Teacher s Notes Problem of the Month: Courtney s Collection Overview: In the Problem of the Month, Courtney s Collection, students use number theory, number operations, organized lists and counting methods to solve problems. The mathematical topics that underlie this POM are knowledge of number sense, number properties, comparison subtraction, division, factors and divisibility, counting principles, systematic charting and generalizations. The mathematics that includes counting principles and systemic charting is often referred to as discrete mathematics. In the first level of the POM, students are presented with a problem that involves summing three values. Their task involves a situation where they are to select three coins from a large set of pennies, nickels and dimes. The students are asked to select three coins and determine how much money they have altogether. In level B, students start to examine how many different ways they can select three coins and the sum of each set of three coins. In level C, students consider a collection of 5 cents, 6 cents and 7 cents stamps. They are asked to determine if they had an unlimited number of each stamp, which postage amounts they could make and which postage amounts they can t make. In level D, the students examine three different sets of stamps. They are asked to determine whether the sets produce a finite set of impossible values or an infinite set of impossible values. For the stamps with finite amount of impossible values, the students are asked to find the largest impossible value. In level E, students are asked to generalize their findings from level D. Students are asked to predict and justify whether a set of stamps (cardinality 3) has a finite set of impossible values or an infinite set of impossible values. For the stamps with finite amount of impossible values, the students are asked to describe a process for finding the largest impossible value. Mathematical Concepts: Number theory is one of the oldest branches of mathematics. It involves examining and generalizing numbers and their operations. Exploring the real number system and its subsets along with mathematical operations is central to number theory. Justification and proof play an important role in solving problems and developing theorems about the number system. These generalizations and theorems are often represented through algebraic notation and involve algebraic properties and axioms. Another important aspects of number theory that has ties to elementary and middle school math is the composition of numbers including factors, multiples, primes, composite, odds and evens and other classification or sets of numbers. Discrete mathematics is the study of the sets of and operations on discrete objects. Discrete means distinct from others, separate or discontinuous. Basically discrete is not continuous mathematics. Traditionally, K-12 mathematics starts with discrete topics in POM Teacher s Notes Courtney s Collection Page 1
lower grades, but by secondary mathematics the topics usually focus on continuous mathematics as a preparation for calculus. The real number line is an example of a continuous set of points. But, discrete mathematics plays an important role in higher mathematics courses. Much of the mathematics that come from real life come in the form of distinct objects and values. Making sense of this distinct information and quantities is important, thus discrete mathematics has numerous applications to real life. A more formal definition of discrete mathematics may be stated as the study of sets of distinct data. Tasks usually performed while doing discrete mathematics involve categorizing, counting, classifying, organizing and optimizing sets of information. Discrete mathematics can be put into three broad categories: existence problems, counting problems, and optimization problems. Questions involving typical discrete mathematics include: Is there a way to.? How many ways.? Out of all of these possibilities, which is the best? How can this information be organized to best? Discrete mathematics has been referred to as optimization analysis. The branch of mathematics known as discrete mathematics is comprised of a number of mathematical topics. Although opinions vary on what topics are discrete mathematics, the following topics are usually listed under the discrete mathematics umbrella: Combinatorics Graph theory Game theory Boolean algebra Linear algebra Theory of algorithms Matrices Recurrence relations Set theory Some of these topics have been traditionally taught as part of other mathematical topics. For example, counting principles from combinatorics is usually taught at the start of a probability chapter. Combinatorics Counting principles is a major topic in combinatorics. There are two basic counting functions, permutation and combination. A permutation is the number of ways one can order r items from a set of n objects. For example, how many different batting orders (1 9) can you makes from a team of 12 players? The order matters because the permutation specifies the order as well as the size of the set. Therefore there are 12 ways to pick the first player, times 11 ways to pick the POM Teacher s Notes Courtney s Collection Page 2
second player, times 10 ways to pick the third play and so forth down to 4 ways to pick the ninth and final player. Therefore the permutation of 9 players in order out of 12 is 12 11 10 9 8 7 6 5 4 = 79,833,600 ways. A combination is a selection of r items without order from a set of n objects. An example in a similar context is how can you choose 9 players to play from a group of 12 players. The answer is 220 unique teams, but position or order does not matter; it is just the different make up of the team. Factorials Factorial is an operation that is indicated by a natural number followed by the exclamation point (!). A factorial is the product that results from multiplying a natural number by each of its subsequent natural numbers down to one. For example, 5! is 5 4 3 2 1 or 120. Factorials are useful in calculating permutations and combinations. As shown in the example of a permutation, the permutation expression of 12 11 10 9 8 7 6 5 4 could have been written in factorial notations such as 12!/3! Tree Diagrams A tree diagram is a type of organizational chart for displaying data. The tree begins with a node that contains some information. The node is linked to other nodes containing information by branches usually symbolized by line segments. Each node may have several links that branch off to other nodes. The tree diagram often conveys a top down data structure. A company s employees reporting structure is often described in a tree diagram. Many counting problems can be illustrated using a tree diagram. For example, if one were asked to illustrate the number of ice cream sundaes that can be made from three flavors of ice cream, two toppings and whether one wanted nuts or no nuts, the following diagram could be constructed. Sundaes Chocolate Vanilla Strawberry Caramel Fudge Caramel Fudge Caramel Fudge Nuts None Nuts None Nuts None Nuts None Nuts None Nuts None If one follows the branches from the top to the bottom, every type of possible sundae is listed. The total number of possible sundaes is the number of nodes on the bottom line of the tree diagram. In this case there are 12 possible sundaes. POM Teacher s Notes Courtney s Collection Page 3
Systematic Lists A systematic list is another data structure. A list is made following some process. A simple systematic list might be a list of names in alphabetical order. When data has several sets of attributes, a list might be made in various ways. One method is to pick a single attribute and arrange the list using that attribute first. Then a second attribute is considered and the list is continued, and so forth. For example, suppose you were asked how many ways you can have a 50 in change using quarters, dimes and nickels. One method would be to use a systematic list. Quarters Dimes Nickels 2 0 0 1 2 1 1 1 3 1 0 5 0 5 0 0 4 2 0 3 4 0 2 6 0 1 8 0 0 10 Systematic lists are powerful tools in helping to answer the questions of have we found every way. It also requires the important reasoning skills of classifying attributes. Factors and Multiples Factors and multiple are considered in the set of natural numbers. All natural numbers have at least two natural number factors. The number itself and one. The number may also have other factors. Those are numbers when multiplied together produce a product equal to the original number. A factor of a number is said to be a multiple of the factor. A factor can be paired with another factor to produce a product equal to its multiple. A perfect square will have one multiple that factor pair is the factor itself. For example 36 has a factor 6. The factor pair of 6 for the multiple 36 is also 6. 36 has other factors such as 9 that has a factor pair 4. Primes and Composites A is a natural number whose only factors are 1 and itself. A composite is a number who has more than one set of factor pairs. All natural numbers are either primes or composites but not both. A prime factor is a factor that is also a prime number. One is not consider a prime, since it has an infinite of factor pairs that define a prime. Composites can be decomposed into a product of prime factors. POM Teacher s Notes Courtney s Collection Page 4
Solutions: Level A: Courtney has a bank of pennies, nickels and dimes. Courtney pulled out three coins from the bank. Name the coins she picked. How much money does she have? Show how you figured it out. Student will show various examples such as: 1 + 5 + 10 = 16 1 + 1 + 1 = 3 Show a different way that Courtney could pick three coins. How much does Courtney have now? Show how you figured it out. There are ten different combinations 1 + 1 + 1 = 3 1 + 1 + 5 = 7 1 + 1 + 10 = 12 1 + 5 + 5 = 11 1 + 5 + 10 = 16 5 + 5 + 5 = 15 5 + 5 + 10 = 20 10 + 10 + 1 = 21 5 + 10 + 10 = 25 10 + 10 + 10 = 30 POM Teacher s Notes Courtney s Collection Page 5
Level B: How many different ways can Courtney pick three coins out of her bank? Show all the ways you can find. Ways First Coin Second Coin Third Coin Total 1 Penny Penny Penny 3 2 Penny Penny Nickel 7 3 Penny Penny Dime 12 4 Penny Nickel Nickel 11 5 Penny Nickel Dime 16 6 Penny Dime Dime 21 7 Nickel Nickel Nickel 15 8 Nickel Nickel Dime 20 9 Nickel Dime Dime 25 10 Dime Dime Dime 30 What are the different amounts of money Courtney would have by picking any three coins? 3 7 12 11 16 20 21 15 25 30 How do you know you found all the possible ways? Explain your answer. I made a systematic list and exhausted all the possibilities. Another way to reason about the problem is to break the problem into cases and consider the different sets of coins. There are three cases. Case 1: Each coin is the same type. There are 3 sets: {Penny, Penny, Penny} {Nickel, Nickel, Nickel} {Dime, Dime, Dime} Case 2: Each coin is a different type: There is only one set and of course order does not matter, because the sum is always the same. {Penny, Nickel, Dime} Case 3: Two coins are the same and one is different. For each single coin type (3), there are two other choices of coin type for a total of 6 possible sets. {Penny, Nickel, Nickel} {Penny, Dime, Dime} {Nickel, Penny, Penny} {Nickel, Dime, Dime} {Dime, Penny, Penny} {Dime, Nickel, Nickel} So adding the sets in all three cases gives you a total of 10 sets (or combinations of coins). POM Teacher s Notes Courtney s Collection Page 6
Level C: Courtney visited her grandmother. Her grandmother used to collect stamps. She had a shoebox full of 5 cents, 6 cents and 7 cents stamps. Courtney thought, I could mail a lot of different size letters and postcards with these stamps. She tried to figure out the different amount of postage she could make with a combination of those stamps. What totals could she make and what totals are impossible? Explain how you found your solution. How do you know your solution is correct? The postage values that can t be made are 1, 2, 3, 4, 8, 9. Starting with 10, all postage values can be made from combinations of 5, 6, and 7 stamps. A method to justify the solution is to examine the first five sum values generated by adding two of the stamp types: 10 = 5 + 5, 11 = 5 + 6, 12 = 5 + 7, 13 = 6 + 7, 14 = 7 + 7 There is a pattern where each consecutive sum increases by 1. Considering the entire set of integers, the set of values that are divisible by 5 can be generated by summing an n number of 5 stamps. The values in this set, if arranged in order, have a consistent difference of five. All values that lie between the two consecutive values divisible by five following the pattern of: 5n, 5n+6, 5n+7, 5(n-1) + 6 +7, 5(n-1) + 7 + 7, 5(n+1), etc. POM Teacher s Notes Courtney s Collection Page 7
Level D: Courtney s grandmother said, Your grandpa has different shoe boxes with other stamps. His shoebox has just 4 cents, 6 cents and 8 cents stamps. Which totals can you make with these stamps? Since all three of the stamps are even, their sums must always be even. The fact that 4 will sum to every other even number and you have two consecutive even numbers, you can find all other even numbers greater than the two consecutive numbers. So you can generate every even number except 2, Or If x is the number of 4 stamps, y the number of 6 stamps and z the number of 8 stamps, then for any postage P that can be made, it would have to satisfy the following equation 4x + 6y + 8z = P. Therefore, 2 (2x+3y+4z) = P which shows that P must always be an even number and the smallest possible non-zero postage is 4 Courtney said, I wonder why there is a difference between Grandma s and Grandpa s shoeboxes. I like finding the largest total I can t make, but Grandpa s box has an unlimited amount of totals I can t make. I wonder why it works for some and not for others? She continues, I am going to investigate which three stamp amounts are like the Grandma s shoebox and which three stamp amounts are like the Grandpa s shoebox. I am going to compare three different sets. I will try: 6 cents, 7 cents and 8 cents. Then I will try: 6 cents, 9 cents and 12 cents. Finally I will try: 6 cents, 8 cents and 9 cents. Explain how the three different sets are like either Grandma s shoebox or Grandpa s shoebox. Two of the sets of stamps has a finite sets of postage with a largest impossible value. Those two sets are 6, 7, 8 and 6, 8, 9. The third set, 6, 9, 12, has a common factor 3. Since three is a common factor, every sum made by those stamps or their multiples also has a common factor of 3. That means every sum that is created is a multiple of three, leaving non-multiple of three as amounts that can t be made. Therefore this set has an infinite amount of impossible values. The amounts 6, 7, 8 are relatively prime between all three amounts. Therefore there is a finite set of amounts that cannot be made. All other postage values can be made. A method I used to find the greatest impossible value is to construct tables comparing one stamp to a second stamp and the amount of postage that can be made by the combinations of the two stamps. I made a table for 6 and 8 stamps, a second table for 6 and 7 stamps and a third table for 8 and 7 stamps. Using those three tables, I made a list of all the impossible postage amounts. This list was as follows: 1,2,3,4,5,9,10, 11,13,17. At this point, 17 appears to be the largest impossible number, but I had not considered if I used all three stamps. I started with 17 and tested it by repeatedly subtracting 6 from 17 POM Teacher s Notes Courtney s Collection Page 8
to see if any number in the 8 and 7 table appear. I found that I could not make a combination with all three stamps, therefore 17 is the largest impossible number 7 0 1 2 3 4 5 6 7 6 0 0 7 14 21 28 35 42 49 1 6 13 20 27 34 41 48 55 2 12 19 26 33 40 47 54 61 3 18 25 32 39 46 53 60 67 4 24 31 38 45 52 59 66 73 5 30 37 44 51 58 65 72 79 6 36 43 50 57 64 71 78 85 7 42 49 56 63 70 77 84 91 8 48 55 62 69 76 83 90 97 9 54 61 68 75 82 89 96 103 10 60 67 74 81 88 95 102 109 11 66 73 80 87 94 101 108 115 8 0 1 2 3 4 5 6 7 6 0 0 8 16 24 32 40 48 56 1 6 14 22 30 38 46 54 62 2 12 20 28 36 44 52 60 68 3 18 26 34 42 50 58 66 74 4 24 32 40 48 56 64 72 80 5 30 38 46 54 62 70 78 86 6 36 44 52 60 68 76 84 92 7 42 50 58 66 74 82 90 98 8 48 56 64 72 80 88 96 104 9 54 62 70 78 86 94 102 110 10 60 68 76 84 92 100 108 116 11 66 74 82 90 98 106 114 122 7 0 1 2 3 4 5 6 7 8 0 0 7 14 21 28 35 42 49 1 8 15 22 29 36 43 50 57 2 16 23 30 37 44 51 58 65 3 24 31 38 45 52 59 66 73 4 32 39 46 53 60 67 74 81 5 40 47 54 61 68 75 82 89 6 48 55 62 69 76 83 90 97 7 56 63 70 77 84 91 98 105 8 64 71 78 85 92 99 106 113 9 72 79 86 93 100 107 114 121 10 80 87 94 101 108 115 122 129 11 88 95 102 109 116 123 130 137 POM Teacher s Notes Courtney s Collection Page 9
The 6, 9, 12 has an infinite number of postage that can t be made. This is true because they are not all three relatively prime. They all have 3 as a common factor, therefore when you add any combination of stamps, the value will always be divisible by three. Since there are many stamps that are not divisible by three, those amounts cannot be made. The 6, 8, 9 has a finite set of postage that cannot be made. The amounts of 1,2,3,4,5,7, 10, 11, 13 and 19 cannot be made. All other postage values can be made. A method I used to find the greatest impossible value was to construct tables comparing one stamp to a second stamp and the amount of postage that can be made by the combinations of the two stamps. I made a table for 6 and 8 stamps, a second table for 6 and 9 stamps and a third table for 8 and 9 stamps. Using those three tables I made a list of all the impossible postage amounts. This list was as follows: 1,2,3,4,5, 7, 10, 11,13,19, 23,29,31, 37,47, 55. At this point, 55 appears to be the largest impossible number, but I had not considered if I used all three stamps. I started with 55 and tested it by repeatedly subtracting 6 from 55 to see if any number in the 8 and 9 table appear. When it did, I cancelled it from the list. As a result 55, 47, 37, 31, 29 and 23 could all be made using the three stamps in some configuration. When I got to 19, I found that I could not make a combination with all three stamps, therefore 19 is the largest impossible number. 9 0 1 2 3 4 5 6 7 8 0 0 9 18 27 36 45 54 63 1 8 17 26 35 44 53 62 71 2 16 25 34 43 52 61 70 79 3 24 33 42 51 60 69 78 87 4 32 41 50 59 68 77 86 95 5 40 49 58 67 76 85 94 103 6 48 57 66 75 84 93 102 111 7 56 65 74 83 92 101 110 119 8 64 73 82 91 100 109 118 127 9 72 81 90 99 108 117 126 135 10 80 89 98 107 116 125 134 143 11 88 97 106 115 124 133 142 151 6 0 1 2 3 4 5 6 7 8 0 0 6 12 18 24 30 36 42 1 8 14 20 26 32 38 44 50 2 16 22 28 34 40 46 52 58 3 24 30 36 42 48 54 60 66 4 32 38 44 50 56 62 68 74 5 40 46 52 58 64 70 76 82 6 48 54 60 66 72 78 84 90 7 56 62 68 74 80 86 92 98 8 64 70 76 82 88 94 100 106 POM Teacher s Notes Courtney s Collection Page 10
9 72 78 84 90 96 102 108 114 10 80 86 92 98 104 110 116 122 11 88 94 100 106 112 118 124 130 12 96 102 108 114 120 126 132 138 6 0 1 2 3 4 5 6 7 9 0 0 6 12 18 24 30 36 42 1 9 15 21 27 33 39 45 51 2 18 24 30 36 42 48 54 60 3 27 33 39 45 51 57 63 69 4 36 42 48 54 60 66 72 78 5 45 51 57 63 69 75 81 87 6 54 60 66 72 78 84 90 96 7 63 69 75 81 87 93 99 105 8 72 78 84 90 96 102 108 114 9 81 87 93 99 105 111 117 123 10 90 96 102 108 114 120 126 132 11 99 105 111 117 123 129 135 141 POM Teacher s Notes Courtney s Collection Page 11
Level E: Determine a method for predicting whether a given set of any three stamps would have a largest impossible total or there isn t a largest impossible total, rather there are infinite impossible totals. Justify your method using mathematical reasoning. A set of three stamps will have an infinite set of impossible values if the three are not relatively prime with each other. In other words, there is a common factor other than 1 that is divisible into all three values. Since that common factor exists, every sum made by those stamps or their multiples also has that common factor. That means every sum that is created is a multiple of that common factor, leaving non-multiple of the common factor as amounts that can t be made. Therefore this set has an infinite amount of impossible values. Algebraically one can explain it as follows: Suppose you have three numbers p, q, r that are not relatively prime. Therefore they have a common factor k, such that: p=kx, q= ky, and r = kz; for any postage amount n there is an a, b, c such that: n = ap + bq + cr substituting the common factor equivalence for p, q, r one gets n = akx + bky + ckz n = k( ax+by+cz) Therefore n is always a multiple of k, so non-multiple of k cannot be made by adding the stamps. For those sets of three stamps that have a largest impossible total, make a conjecture about how to find that total. There is a process to follow to find the greatest impossible value of three stamps. This can only be found if all three stamps are relatively prime, otherwise there is no greatest impossible factors. If the two stamps are relatively prime, then the largest impossible value would be the product of the two stamps amounts minus the sum of those two stamps. This conjecture is often determined through experimentation, but its proof requires a higher level of mathematical sophistication. So we will base this process on an assumption that this conjecture can be proved. POM Teacher s Notes Courtney s Collection Page 12
Since there are three stamps (a, b, c ), there are three tables that need to be constructed (a and b, a and c, b and c ). For each table with relatively prime factors, find the largest impossible value p, (p = ab-(a+b)). If a table does not have relatively prime factor, then p doesn't exist, so move to examine the other tables. When you find a table with a largest impossible value, you will still need to consider the case where all three stamps are factors in finding a postage value (postage value = ax+by+cz, where x,y,z are numbers of stamps). This process involves using the third stamps not one of the factors in the table (c if the table is a and b ). Using repeated subtraction, check all outcomes of p, where p = ab-(a+b) - cn and n is a natural number. If on any successive iteration p appears as a combination of postage in any table, then p is ruled out. This process continues with all values in decreasing order that do not appear in any of the three tables. When p is found not to a combination of the three factors, then it can be determined as the largest impossible value. POM Teacher s Notes Courtney s Collection Page 13