problem, David Barnette 9. Mathematical Gems III, Ross Honsberger 10. More Mathematical Morsels, Ross Honsberger PREFACE The technical background required for the enjoyment of the essays in this collection seldonl goes beyond the level of the college freshman. It is renlarkable ho",- nluch exciting nlathenlatics exists at this ele­ mentarv level. While \ye can't help learning something from each new problenl, these topics are presented solely for your el{;oy111enf, and al­ though a certain anlount of drive and concentration is a price that cannot be avoided, I hope you \yill find that these little genls are well worth the effoli necessary for their appreciation. A glossary of technical ternlS and ideas is given at the end of the es­ says; the ""Yords that are explained there are nlarked in the text with an asterisk (*). I '" ould like to take this opportunity to thank the menlbers of the Subconlnlittee on the Dolciani Expositions-Joe Malkevitch, Ken­ neth Rebnlan, Alan Tucker. and Donald Snlall-for a careful review of the manuscript; their constructive criticisnl led to nlany inl­ provements. Ross HONSBERGER CONTENTS CllAfYfER PAGE 1. Gleanings from Combinatorics 1 2. Gleanings from Geometry 18 3. Two Problems in Combinatorial Geometry 36 4. Sheep Fleecing with Walter Funkenbusch 48 5. Two Problems in Graph Theory 56 6. Two Applications of Generating Functions 64 7. Some Problems from the Olympiads 76 8. A Second Look at the Fibonacci and Lucas Numbers 102 9. Some Problems in Combinatorics 139 10. Four Clever Schemes in Cryptography 151 11. Gleanings from Number Theory 174 12. Schur's Theorem: An Application of Ramsey's Theorem 183 13. Two Applications of ReIly's Theorem 186 14. An Introduction to Ramanujan's Highly Composite Numbers 193 15. On Sets of Points in the Plane 201 16. Two Surprises from Algebra 208 17. A Problem of Paul Erdos 215 18. Cai Mao-Cheng's Solution to Katona's Problem on Families of Separating Subsets 224 Solutions to Selected Exercises 241 Glossary 247 Index 249 SOLUTIONS TO SELECTED EXERCISES Two Problems on Generating Functions 1. The generating function for the partitions in question are (i) no even part repeated: IIk~l (1 - x 2k - 1)-1(1 + x 2k ); k 2k (ii) no part occurs more than 3 times: rrk~ 1 (1 + x + x + x 3k ); (iii) no part is divisible by 4: IIk~ 1 (1 - x 4k)/(1 - x k ). Easy manipulations show that (i) and (ii) are the same as (iii). 2. The generating function for the number of partitions in which no part occurs more than d times is 1 - x(d+ 1)k dk j(x) = IT (1 + x k + x 2k + ... + x ) = IT 1 _ x k~l k~l k The generating function for the number of partitions in which no term is a multiple of d + 1 is clearly the same as this latter expression forj(x). 3. The generating function for the number of partitions in which every part occurs either 2, 3, or 5 times is II (1 + x 2k + x 3k + x Sk ) = IT (1 + x 2k )(1 + x 3k ) k~l k~l 1 - x 4k 1 - x 6k = IT --­ k~l 1 - x 2k 1 - x 3k • 241 242 SOLUTIONS In the denominator, the factor 1 - x 2k provides a term 1 - xi for i = 2, 4, 6, 8, 10, 12, 14, ... , and 1 - x 3k provides a term 1 - xi for i = 3, 6, 9, 12, 15, .... The 1 - x 4k in the numerator cancels terms 1 ­ xi for i = 4, 8, 12, 16, ... , and the 1 - x 6k cancels 1 - xi for i = 6, 12, 18, .... Therefore the ternlS remaining in the denominator are 1 - xi for i = 2, 6, 10, 14, ... , and i = 3, 9, 15, ... , that is, for i = 2,3,6,9, 10 (nlod 12). Hence the conclusion. 4. The generating function for the partitions in \vhich no part oc­ curs exactly once is II (1 + x 2k + x 3k + ...) = II [1 + x 2k (1 + x k + x 2k + ... )] k~1 k~1 = II [1 + x 2k (1 - x k )-I] = II [(1 - x k )-1 (1 - x k + x 2k)] k~1 k~1 1 + x 3k 1 + x 3k 1 + x 6k = II (1 - x k )-1 • = II = IT ------ k~l 1 + x k k~l 1 - x 2k k~l (1 - x 2k)(1 - x 3k)• In the denonlinator, the factor 1 - x 2k provides ternlS 1 - xi for i = 2,4,6,8, ... , and 1 - x 3k provides terms 1 - xi for i = 3,6,9, . The 1 - x 6k in the nUlnerator cancels 1 - xi for i = 6, 12, 18, , leaving ternlS for i = 2,3, 4, 6, 8, 9, 10, 12, ... , that is, those for i = 0, 2, 3, 4 (mod 6). Hence the conclusion. A Second Loob~ at the Fibonacci and Lucas Numbers Posed in section on divisibility. For odd111 we have (17,111) = (34,111) = (19,111) = 1(9,11) =Ib13, or19, that is, 1, 2, or 34. Since (17,111) must be odd, then (17, ill) = 1, and 171'111. Posed in the set of exercises at the end. 2. Since!" = 1,,+2 - !"+b we have S = E I" = E 111+2 - 111+1 11=1 111+1111+2 ,,=1 1,,+11,,+2 SOLUTIONS 243 = E[_1 1_J (in which both terms have 1l=1 1,,+1 111+2. thesamejornl) = ()2 - )J + ()3 - )J + ()4 - )J + ..., having partial sums SIl = (1/12) - (11In+2). Since 11111+2 ~ 0 as ~ n 00, we have S = limll-+ oo SIl = 1112 = 1. 3. We know that L m 1111 if and only if 111 divides into 1l an even number of times. SinceL4 = 7, we have 7111l if and only if 4 divides into n an even number of times, which requires 1l to be even. If n is odd, we never have 71/". s. We have hll(L~1I - 1) = (a: =:211) [(a211 + ~211)2 - 1] 1 = (a2n - {32n)(a 4n + 2a2n {32n + (3411 - 1) a-{3 1 = (a2n - {32n)(a 4n + 1 + (34n) (since cx{3 -1) a-{3 = 1 - ---(a6n + a.2n + {32n - a. 2n - (32n - (3611) a-{3 (since a{3 = -1, then a. 2n {34n = (32n, for example) 1 - --{3- (a61l - (36n) = 16n· a- This givesj211 + 1611 = 12nL~n, and we have IlII + 81211(1211 + 1611) = Iln + 812n ·/2nL 'tz = Il11 + 81~IIL~1l = Iln + 81ln (since/2n L 2n = 1411) = 9/111 = (3/411)2, a perfect square. 244 SOLUTIONS 6. (ii). We have 0·_0 (an - {3n)llll L == linl {fn == linl no~oo n--'OO a - {3 . a[l - ({3/a)l1.Jllll == hnl . n·-"oo (a - (3) 1//1 Now, a == (1 + -)"5)/2 == 1.6 approximately, and {3 == (1 - -JS)/2 == -.6 approximately, and we have 1{3lal < 1, implying -' linl 11 00 ({31 CI.)l1 == O. Also a - (3 == -J"S > 1, implying -. linlll oo (-JS) 1//1 == 1. Thus L == CI.. 7. We know that,l2n+1 == ,l~+1 + .l~; also L n == .l~+1 + .1',,-1 by definition. Therefore we have fn+l L n+2 - j~+2Ln == .l~+tC/~+J + ,In+l) - .fn+2(l'n+1 + ,I'll-I) == ,f~+ I + j~ +l,ln+J - .l~ +2.t~ -+ I- ,fn+2.lll-1 == .l~+1 + .In-+I(j~+J - .In+2) - Cln+1 +j~)(ln+1 - .l~) 2 0 -,- /,2n+l +.
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