CHAPTER 1 Introduction to Problem Solving 1.1. The Problem-Solving Process and strategies An exercise is where a routine procedure is applied to arrive at an answer. A problem requires reflection and perhaps some original step. George Polya s 4-step process for solving problems: (1) Understand the problem (2) Devise a plan (3) Carry out the plan (4) Look back In devising a plan, we often use a strategy an artful means to an end. Mathematical Modeling: 1
2 1. INTRODUCTION TO PROBLEM SOLVING Problem Solving Strategies Strategy 1 Guess and test Problem (Page 20 #14). Find digits A, B, C, and D that solve the following cryptarithm. ABCD DCBA (1) Understand the problem could two digits be the same? (2) Devise a plan could use random guess and test or systematic guess and test, but both are ine cient. We will use an inferential guess and test. (3) Carry out the plan: Since the answer has 4 digits, A = 1 or A = 2. Since A = 4 D, A is even, so A = 2. 2BCD DCB2 Since D is the left digit of the answer, D = 8 or D = 9. Since 8 = 32 and 9 = 36, D = 8. 2BC8 8CB2
1.1. THE PROBLEM-SOLVING PROCESS AND STRATEGIES 3 2BC8 8CB2 Since 4 B+ a carry = C, B = 0 or B = 1 or B = 2. Since 4 C is even, 4 C + 3 is odd, so B = 1. 21C8 8C12 Since 4 C has 8 as its units digit, C = 2 or C = 7. Since 2128 2512 and 2178 8712, C = 7.
4 1. INTRODUCTION TO PROBLEM SOLVING (4) Look back. A = 2, B = 1, C = 7, D = 8 is the only correct answer. We had to use basic number properties. Are there other starting points? A is even. The largest possible carry is 3. Example. If it takes 20 minutes to cut a log into 5 pieces, how long would it take to cut the log into 10 pieces? Did you get 40 minutes as an answer? If so, you are wrong. Strategy 2 Draw a picture. (1) Understand the problem. Does each cut take the same amount of time? We need to assume this to solve the problem. Do all pieces need to be the same length? No need to assume this. (2) Devise a plan. First cut gives 2 pieces. Each additional cut gives 1 more piece. (3) Carry out the plan. Five pieces requires 4 cuts. 20 4 = 5 minutes per cut.
1.1. THE PROBLEM-SOLVING PROCESS AND STRATEGIES 5 Ten pieces requires 5 more cuts, or 9 in all. (4) Look back. This seems to be the only solution. 9 cuts 5minutes/cut = 45 minutes. A second approach: 1 cut gives 2 pieces, and each of these is cut into 5 pieces In general, n pieces requires n 1 + 4 + 4 = 9 cuts. 1 cuts. Problem (Page 20 # 13). 3 for 2 shoe sale. Pat wants a $56 pair and a $39 pair, Chris a $45 pair. Neither wants 3 pair. What s the fair price for each to pay of the $101 bill? (1) Understand the problem? What does fair mean? Fair means that each pays the same per cent or fraction of their bill if the pairs were not joined to make 3: $95 for Pat and $45 for Chris. (2) Devise a plan. Strategy 3 Use a variable. Let x = the fraction of the overall original price paid due to the sale. The total paid is $101. Use algebra. (3) Carry out the plan. Pat s part + Chris poart = 101 95x + 45x = 101 140x = 101 x = 101 140
6 1. INTRODUCTION TO PROBLEM SOLVING 101 Pat s part = 95 = $68.54 and 140 101 Chris part = 45 = $32.46. 140 (4) Look back. Could this be done without algebra? 1.2. Three Additional Strategies Problem (Page 35 #10). The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987,... where, after the first two terms, each term is the sum of the two previous terms. Strategy 4 Look for a pattern. Observe: 1 + 1 = 3 1 1 + 1 + 2 = 5 1 1 + 1 + 2 + 3 = 8 1 1 + 1 + 2 + 3 + 5 = 13 1 Predict the sum of the first 12 numbers of the Fibonacci sequence without adding. (1) Understand the problem. We are to predict a sum based on the above information. (2) Devise a plan. We see a pattern: To find the sum of the terms to a given point in the sequence, go two terms beyond and subtract 1 (if F n is the nth term of the sequence, F 1 + F 2 + F n = F n+2 1. (3) Carry out the plan. 1 + 1 + 2 + 3 + 5 + + 144 = 377 1 = 376
(4) Look back. The prediction is correct. Why does this work? Example. But F 2 = 1, so 1.2. THREE ADDITIONAL STRATEGIES 7 F 6 = F 4 + F 5 = F 4 + F 3 + F 4 = F 4 + F 3 + F 2 + F 3 = F 4 + F 3 + F 2 + F 1 + F 2 F 6 1 = F 4 + F 3 + +F 2 + F 1. Problem (Page 35 #8). How many cubes are in the 10th collection of cubes in this sequence? Strategy 5 Make a list (often combined with Look for a Pattern ). (1)Understand the problem. We assume there are hidden cubes. Our current sequence is (2) Devise a plan. 1, 4, 10 We focus on the columns. For instance, collection 3 has 3 columns one block high, 2 columns 2 blocks high, and 1 column 3 blocks high, suggesting a pattern. We form a list.
8 1. INTRODUCTION TO PROBLEM SOLVING (3) Carry out the plan. Number of Blocks Collection Column Height 1 2 3 4 5 6 7 8 9 10 1 block 1 2 3 4 5 6 7 8 9 10 2 blocks 2 4 6 8 10 12 14 16 18 3 blocks 3 6 9 12 15 18 21 24 4 blocks 4 8 12 16 20 24 28 5 blocks 5 10 15 20 25 30 6 blocks 6 12 18 24 30 7 blocks 7 14 21 28 8 blocks 8 16 24 9 blocks 9 18 10 blocks 10 Total Blocks in Collection 1 4 10 20 35 56 84 120 165 220 (4)Look back. The column total is the number of blocks in each collection. The 10th collection has 220 blocks. Do you see any patterns in the listing? Each row and each diagonal is an arithmetic progression and each column is symmetric. How many blocks would be in collection 11? List the column increments: 3, 6, 10, 15, 21, 28, 36, 45, 55 List these increments: 3, 4, 5, 6, 7, 8, 9, 10 Then column 11 would increase by 11 more blocks than column 10 did, or 66 blocks. Collection 11 would have 220 + 66 = 286 blocks.
1.2. THREE ADDITIONAL STRATEGIES 9 Example. There are 22 people at a party. Each person shakes hands with all the others exactly once. How many handshakes are there? (1) Understanding the problem is pretty clear. (2) Devise a plan. Strategy 6 Solve a simpler problem. Then generalize to the original (often used in conjuction with Look for a pattern ). Suppose there are just 5 people; or 6 people. (3) The first person shakes hands with 4 people, the second with 3 others besides, the third with 2 others besides, and the fourth with 1 other besides. These are 4 + 3 + 2 + 1 = 10 handshakes. For 6 people: 5 + 4 + 3 + 2 + 1 = 15 handshakes. 10 and 15 are triangular numbers - the sum of consecutive integers beginning with 1. Notice: 1 + 2 + 3 {z } +4 {z } Each pair averages 5 2 and 4 5 2 = 10. 1 + 2 + {z} 3 {z +4 } +5 {z } Each pair averages 3 and 5 3 = 15.
10 1. INTRODUCTION TO PROBLEM SOLVING 1 + 2 + 3 + + (n 1) + n = n n + 1 n(n + 1) =. 2 2 So, for 22 people, we add 21 22 1 + 2 + 3 + + 21 + 22 = = 231 2 (4) Look back. n(n + 1) For n people, there are handshakes. 2 From Finite Math, C 22,2 = 231 also gives the answer.