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i i “main” — 2008/10/2 — 18:29 — page 1 — #13 i i 1 Elementary Problems Let’s begin with some relatively easy problems. The challenges become gradually more difficult as you go through the book. The problems in this chapter can be solved without advanced mathematics. Knowledge of basic arithmetic, algebra, and geometry willbe help- ful, as well as your own creative thinking.I recommend that you attempt all the problems, even if you already know the answers, because you may discover new and interesting as- pects of the solutions. A bonus after each solution discusses a related mathematical topic. Remember, each problem has an aha! solution. 1.1 Arithmetic Fair Division Abbyhas fifteen cookies and Betty has nine cookies. Carly, who has no cookies,pays Abby and Betty 24 cents to share their cookies. Each girl eats one-third of the cookies. Betty says that she and Abby should divide the 24 cents evenly, each taking 12 cents. Abby says that since she supplied fifteen cookies and Betty only nine, she should take 15 cents and Betty 9 cents. What is the fair division of the 24 cents between Abby and Betty? Solution The key is to determine the worth of one cookie. Each girl eats eight cookies. Since Carly pays 24 cents for eight cookies, each cookie is worth 3 cents. Thus Abby, who starts with fifteen cookies and sells seven to Carly, should receive 21 cents, and Betty, who starts with nine cookies and sells one to Carly, should receive 3 cents. 1 i i i i i i “main” — 2008/10/2 — 18:29 — page 2 — #14 i i 2 1 Elementary Problems Bonus: Cookie Jar Division Abby, Betty, and Carly have 21 cookie jars (all the same size). Seven are full, seven are half-full, and seven are empty. The girls wish to divide the cookie jars among themselves so that each girl gets the same number of jars and the same amount of cookies. How can they do this without opening the jars? There are two different ways, as shown below. Abby FFFHEEE Abby FFHHHEE Betty FFFHEE E Betty FFHHHEE Carly FHHHHHE Carly FFF HEEE We have designated the full cookie jars by F, the half-full ones by H, and the empty ones 1 by E. In both solutions, each girl receives seven jars and 3 2 jars’ worth of cookies. We have not counted as different the same solution with the girls’ names permuted. As we will see later (in “Integer Triangles” on p. 165), each solution corresponds to a triangle with integer sides and perimeter 7, as shown below. In a given triangle, the side lengths equal the number of full cookie jars for each girl in the corresponding cookie jar solution. @ 3 ( 2 @ 2 (((( ((( @ h(h 1 hhhh @ hhh @ 3 3 A Mere Fraction (a) Find an integral fraction between 1=4 and 1=3 such that the denominator is a positive integer less than 10. (b) Find an integral fraction between 7=10 and 5=7 such that the denominatoris a positive integer less than 20. Solution 1 (a) Since 3<3 2 <4, we have, upon taking reciprocals, the inequalities 1 2 1 < < : 4 7 3 We double-check, by cross-multiplication: 1 2 < since 1 7<4 2 4 7 and 2 1 < since 2 3<7 1: 7 3 One can check by exhaustion that 2=7 is the only solution. i i i i i i “main” — 2008/10/2 — 18:29 — page 3 — #15 i i 1.1 Arithmetic 3 (b) Notice that the fraction found in (a) can be obtained by adding the numerators and denominators of 1=4 and 1=3: 1 1 2 C : 4 3 D 7 C Would the same trick work for 7=10 and 5=7? Let’s try the plausible answer 7 5 12 C : 10 7 D 17 C We verify the inequalities 7 12 5 < < 10 17 7 by cross-multiplication: 7 12 < since 7 17 < 10 12 10 17 and 12 5 < since 12 7<17 5: 17 7 One can check by exhaustion that 12=17 is the only solution. Bonus: Mediant Fractions The answers to (a) and (b) are called mediant fractions.1 The mediant fraction of a=b and c=d is .a c/=.b d/. Thus, the mediant fraction of 1=4 and 1=3 is 2=7, and the mediant C C fraction of 7=10 and 5=7 is 12=17. If a=b < c=d (with b and d positive), then a a c c < C < : b b d d C I encourage the reader to prove these inequalities using cross-multiplication. Here is an aha! proof of the mediant fraction inequalities. Assuming that a, b, c and d are all positive, we will interpret the fractions as concentrations of salt in water. Suppose that we have two solutions of salt water, the first with a teaspoons of salt in b gallons of water, and the second with c teaspoons of salt in d gallons of water. The concentration of salt in the first solution is a=b teaspoons/gallon, while the concentration of salt in the second solution is c=d teaspoons/gallon. Suppose that the first solution is less salty than the second, i.e., a=b < c=d. Now, if we combine the two solutions, we obtain a solution with a c teaspoons of salt in b d gallons of water, so that the salinity of the new C C solution is .a c/=.b d/ teaspoons/gallon. Certainly, the new solution is saltier than the C C first solution and less salty than the second. That is to say, a a c c < C < : b b d d C Our proof can truly be called a saline solution! 1Mediant fractions arise in the study of Farey sequencesand in the “solution” of Simpson’s Paradox. i i i i i i “main” — 2008/10/2 — 18:29 — page 4 — #16 i i 4 1 Elementary Problems A Long Sum What is the sum of the first 100 integers, 1 2 3 100‹ C C CC Of course, we could laboriously add up the numbers. But we seek instead an aha! solu- tion, a simple calculation that immediately gives the answer and insight into the problem. Solution Observe that the numbers may be paired as follows: 1 and 100, 2 and 99, 3 and 98, ..., 50 and 51. We have 50 pairs, each pair adding up to 101,so oursum is 50 101 5050. D This solution works in general for the sum 1 2 3 n; C C CC where n is an even number. We pair the numbers as before: 1 and n, 2 and n 1, 3 and n 2, ..., n=2 and n=2 1. C We have n=2 pairs, each adding up to n 1,so oursum is n=2 .n 1/ n.n 1/=2. C C D C What if n is odd? We can no longer pair all the numbers (as the middle term has no mate). However, throwing in a 0 doesn’t change the total: 0 1 2 3 n: C C C CC Now we have an even number of terms, and they may be paired as 0 and n, 1 and n 1, 2 and n 2, ..., .n 1/=2 and .n 1/=2 1. C We have .n 1/=2 pairs, each adding up to n, so oursum is n.n 1/=2 (again). C C A “duplication method” works for both n even and n odd. Let S be the sum, and intro- duce a duplicate of S, written backwards: S 1 2 3 n: D C C C C S n .n 1/ .n 2/ 1: D C C C C i i i i i i “main” — 2008/10/2 — 18:29 — page 5 — #17 i i 1.1 Arithmetic 5 Add the two expressions for S, summing the first terms, then the second terms, and so on: 2S .n 1/ .n 1/ .n 1/ .n 1/: D C C C C C CC C (The term n 1 occurs n times.) This simplifies to C 2S n.n 1/; D C or n.n 1/ S C : D 2 Bonus: Sum of an Arithmetic Progression Carl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time, is said to have solved our problem for n 100 when he was a 10-year-old school pupil. However, D according to E. T. Bell [2], the problem that Gauss solved was actually more difficult: The problem was of the following sort, 81297 81495 81693 : : : 100899, C C C C where the step from one number to the next is the same all along (here 198), and a given number of terms (here 100) are to be added. The problem mentioned by Bell is the sum of an arithmetic progression. We can evaluate the sum using our formula for the sum of the first n integers. The calculation is 81297 81495 100899 81297 100 198 .1 99/ C CC D C CC 99 100 8129700 198 D C 2 9109800: D Sums of Consecutive Integers Behold the identities 1 2 3 C D 4 5 6 7 8 C C D C 9 10 11 12 13 14 15 C C C D C C 16 17 18 19 20 21 22 23 24 C C C C D C C C 25 26 27 28 29 30 31 32 33 34 35 C C C C C D C C C C 36 37 38 39 40 41 42 43 44 45 46 47 48: C C C C C C D C C C C C What is the pattern and why does it work?2 2Roger B.
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