ELEMENTARY PROBLEMS AND SOLUTIONS


 Jasmine Alexander
 1 years ago
 Views:
Transcription
1 Edited by A. P. HILLMAN Assistant Editors GLORIA C. PADILLA CHARLES R. WALL Send all communications regarding ELEMENTARY PROBLEMS SOLUTIONS to PROFESSOR A. P. HILLMAN; 709 Solano Dr., S.E.,; Albuquerque, NM Each solution or problem should be on a separate sheet (or sheets). Preference will be given to those typed with double spacing in the format used below. Solutions should be received within four months of the publication date. Proposed problems should be accompanied by their solutions. DEFINITIONS The Fibonacci numbers F n the Lucas numbers L n satisfy F n+2 L n+2 = Fn f n» ^0 = > ^ 1 = 1 = L n+1 + L n> L 0 = 2 > L i = 1 ' Also, a 3 designate the roots (1 + v5)/2 (1  v5)/2, respectively, of x 2  x  1 = 0. PROBLEMS PROPOSED IN THIS ISSUE B520 Proposed by Charles R. Wall, Trident Technical College, Charleston, SC (a) Suppose that one has a table for multiplication (mod 10) in which a, b,,.., j have been substituted for 0, 1,..., 9 in some order. How many decodings of the substitution are possible? (B) Answer the analogous question for a table of multiplication (mod 12). B521 Proposed by Charles R. Wall, Trident Technical College, Charleston, SC See the previous problem. Find all moduli m > 1 for which the multiplication (mod m) table can be decoded in only one way. B522 Proposed by loan Tomescu, University of Bucharest, Romania Find the number A(n) of sequences (a l5 a 2,..., a k ) of integers ai satisfying 1 < a t < a i + 1 < n a i ai E 1 (mod 2) for i = 1, 2,..., k  1. [Here k is variable, but of course 1 < k < n. For example, the three allowable sequences for n = 2 are (1), (2), (1,2).] B523 Proposed by Laszlo Cseh Imre Merenyi, Cluj, Romania Let p, a Q, at,..., a n be integers with p a positive prime such that gcd(a 0, p) = 1 = gcd(a n, p). 1984] 183
2 Prove that in {0, 1,..., p  1} there are as many solutions of the congruence as there are of the congruence a n x n + a n _ 1 x n ~ ciyz + a 0 = 0 (mod p) a 0 x n + a x x n ~ x + '" + (ZniX + a = 0 (mod p ). B524 Proposed by Herta T. Freitag, Roanoke, VA Let S n = Ff n _ 1 + F n F n. 1 (F 2n _ 1 + F n 2 ) + 3F n F n + 1 (F 2n _ 1 + F F _i). Show that S n is the square of a Fibonacci number. B525 Proposed by Walter Blumberg, Coral Springs, FL Let x,y, z be positive integers such that 2 X  1 = / s # > 1. Prove that z = 1. SOLUTIONS FibonacciLucas Centroid B496 Proposed by Stanley Rabinowitz, Digital Equip. Corp., Merrimack, NH Show that the centroid of the triangle whose vertices have coordinates is (F n + h, L n+h ). (F n, L n ), (F n+1, L n+1 ), (F n + 6, n + 6 ) Solution, independently, by Walter Blumberg, Coral Springs, FL; Wray G. Brady, Slippery Rock, PA; Paul S. Bruckman, Sacramento, CA; Laszlo Cseh, Cluj, Romania; Leonard Dresel, Reading, Engl; Herta T. Freitag, Roanoke, VA; L. Kuipers, Switzerl; Stanley Rabinowitz, Merimack, NH; Imre Merenyi, Cluj, Romania; John W. Milsom, Butler, PA; Bob Prielipp, Oshkosh, WI; Sahib Singh, Clarion, PA; Lawrence Somer, Washington, CD; Gregory Wulczyn, Lewisburg, PA. The coordinates (x, y) of the centroid are given by 3x = F n + F n+1 + Fn+6 = F n F n + h + F n+5 ~ ^n + 2 "*" ^n+h + ^n+3 ^n + h similarly, Hence, the centroid is (F n+h, L n+h ). 3y = L n + L n+1 + L n+6 = 3L n+ h. Area of a FibonacciLucas Triangle BkS7 Proposed by Stanley Rabinowitz, Digital Equip. Corp., Merrimack, NH For d an odd positive integer, find the area of the triangle with vertices (F n9 L n ), (F n + di L n+d ), (F n + 2d> L n+2d ). 184 [May
3 Solution by Paul S. Bruckman, Sacramento, CA By means of a wellknown determinant formula,the area of the given triangle is given by A =i i i 1 II (In the above expression, the inner bar is the determinant symbol, the outer bar represents absolute value.) Then A = 2\(F n + 2d L n + d " Fn + d^n + ld} ~ ( F n + 2d^n ~ F n L n + 2d^ + ( F n+d^n ~ F n L n + d ) \. Using the relation this becomes: F U L V  F V L U = 2(l) v F u _ v, (2) A = (l) n + d F d  {l) n F 2d + (l) n F d = (.l) d F d  F 2d + F d> which equals F 2d when d is odd ( equals F 2d  2F d when d is even). Also solved by Walter Blumberg, Wray G. Brady, Leonard Dresel, Herta T. Freltag, L 0 Kuipers, Graham Lord, John W. Milsom, Bob Prielipp, Sahib Singh, Gregory Wulczyn, the proposer. Fibonacci Recursions Modulo 10 B498 Proposed by Herta T. Freitag, Roanoke, VA Characterize the positive integers k such that, for all positive integers n, F n + F n + k = F n + 2k (mod 10). When k is odd, we have the identity F m+k  F m k = F m L k. Applying this with m = n + k 9 we have F m = F m Lk (mod 10), this will be satisfied whenever L k = 1 (mod 10). On the other h, when k is even, we have F m + k  F m _ k = L m F k, it is not possible to satisfy the given recurrence for even k. Returning to the case of odd k, the condition L k = 1 (mod 10) is equivalent to L k = 1 (mod 2) L k = 1 (mod 5). The first condition implies that k is not divisible by 3; with the help of the Binet formula for L ki the second condition reduces to 2 " = 1 (mod 5), which gives that k  1 in a multiple of 4. Combining these results, we have k = ov k = 12t + 5 (t = 0, 1, 2, 3,... ). Also solved by Paul S. Bruckman, Laszlo Cseh, L. Kuipers, Imre Merenyi, Bob Prielipp, Sahib Singh, Lawrence Somer, Gregory Wulczyn, the proposer. 1984] 185
4 Lucas Recursions Modulo 12 B499 Proposed by Herta T. Freitag, Roanoke, VA Do the Lucas numbers analogue of B498. We have the identit ies L m + k Lm k L m L k when k is odd, L m + k Lmk 5F m F k when k is even. Hence, putting m = n + k 9 the relation L n + L n + k E L n + 2k (mod 10) leads, for odd k, to L m ~ L m L k (mod 10), so that we require L k = 1 (mod 10), leading to the same values of k as in B498 above. Also solved by Paul S. Bruckman, Laszlo Cseh, L. Kuipers, Imre Merenyi, Bob Prielipp, Lawrence Somer, Gregory Wulczyn, the proposer. Two Kinds of Divisibility B500 Proposed by Philip L. Mana, Albuquerque, NM Let A(n) B(n) be polynomials of positive degree with integer coefficients such that B(k)\A(k) for all integers k. Must there exist a nonzero integer h a polynomial C(n) with integer coefficients such that ha(n) = 5(n)C(n)? Solution by the proposer. Using the division algorithm multiplying by an integer In so as to make all coefficients into integers, one has ha(n) = Q(n)B(n) + R(n), (*) where Q(n) R(n) are polynomials in n with integral coefficients R(n) is either the zero polynomial or has degree less than B(n). The hypothesis that B(n)\A(n) (*) imply that B(n)\R(n) for all integers n. If R(n) is not the zero polynomial, R(n) has lower degree than B(n) so lim [R(n)/B(n)] = 0; also R(n) is zero for only a finite number of integers n. Thus 0 < R(n)/B(n) < 1 for some large enough n 9 contradicting B(n)\R(n). HenceR(n) is the zero polynomial (?v) shows that the answer is "yes." Also solved by Paul S. Bruckman L. Kuipers. Doubling Back on a Sequence B501 Proposed by J. O. Shallit & J. P. Yamron, U.C. Berkeley, CA Let a be the mapping that sends a sequence X = (x l9 x 2,..., x, ) of length 2k to the sequence of length k, (X\A ) \X ^X 2k 9 2 2k  1 ' > k k + 1 ) * Let V = (1, 2, 3,..., 2 h ), a 2 (7) = a(a(f)), etc. Prove that a(7), a 2 (7),..., a ~ (7) are all strictly increasing sequences. 186 [May
5 Suppose the nubmers a ls a 2, a 3, a h form a strictly increasing sequence, subject to the condition a 1 + a h = a 2 + a 3 = S, then gives hence, (a h  a x ) 2 > (a 3  a 2 ) 2 (a h  a x ) 2 = (a 3 + a 2 ) 24a r+ a 1 > 4a 3 a 2 a^^. < a 2 a 3. Now any two consecutive terms of a(7) are of the form a^^, a 2 a 3, with a x + a^ = a 3 + a 2 = h, so that it follows that a(7) is a strictly increasing sequence. Next, consider a (7). To avoid a notational forest, we shall apply our method to the specific case where h = 4, with 7 = (1, 2, 3,..., 16). Then, using a dot to denote multiplication, we have where a(7) = (116, 2 15, 314,..., 89) a 2 (T) = ( , ,..., ) = ( , ,..., ) = (b 1 c ls b 2 e 2,..., 2^ e 4 ) (b is i 2, Z> 3, \ ) = a(l, 2, 3,..., 8) (o 13 c 2, c 3, c h ) = a(9, 10, 11,..., 16). By our previous argument, b± is strictly increasing, similarly c^ is. Thus a 2 (7) = (b%ci) is a strictly increasing sequence. Similarly, we can show that a 3 (7) = (d 1 e 1 f 1 g 19 d 2 e 2 f 2 g 2 ), where (d 2, d 2 ) = a(l, 2, 3, 4), (e l5 e 2 ) = a(5, 6, 7, 8), etc., is strictly increasing. The above arguments can be generalized to apply to any value of h, Also solved by Paul S. Bruckman, L. Kuipers, the proposers. 1984] 187
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory
More informationSOLUTIONS FOR PROBLEM SET 2
SOLUTIONS FOR PROBLEM SET 2 A: There exist primes p such that p+6k is also prime for k = 1,2 and 3. One such prime is p = 11. Another such prime is p = 41. Prove that there exists exactly one prime p such
More informationSolutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014
Solutions to Homework 6 Mathematics 503 Foundations of Mathematics Spring 2014 3.4: 1. If m is any integer, then m(m + 1) = m 2 + m is the product of m and its successor. That it to say, m 2 + m is the
More information1 Sets and Set Notation.
LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most
More informationSUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by
SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples
More informationRevised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)
Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.
More informationMEP Pupil Text 12. A list of numbers which form a pattern is called a sequence. In this section, straightforward sequences are continued.
MEP Pupil Text Number Patterns. Simple Number Patterns A list of numbers which form a pattern is called a sequence. In this section, straightforward sequences are continued. Worked Example Write down the
More informationAn Introductory Course in Elementary Number Theory. Wissam Raji
An Introductory Course in Elementary Number Theory Wissam Raji 2 Preface These notes serve as course notes for an undergraduate course in number theory. Most if not all universities worldwide offer introductory
More informationPROOFS BY DESCENT KEITH CONRAD
PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the
More informationA Course on Number Theory. Peter J. Cameron
A Course on Number Theory Peter J. Cameron ii Preface These are the notes of the course MTH6128, Number Theory, which I taught at Queen Mary, University of London, in the spring semester of 2009. There
More information= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without
More informationAlgebra & Number Theory. A. Baker
Algebra & Number Theory [0/0/2009] A. Baker Department of Mathematics, University of Glasgow. Email address: a.baker@maths.gla.ac.uk URL: http://www.maths.gla.ac.uk/ ajb Contents Chapter. Basic Number
More informationINTRODUCTORY SET THEORY
M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H1088 Budapest, Múzeum krt. 68. CONTENTS 1. SETS Set, equal sets, subset,
More informationBasic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011
Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely
More information784: algebraic NUMBER THEORY (Instructor s Notes)*
Math 784: algebraic NUMBER THEORY (Instructor s Notes)* Algebraic Number Theory: What is it? The goals of the subject include: (i) to use algebraic concepts to deduce information about integers and other
More informationNumber Theory for Mathematical Contests. David A. SANTOS dsantos@ccp.edu
Number Theory for Mathematical Contests David A. SANTOS dsantos@ccp.edu August 13, 2005 REVISION Contents Preface 1 Preliminaries 1 1.1 Introduction.................. 1 1.2 WellOrdering.................
More informationIf A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?
Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question
More informationGroup Fundamentals. Chapter 1. 1.1 Groups and Subgroups. 1.1.1 Definition
Chapter 1 Group Fundamentals 1.1 Groups and Subgroups 1.1.1 Definition A group is a nonempty set G on which there is defined a binary operation (a, b) ab satisfying the following properties. Closure: If
More informationALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY
ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY HENRY COHN, JOSHUA GREENE, JONATHAN HANKE 1. Introduction These notes are from a series of lectures given by Henry Cohn during MIT s Independent Activities
More informationTAKEAWAY GAMES. ALLEN J. SCHWENK California Institute of Technology, Pasadena, California INTRODUCTION
TAKEAWAY GAMES ALLEN J. SCHWENK California Institute of Technology, Pasadena, California L INTRODUCTION Several games of Tf takeaway?f have become popular. The purpose of this paper is to determine the
More information4. FIRST STEPS IN THE THEORY 4.1. A
4. FIRST STEPS IN THE THEORY 4.1. A Catalogue of All Groups: The Impossible Dream The fundamental problem of group theory is to systematically explore the landscape and to chart what lies out there. We
More informationTHE CONGRUENT NUMBER PROBLEM
THE CONGRUENT NUMBER PROBLEM KEITH CONRAD 1. Introduction A right triangle is called rational when its legs and hypotenuse are all rational numbers. Examples of rational right triangles include Pythagorean
More informationTopics in Number Theory, Algebra, and Geometry. Ambar N. Sengupta
Topics in Number Theory, Algebra, and Geometry Ambar N. Sengupta December, 2006 2 Ambar N. Sengupta Contents Introductory Remarks........................... 5 1 Topics in Number Theory 7 1.1 Basic Notions
More informationOn the numbertheoretic functions ν(n) and Ω(n)
ACTA ARITHMETICA LXXVIII.1 (1996) On the numbertheoretic functions ν(n) and Ω(n) by Jiahai Kan (Nanjing) 1. Introduction. Let d(n) denote the divisor function, ν(n) the number of distinct prime factors,
More informationWHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?
WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly
More informationThe number field sieve
The number field sieve A.K. Lenstra Bellcore, 435 South Street, Morristown, NJ 07960 H.W. Lenstra, Jr. Department of Mathematics, University of California, Berkeley, CA 94720 M.S. Manasse DEC SRC, 130
More information
Number Theory Naoki Sato
Number Theory Naoki Sato 0 Preface This set of notes on number theory was originally written in 1995 for students at the IMO level. It covers the basic background material
More informationFigure 2.1: Center of mass of four points.
Chapter 2 Bézier curves are named after their inventor, Dr. Pierre Bézier. Bézier was an engineer with the Renault car company and set out in the early 196 s to develop a curve formulation which would
More information( 1)2 + 2 2 + 2 2 = 9 = 3 We would like to make the length 6. The only vectors in the same direction as v are those
1.(6pts) Which of the following vectors has the same direction as v 1,, but has length 6? (a), 4, 4 (b),, (c) 4,, 4 (d), 4, 4 (e) 0, 6, 0 The length of v is given by ( 1) + + 9 3 We would like to make
More informationAn example of a computable
An example of a computable absolutely normal number Verónica Becher Santiago Figueira Abstract The first example of an absolutely normal number was given by Sierpinski in 96, twenty years before the concept
More information