Competitive math for middle school

The 39 selfcontained sections in this book present workedout examples as well as many sample problems categorized by the level of difficulty as Bronze, Silver, and Gold in order to help the readers gauge their progress and learning. Detailed solutions to all problems in each section are provided at the end of each chapter. The book can be used not only as a text but also for selfstudy. The text covers algebra (solving single equations and systems of equations of varying degrees, algebraic manipulations for creative problem solving, inequalities, basic set theory, sequences and series, rates and proportions, unit analysis, and percentages), probability (counting techniques, introductory probability theory, more set theory, permutations and combinations, expected value, and symmetry), and number theory (prime factorizations and their applications, Diophantine equations, number bases, modular arithmetic, and divisibility). It focuses on guiding students through creative problemsolving and on teaching them to apply their knowledge in a wide variety of scenarios rather than rote memorization of mathematical facts. It is aimed at, but not limited to, highperforming middle school students and goes further in depth and teaches new concepts not otherwise taught in traditional public schools.

Competitive Math for Middle School Pan Stanford Series on Renewable Energy — Volume Competitive Math for Middle School Algebra, Probability, and Number Theory editors Vinod Krishnamoorthy Preben Maegaard Anna Krenz Wolfgang Palz The Rise of Modern Wind Energy Wind Power for the World February 9, 2018 14:42 PSP Book - 9in x 6in Published by Pan Stanford Publishing Pte Ltd Penthouse Level, Suntec Tower Temasek Boulevard Singapore 038988 Email: editorial@panstanford.com Web: www.panstanford.com British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Competitive Math for Middle School: Algebra, Probability, and Number Theory Copyright c 2018 Pan Stanford Publishing Pte Ltd Cover image: Courtesy of Nirmala Moorthy All rights reserved This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA In this case permission to photocopy is not required from the publisher ISBN 978-981-4774-13-0 (Paperback) ISBN 978-1-315-19663-3 (eBook) 00-Prelims February 6, 2018 13:35 PSP Book - 9in x 6in 00-Prelims Contents ix Preface Algebra Part Part Part Part Part Part Part Part Part Part 10 Part 11 Part 12 Part 13 Part 14 Part 15 Part 16 Part 17 Part 18 Part 19 Part 20 Part 21 Part 22 Part 23 Linear Equations Cross Multiplication Systems of Equations Exponents and Roots Simplifying Radical Expressions Proportions Inequalities Counting Numbers Sequences and Series Distance, Rate, and Time Rates Unit Conversion and Analysis Percentages Mixtures Polynomial Expansions Equivalent Expressions Factoring Special Factorizations The Quadratic Formula Solving for Whole Expressions Infinite Series Sets Mean, Median, Mode, and Range 1 13 21 23 28 31 32 36 38 42 45 48 50 53 54 59 60 65 66 70 74 Solutions Manual 76 Part Part 76 82 Linear Equations Cross Multiplication February 6, 2018 13:35 PSP Book - 9in x 6in 00-Prelims vi Contents Part Part Part Part Part Part Part Part 10 Part 11 Part 12 Part 13 Part 14 Part 15 Part 16 Part 17 Part 18 Part 19 Part 20 Part 21 Part 22 Part 23 Systems of Equations Roots and Exponents Simplifying Radical Expressions Proportions Inequalities Counting Numbers Sequences and Series Distance, Rate, and Time Rates Unit Conversion and Analysis Percentages Mixtures Polynomial Expansions Equivalent Expressions Factoring Special Factorizations The Quadratic Formula Solving for Whole Expressions Infinite Series Sets Mean, Median, Mode and Range 84 89 94 95 100 104 105 110 113 119 120 123 124 129 130 135 137 142 144 146 147 Counting and Probability Part Basic Probability Part Basic Counting Part Multiple Events Part Orderings Part Dependent Events Part Subsets Part Organized Counting Part Permutations and Combinations Part Expected Value Part 10 Weighted Average Part 11 Generality 151 151 155 157 160 162 168 169 171 176 178 180 Solutions Manual 183 Part Part Part 183 185 187 Basic Probability Basic Counting Multiple Events February 6, 2018 13:35 PSP Book - 9in x 6in 00-Prelims Contents Part Part Part Part Part Part Part 10 Part 11 Orderings Dependent Events Subsets Organized Counting Permutations and Combinations Expected Value Weighted Average Generality Number Theory Part Basic Number Theory Part Counting Factors, GCF, LCM Part Bases Part Modular Arithmetic Part Divisibility Tricks 189 192 200 202 206 211 212 214 219 219 224 229 231 235 Solutions Manual 237 Part Part Part Part Part 237 239 243 247 252 Index Basic Number Theory Counting Factors, GCF, LCM Bases Modular Arithmetic Divisibility Tricks 255 vii February 6, 2018 13:35 PSP Book - 9in x 6in 00-Prelims February 6, 2018 13:35 PSP Book - 9in x 6in 00-Prelims Preface I originally began writing this textbook after teaching creative math to middle school students, who were endlessly fascinated, just as I had been, with the field of mathematics This book is a compilation of important concepts used in competition mathematics in Algebra, Counting/Probability, and Number Theory Over 420 problems are provided with detailed solutions found at the end of each chapter These solutions are intended to guide students to identify a promising approach and to execute the necessary math I recommend that students try all of the problems, even if they seem intimidating, and use the solutions to the problems as part of the learning process; they are as essential to learning as the teachings and examples given in the body of the text After reading the solution, students should try to reproduce it themselves My hope is to provide not only mathematical facts and techniques but also examples of how they may be applied, so that the student gains a thorough understanding of the material and confidence in their problem-solving abilities Creative math is becoming increasingly important in schools all over the world The new trend is conspicuous in the remodeling of standardized tests such as the American SAT, the standard entrance exam for U.S colleges Advanced middle school students and high school students can use this book to gain an advantage in school and develop critical thinking skills Vinod Krishnamoorthy February 6, 2018 13:36 PSP Book - 9in x 6in 242 Number Theory 11 Gold • 25, the greatest common factor of the two numbers, equals 52 Therefore, the term containing the lowest power of in either of the two numbers’ prime factorizations must be 52 , and there must be no other common factor to the two • 825, the least common multiple of the two numbers, equals × 52 × 11 Therefore, the highest power of in either of the two numbers’ prime factorizations must be 52 , the highest power of must be 31 , and the highest power of 11 must be 111 • We can now conclude that both of the numbers must have 52 in their prime factorization, as the highest and lowest powers of in either of the prime factorizations are both 52 • We can also conclude that exactly one of the numbers has 31 in its prime factorization (if both had a 31 , 31 would be in the greatest common factor) and exactly one of the numbers has a 111 in its prime factorization • Going through all of the possible cases, the two numbers can be either 52 and × 52 × 11 or × 52 and 52 × 11 The latter pair has a sum of 350, so the two numbers are 75 and 275 12 Gold • Notice that 950/95 = 10 Therefore, 95 is the greatest common factor of the two numbers • What is the prime factorization of 95? 95 = × 19 Every subset of one and one 19 will be a common factor These four subsets are (50 × 190 ), (51 × 190 ), 19 (50 × 191 ), and 95 (51 × 191 ) 13 Silver • The numbers have as common factor (They have many, many more common factors as well) Therefore, they are not relatively prime 03-chapter-03 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Bases 14 Silver 22 3311 = 311 ì 1111 , and 1622 = (24 ) = 288 These prime factorizations not have any factors in common, so 3311 and 1622 are relatively prime 15 Silver • We not know the three positive integers right away, so we take the prime factorization of 495 first 495 = 32 ì ì 11 To split 495 into the product of relatively prime positive integers, we must split the prime factorization into relatively prime pieces If the two 3’s are separated, two or more of the relatively prime pieces will be divisible by 3, so the three positive integers will not be relatively prime • Therefore, one of the positive integers has to be 32 or It is easy to see that the other two integers are and 11 16 Silver • Start with the two-element subsets Those that work are {1,2}, {1,3}, {1,4}, {2,3}, and {3,4}, as the greatest common factor of the elements in these sets is • The only three-element subsets that work are {1,2,3} and {1,3,4} In {1,2,4} and {2,3,4}, and are not relatively prime • The one four-element subset, {1,2,3,4}, does not work, as and have a common factor of Part 3: Bases Silver • Let us call the tens digit of the number x and the units digit y The number equals 10x + y, and the sum of its digits is simply x + y We have that 10x + y = 3(x + y) Expanding yields 10x + y = 3x + 3y 243 February 6, 2018 13:36 PSP Book - 9in x 6in 244 Number Theory • In multi-variable problems, it is often helpful to solve for one variable in terms of the other(s) Let us solve for x in terms of y Doing so, we find that x = y • How we solve this? We use the fact that both x and y have to be one-digit whole numbers, since they are digits For y to be a whole number, y has to equal either or (remember that y cannot be greater than 9) • 00 is not a two-digit number, so y has to be It follows that x = (2/7) × = Therefore, the number we are looking for is 27 The product of and is 14 Silver • 10102 can be written as (20 × 0) + (21 × 1) + (22 × 0) + (23 × 1) This equals + + + 8, or 10 Silver • 23315 can be written as (50 × 1) + (51 × 3) + (52 × 3) + (53 × 2) This equals + 15 + 75 + 250, or 341 Silver • The digit A has a value of 10 and the digit B has a value of 11 3AB313 can be written as (130 × 3) + (131 × 11) + (132 × 10) + (133 × 3) This equals + 143 + 1690 + 6591, or 8427 Silver • To convert the base 10 number 135 to base 4, we repeatedly divide 135 by and use the remainders as the digits of our result • 135/4 = 33 remainder 3, so the units digit is 33 divided by yields remainder 1, so the tens digit is 8/4 = remainder 0, so the hundreds digit is Lastly, we divide 2/4, finding that the thousands digit is • Putting these together yields 20134 Silver • Dividing 3907 by 16, we receive a remainder of and a quotient of 244 Dividing 244 by 16, we receive a remainder 03-chapter-03 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Bases of and a quotient of 15 Dividing 15 by 16, we receive a remainder of 15 and a quotient of • Putting the digits together yields F4316 , as the digit F represents 15 Silver • First, we convert the base number to base 10, and then we convert the base 10 number to base • 5437 can be written as (70 × 3) + (71 × 4) + (72 × 5) or 276 • Now, we repeatedly divide 276 by The first result is 30 remainder 6, so the units digit of the base number is Next, we divide 30 by 9, obtaining remainder Therefore, the tens digit is Lastly, we divide by 9, obtaining remainder • Putting them together yields 3369 Silver • First, we convert the base number to base 10, and then we convert the base 10 number to base 12 • 2121013 can be expanded into (30 × 1) + (31 × 0) + (32 × 1) + (33 × 2) + (34 × 1) + (35 × 2) This equals + + + 54 + 81 + 486, or 631 • Next, we convert this to base 12 631/12 equals 52 remainder 7, so the units digit is 52/12 equals remainder 4, so the tens digit is Finally, 4/12 equals remainder 4, so the hundreds digit is as well • Putting the digits together yields 44712 Silver • Our strategy will be to convert both of the numbers to base 10, find their sum, and then convert the result back to base 10102 = (20 ì 0) + (21 ì 1) + (22 × 0) + (23 × 1), or 10 111012 = (20 × 1) + (21 × 0) + (22 × 1) + (23 × 1) + (24 × 1), or 29 10 + 29 equals 39 • Now for converting 39 to base 39/2 equals 19 remainder 1, so the units digit is 19/2 = remainder 1, so the next digit 245 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 246 Number Theory to the left is 9/2 = remainder 1, 4/2 = remainder 0, 2/2 = remainder 0, and 1/2 equals remainder Putting the remainders together, we obtain our answer, 1001112 10 Silver • To solve the problem, we convert both numbers to base 10, multiply them, and convert the result to base 2334 = (40 ì 3) + (41 ì 3) + (42 × 2) = + 12 + 32 = 47 1223 = (30 × 2) + (31 × 2) + (32 × 1) = 17 47 × 17 = 799 • Now for converting 799 to base 799/7 = 144 remainder 1, 114/7 = 16 remainder 2, 16/7 = remainder 2, and 2/7 = remainder Putting the remainders together, we obtain the answer, 22217 11 Bronze • The digit B is in the tens place, so it adds ten times its value to the overall number The digit A is in the ones place, so it adds times its value to the overall number Therefore, the number’s value is 10B + A 12 Gold • Let us call the original three-digit positive integer XYZ, where X is the hundreds digit, Y is the units digit, and Z is the ones digit The integer is equivalent to 100X + 10Y + Z when written in terms of the values of its digits • Reversing the digits of XYZ produces the three-digit positive integer ZYX ZYX is equivalent to 100Z + 10Y + X We know from the problem that 100X + 10Y + Z − 198 = 100Z + 10Y + X • Subtracting 10Y from both sides yields 100X + Z − 198 100Z + X We are looking to find the value of X − Subtracting 100Z and then X from both sides yields 99X 99Z − 198 = 0, and adding 198 to both sides yields 99X 99Z = 198 = Z − − February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Modular Arithmetic 247 • 99X − 99Z can be factored into 99(X − Z ), leaving us with the equation 99(X − Z ) = 198 Dividing both sides by 99, we find that X − Z = Part 4: Modular Arithmetic Bronze • Subtracting from both sides, we find that x ≡ (mod 5) Bronze • Subtracting from both sides, we find that x ≡ −3 (mod 5) Since −3 is congruent to in mod 5, the simplest form of the solution to the equation is x ≡ (mod 5) Bronze • Subtracting from both sides, we find that x ≡ 84 (mod 6) 84/6 yields a remainder of 0, so 84 is congruent to The simplest solution to the equation is x ≡ (mod 6) Bronze • We subtract from both sides of the equation, obtaining x ≡ (mod 12) Silver • Whether a number is even or odd is depends on its value in mod If the number is congruent to in mod 2, it is even, but if it is congruent to 1, it is odd • The sum of 42 even numbers will always be even Every even number can be replaced with when in mod 2, and the sum of any number of 0’s is • Every odd number can be replaced with in mod The sum of 23 odd numbers in mod is congruent to 1(23) mod 2, and 23 ≡ mod Hence, the first sum is odd • From all of this, it follows that the sum of even numbers is always congruent to in mod 2, the sum of an odd number February 6, 2018 13:36 PSP Book - 9in x 6in 248 Number Theory of odd numbers is congruent to in mod 2, and the sum of an even number of odd numbers is congruent to in mod • By this logic, the second sum is even Silver • Let us call the number n When n is divided by 17, the remainder is 11, so n ≡ 11 (mod 17) • If n is tripled and then added to 11, 11 + 3n is the result How we obtain 11 + 3n from n ≡ 11? We first multiply both sides by 3, which yields 3n ≡ 33(mod 17), and then we add 11 to both sides, obtaining 3n + 11 ≡ 44 (mod 17) • 44/17 yields a remainder of 10, so the equation can be simplified to 3n + 11 ≡ 10 (mod 17) Therefore, the desired remainder is 10 Silver • Adding and subtracting terms from both sides, we simplify the equation to 13x ≡ 19 (mod 7) • 13 is congruent to in mod and 19 is congruent to in mod 7, so we can further simplify the equation into 6x ≡ (mod 7) x ≡ 5/6 is not a valid answer, but replacing with 12 yields that x ≡ (mod 7) The division is valid because and are relatively prime Silver • Let us analyze this sequence The 1st, 4th, 7th, 10th, 13th, terms are 1, the 2nd, 5th, 8th, 11th, 14th, terms are 2, and the 3rd, 6th, 9th, 12th, 15th, terms are • 1, 4, 7, 10, and 13 all are one more than a multiple of 3, so they are all congruent to in mod Similarly, the terms equivalent to are congruent to in mod 3, and the terms equivalent to are congruent to in mod 9010 • What is 9010 in mod 3? equals 3003 remainder 1, so the 9010th term of this series is 03-chapter-03 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Modular Arithmetic 249 Gold • We definitely are not going to compute this value and then see what its units digit is Instead, we must find the value of the expression in mod 10 This will equal the units digit of the expression • 4929 (mod 10) ≡ (mod 10), so 4929732 (mod 10) ≡ 9732 (mod 10) • The units digit of 91 is 9, the units digit of 92 is 1, the units digit of 93 is 92 × (mod 10) ≡ × (mod 10) which is congruent to 9, the units digit of 94 is 93 × (mod 10) ≡ × (mod 10) which is congruent to 1, and this pattern – 9, 1, 9, 1, 9, – continues • Every even power of has units digit 1, and every odd power of has units digit 9, as in the pattern Therefore, 4929732 has units digit 10 Gold • Just as we did in our previous units digit problem, we will look for a pattern among the units digits of the powers of • 21 (mod 10) ≡ 2, 22 (mod 10) ≡ 4, 23 (mod 10) ≡ 8, 24 (mod 10) ≡ 6, 25 (mod 10) ≡ 2, and since we have obtained another 2, we know that the pattern can nothing else but repeat itself over and over again • The pattern of the units digits of the powers of goes 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, Each cycle of the pattern is terms long • When two is raised to an exponent that is congruent to mod 4, the units digit of the result is 2, when two is raised to an exponent that is congruent to mod 4, the units digit of the result is 4, when two is raised to an exponent that is congruent to mod 4, the units digit of the result is 8, and when two is raised to an exponent that is congruent to mod 4, the units digit of the result is • 401 has a remainder of when divided by 4, so the units digit of 2401 is February 6, 2018 13:36 PSP Book - 9in x 6in 250 Number Theory 11 Silver • There are days in a week, and there are 365 days in a year that is not a leap year Leap years have 366 days, and they occur every year that is a multiple of ( 1996, 2000, 2004, 2008 ) These are facts that should just be memorized • Since there are days in a week, days after this Wednesday will also be a Wednesday 14 days after a Wednesday will also be a Wednesday, and so will 21 days after, 28 days after, and so forth with all multiples of • We are looking for the day of the week of 365 days after this Wednesday • When 365 is divided by 7, the remainder is Therefore, 365 is one more than a multiple of One day after Wednesday is Thursday 12 Gold • We must find how many days elapse from November 29, 1972, to February 17, 1973 There are 30 days in November and 31 days in December, so from November 29 to December 31, 32 days elapse • There are 31 days in January, so from December 31 to January 31, 31 days elapse • Lastly, from January 31 to February 17, 17 days elapse Therefore, 80 days elapse from November 29, 1972, to February 17, 1973 • 80/7 yields a remainder of 3, and therefore 80 is three more than a multiple of It follows that February 17, 1973 is three days past a Wednesday, so it is a Saturday • From February 17, 1973, to February 17, 1974, 365 days elapse, as there are 365 days in a year 365 is more than a multiple of Therefore, February 17, 1974 is one day after a Saturday, which is a Sunday • In 1974, there are 28 days in February, so from February 17 to February 28, 11 days elapse From February 28 to March 30, 03-chapter-03 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Modular Arithmetic 251 30 days elapse This is a total of 41 days that elapse between February 17 and March 30 of 1974 • 41 is more than a multiple of 7, so March 30 is days past a Sunday, which is a Saturday 13 Gold • Taking the value of a positive integer in mod 100 will yield the last two digits of the integer 10310 mod 100 can be simplified to 310 mod 100, since 103 is congruent to in mod 100 • The first four powers of are 3, 9, 27, and 81 35 , or 81 × 3, has 43 as its last two digits (we not have to worry about the hundreds digit since we are in mod 100) • The last two digits of 36 are the last two digits of 43 × 3, which are 29 Going through with this process until we hit 310 , we find that its last two digits are 49 Therefore, the tens digit of 310 is 14 Silver • What is 33 mod 15 in simplest form? It is 3, as the remainder when 33 is divided by 15 is Therefore, x is not congruent to 33 in mod 15 • What is 1589 mod 15 in simplest form? Dividing 1589 by 15 yields a remainder of 14, so 1589 is congruent to 14 in mod 15 Therefore, x is not congruent to 1589 in mod 15 • How we tell whether or not x is congruent to mod 5? We know that x is greater than a multiple of 15 We are trying to determine whether or not x is two greater than a multiple of • Every multiple of 15 is also a multiple of 5 past a multiple of 15 is also a multiple of This multiple of is less than past a multiple of 15, so if x is congruent to in mod 15, x is congruent to in mod 15 Platinum • Any base number with digits A, B, C, D, E etc equals 80 × A + 81 × B + 82 × C + 83 × D + 84 × E February 6, 2018 13:36 PSP Book - 9in x 6in 252 Number Theory • All powers of above 82 are divisible by 83 In mod 83 , the number will simplify to 80 × A + 81 × B + 82 × C These are its last three digits • = 43/2 , so 83 = 49/2 It follows that n = Part 5: Divisibility Tricks Silver • The square root of 161 is somewhere between 12 and 13 Therefore, we only have to test divisibility by the primes up to 12 to determine whether 161 is prime 161 is odd, so it is not divisible by • + + = 8, so it is not divisible by 161 does not end in a or a 0, so it is not divisible by However, 161/7 does turn out to be a whole number Therefore, 161 is composite Silver • What is an approximation of the square root of 1001? 302 is easily identified as 900, but we still have to go a little bit further • 312 = 961, and 322 = 1024, so the square root of 1001 is somewhere in between 31 and 32 To determine whether 1001 is prime or composite, we must test divisibility by all prime numbers up to 31 • 1001 is not divisible by 2, 3, or 5, but 1001/7 yields a result of 143 It follows that 1001 is composite Gold • What is an approximation of the square root of 2011? 442 = 1936, and 452 = 2025, so the square root of 2011 is somewhere between 44 and 45 Therefore, we must test divisibility by all prime numbers less than 44 to determine whether 2011 is prime • This leaves us with 14 tests to complete In the end, we find that 2011 is prime 03-chapter-03 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 Divisibility Tricks Bronze • For a number to be divisible by 4, its last two digits have to be divisible by 09 is not divisible by 4, so 1009 is not divisible by Bronze • For a number to be divisible by 5, it has to end in or in a It follows that 11,301 is not divisible by Silver • For a number to be divisible by 6, it must be divisible by both and • 97, 884 is even, so it is divisible by • For a number to be divisible by 3, the sum of its digits has to be divisible by + + + + = 36, and 36 is divisible by 3, so 97,884 is divisible by • Therefore, 97,884 is divisible by Silver • We must split 99 into two or more relatively prime factors and test the given number for divisibility by each of them The best way to this is to split 99 into ì 11 If this number is divisible by both and 11, then it is divisible by 99 For a number to be divisible by 9, the sum of its digits has to be divisible by The sum of the digits of the number is 80, so it is not divisible by 99 Silver • Let us create a divisibility rule for 15 Splitting 15 into × 5, we find that for a number to be divisible by 15, it has to be divisible by and • This number ends in 0, so it must be divisible by The sum of this number’s digits is 43 43 is not divisible by 3, so this number is not divisible by 15 253 February 6, 2018 13:36 PSP Book - 9in x 6in 03-chapter-03 January 25, 2018 10:35 PSP Book - 9in x 6in Index algebra 1, 61, 65, 67, 112 bases 15, 229–230, 243, 245 basic counting 155–156, 185 basic number theory 219, 221, 223, 237 coefficients 1, 11, 22, 52, 109, 126 combinations 23, 156, 169, 171–173, 175, 201–202, 206–207, 209 counting factors 224–225, 227–228, 239, 241 counting numbers 31–32, 104 counting principle 155–156 cross multiplication 6–8, 82–84, 123 dependent events 162–165, 167, 192–193, 195, 197, 199 distance 36–38, 43–44, 110–113, 180, 213 divisibility tricks 235–236, 252–253 elimination 10–11, 53, 85–86, 129, 178 equivalent expressions 53–54, 63, 129 exclamation mark 160 expected value 176–178, 211 exponent 13–20, 50, 89, 91–93, 224–225, 241, 249 factoring 54–58, 60, 65, 130–131, 133, 142, 216, 220–221 factorizations 56, 59–60, 131–135, 220–228, 235, 237–243 fractions 6–7, 16, 21, 23, 26, 30–31, 45, 68, 83–84, 104, 109, 226, 232 GCD, see greatest common divisor GCF, see greatest common factor generality 180–182, 214–217 greatest common divisor (GCD) 225 greatest common factor (GCF) 54, 225–227 inequalities 28–30, 100–103 infinite series 66–67, 69, 144–145, 196 inverse operations LCM, see least common multiple least common multiple (LCM) 225, 227, 228, 242 04-chapter-Ind January 25, 2018 10:35 PSP Book - 9in x 6in 256 Index linear equations 1, 3, 5, 13, 76–77, 79, 81 mean 74–76, 147–149, 171, 206 median 74–76, 147–149 mixtures 26, 48–49, 123 mode 74–76, 147, 149, 171, 205–206 modular arithmetic 231–234, 247, 249, 251 multiple events 157–159, 187 ordered pair 12 orderings 160–161, 172, 189, 191 organized counting 169, 170, 202–203, 205 outcomes 151–153, 155, 157–158, 162–164, 169, 183–184, 186, 188, 194–195, 198, 202–203, 209, 211, 215 percentages 45–49, 58, 120–121 permutations 171–173, 175, 206–207, 209 polynomial expansions 50–51, 124–125, 127 probability 151–170, 174–179, 182–189, 192–200, 203, 207–212, 214 proportions 23–27, 42–43, 95–97, 99 quadratic formula 60–61, 63–64, 137–141, 144 radical expressions 21–22, 59, 94 range 74–76, 147–150, 199, 229 rate 36–41, 43, 48, 110–115, 117–118 roots 13, 15–20, 22, 30, 58, 61, 64, 89–94, 102–103, 134, 138, 141, 143 sequences 32–35, 105–107, 109, 172, 208–209 series 32–35, 66–69, 105–107, 109, 144–145, 196, 248 sets 9, 56, 70–76, 146–149, 155, 161, 168–173, 175, 178, 181–182, 200–210, 215–217, 226, 228, 243 subsets 168–170, 173, 175, 200–202, 208, 210, 222, 224, 228, 237, 242–243 substitution 9–10, 53, 84–85, 87–88, 140 systems of equations 8–12, 65, 84–85, 87 time 36–40, 42–44, 46, 110–118, 153, 155, 166, 179–180, 185, 213–214 unit conversion 42–44, 119 variables 1–3, 7–9, 11–13, 15, 24–25, 28, 42–44, 50, 52, 54, 56–57, 65, 70, 77, 80, 83–88, 90, 93, 100, 120, 127, 132–133, 168 weighted average 178–180, 212–213 04-chapter-Ind .. .Competitive Math for Middle School Pan Stanford Series on Renewable Energy — Volume Competitive Math for Middle School Algebra, Probability, and Number... editorial@panstanford.com Web: www.panstanford.com British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Competitive Math for Middle School: ... multiplication, etc Example 1: 2x + = Solve for x • To isolate the variable x, we must eliminate the +3 and the coefficient Competitive Math for Middle School: Algebra, Probability, and Number Theory