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</div></head><body onload="automathjax();init()"><p>PRINT PAGE</p><h6 id="page22">22</h6><h3 id="heading0">SECTION 3 MORE STRATEGIES FOR PROBLEM SOLVING</h3><p>PROBLEM</p><p>Fifteen people are in a room, and each shakes hands once and only once with everyone else. How many handshakes are there?</p><p>OVERVIEW</p><p>Below you will find Polya’s four-step description of the mathematical problem-solving process juxtaposed with the four steps that some psychologists use to describe problem solving in a more general context.</p><p>Psychologists</p><p>1. Preparation</p><p>2. Incubation</p><p>3. Inspiration</p><p>4. Verification</p><p>Polya</p><p>1. Understand the problem</p><p>2. Make a plan</p><p>3. Carry out the plan</p><p>4. Look back</p><p>Both lists suggest that successful problem solving will require significant investments of time and patience. Polya’s list tends to be more active. Polya’s encouragement to “make a plan” and “carry it out” describes what to do while waiting for “inspiration.”</p><p>CCSSM NOTES</p><p>The Common Core State Standards for Mathematics emphasize the importance of problem solving and critical thinking skills throughout all grades. In order to solve a problem, students must apply their knowledge to find the solution of the problem. In fact, “Mathematically proficient students start by explaining. the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures. and plan a solution pathway rather than simply jumping into a solution attempt” (CCSSM, p. 6).</p><p>Polya’s second major step in problem solving is “Make a plan.” The strategies discussed next, alone or in concert, provide the rough outline of plans that will solve a great many problems and use the thinking skills we have discussed in the previous two sections. The first four strategies were discussed in Section 1.</p><p>BEGIN TEXT BOX:</p><p>Strategies for Making a Plan</p><p>1. Make a chart or table.</p><p>2. Draw a picture or diagram.</p><p>3. Guess, test, and revise.</p><p>4. Form an algebraic model.</p><p>6. Try a simpler version of the problem.</p><p>5. Look for a pattern.</p><p>7. Work backward.</p><p>8. Restate the problem.</p><p>9. Eliminate impossible situations.</p><p>10. Use reasoning.</p><p>PRINT PAGE</p><h6 id="page23">23</h6><h4 id="heading1">LOOK FOR A PATTERN: TRY A SIMPLER VERSION</h4><p>One of the key techniques in problem solving is to look for a pattern. Further, if you examine a simpler version of the problem, you may find it is easier to detect a pattern if one exists. This dual strategy of looking for a pattern in a simple version of the problem provides a potent weapon in solving problems similar to the following.</p><p>EXAMPLE 1</p><p>How many people can be seated at ten square, 4-person tables that are lined up end to end?</p><p>SOLUTION</p><p>First look at simpler versions of the problem. As we proceed, we will confirm our thinking with pictures and summarize what we learn in a chart (Figure 1-9).</p><p>FIGURE 1-11</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here is figure 1-11, not figure 1-9. This figure is a table that contains the following information:</p><p>BEGIN TABLE:</p><p>Arrangement and Number of Tables: 1 table</p><p>Number of Seats: 2 on sides<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>+</mo><mn>2</mn></mrow></math></span>on ends<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo><mn>4</mn></mrow></math></span></p><p>Arrangement and Number of Tables: 2 tables</p><p>Number of Seats: <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>⋅</mi><mn>2</mn></mrow></math></span> on sides<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>+</mo><mn>2</mn></mrow></math></span>on ends<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo><mn>6</mn></mrow></math></span></p><p>Arrangement and Number of Tables: 3 tables</p><p>Number of Seats: <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>⋅</mi><mn>3</mn></mrow></math></span> on sides<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>+</mo><mn>2</mn></mrow></math></span>on ends<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo><mn>8</mn></mrow></math></span></p><p>Arrangement and Number of Tables: 4 tables</p><p>Number of Seats: <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>⋅</mi><mn>4</mn></mrow></math></span> on sides<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>+</mo><mn>2</mn></mrow></math></span>on ends<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo><mn>10</mn></mrow></math></span></p><p>Arrangement and Number of Tables: n tables</p><p>Number of Seats: <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>n</mi><mo>+</mo><mn>2</mn></mrow></math></span></p><p>END TABLE.</p><p>END PRODUCER’S NOTE.</p><p>Thus for 10 tables (<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>10</mn></mrow></math></span>),</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mfenced open="(" close=")"><mrow><mn>10</mn></mrow></mfenced><mo>+</mo><mn>2</mn><mo>=</mo><mn>22</mn></mrow></math></span></p><p>Twenty-two people can be seated.</p><p>EXAMPLE 2</p><p>Fifteen people are in a room, and each shakes hands once and only once with everyone else. How many handshakes occur?</p><p>SOLUTION</p><p>We consider simple versions of this problem to see if a pattern emerges.</p><p>Two people:</p><p>A (two way arrow) B (1 handshake)</p><p>Three people:</p><p>A (two way arrow) B</p><p>B (two way arrow) C</p><p>A (two way arrow) C</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>3</mn></mrow></math></span> handshakes</p><p>PRINT PAGE</p><h6 id="page24">24</h6><p>Four people:</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>B</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>C</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>C</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>3</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>6</mn></mrow></math></span> handshakes</p><p>Five people:</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>B</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>C</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>D</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>E</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>C</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>E</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>D</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>E</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><munderover><mo>↔</mo><mrow></mrow><mrow></mrow></munderover><mi>E</mi></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>4</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>10</mn></mrow></math></span> handshakes</p><p>From the pattern, we infer that 15 people shake hands this many times:</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>14</mn><mo>+</mo><mn>13</mn><mo>+</mo><mi>…</mi><mo>+</mo><mn>3</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>105</mn></mrow></math></span></p><p>(See Exercise 33 for some ideas on how to compute this sum.)</p><h4 id="heading2">WORK BACKWARD</h4><p>Some problems do not ask for a numerical answer, but rather ask how one can achieve a specified goal. In these circumstances, it is often useful to work backward from the desired goal.</p><p>EXAMPLE 3</p><p>How can you measure 1 liter of water from a faucet if you have only a 4-liter container and a 7-liter container?</p><p>SOLUTION</p><p>Our goal is to have 1 liter of water in either the 4-liter or 7-liter container. Let us imagine that we have achieved the goal (see Figure 1-12(a)) and then think about steps that could have preceded it.</p><p>FIGURE 1-12 (a)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown in this figure is a container with 1 L of water and to the right an empty much larger container.</p><p>END PRODUCER’S NOTE.</p><p>We would have 1 liter left in the 4-liter container if we had filled the 4-liter container and poured off exactly 3 liters. How could we manage to pour off exactly 3 liters? If the 7-liter container contained exactly 4 liters, it would be easy to accomplish this task. See Figure 1-12(b).</p><p>FIGURE 1-12 (b)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown in this figure is a full container with 4 L of water in it. There’s an arrow pointing from this container to the larger one beside that also has 4 L in it.</p><p>END PRODUCER’S NOTE.</p><p>Now, how could we get exactly 4 liters into the 7-liter container? Clearly, we could fill the 4-liter container and pour it into the 7-liter container (Figure 1-12(c)).</p><p>FIGURE 1-12 (c)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown in this figure is a full container with 4 L of water in it. There’s an arrow pointing from this container to the larger one beside that is empty.</p><p>END PRODUCER’S NOTE.</p><p>In summary, to obtain 1 liter, we (a) fill the 4-liter container, (b) pour it into the 7-liter container, (c) fill the 4-liter container, and (d) pour as much as possible into the 7-liter container. This sequence of steps leaves 1 liter in the 4-liter container.</p><p>As you have noticed, we do not always discuss all four of Polya’s steps on each problem, and you will probably not do so either as you write up your solutions. However, Polya’s steps should serve as a mental outline as you proceed through your problems. Further, as you begin a problem, it is always good practice to identify the unknowns on your paper and to summarize the available information, preferably in a picture or table. After the problem is solved, you should always interpret the result with an English sentence. Finally, mentally review the solution both to check results and to look for generalizations and other problems that your work suggests. As we discuss the next two examples, we will give you guidance by referring more explicitly to Polya’s steps.</p><p>PRINT PAGE</p><h6 id="page25">25</h6><h4 id="heading3">RESTATE THE PROBLEM</h4><p>It is always good practice to restate the problem in your own words, turning and testing the meanings to gain a better understanding. On some occasions, you will restate the problem so clearly that the solution to the problem becomes almost obvious.</p><p>EXAMPLE 4</p><p>To select the best two-person debate team in the United States, 117 teams are selected, with the understanding that when a team loses a debate, the team is dropped from the tournament. How many debates must be completed to select a winner?</p><p>SOLUTION</p><p>UNDERSTAND THE PROBLEM</p><p>What is unknown? How many debates will it take to determine the winner? What are the data? There are 117 teams, and it is a single-elimination tournament. In other words, if a team loses one debate, it is out of the tournament. If Team A competes against Team B and Team A wins, Team B exits from the tournament and Team A competes again.</p><p>DEVISE A PLAN</p><p>Let us try to reword this problem so that the answer is clearer.</p><p>CARRY OUT THE PLAN</p><p>The unknown is “How many debates until a champion is determined?” Rephrased:</p><p>How many teams must lose in order to choose a champion? 116</p><p>How many teams lose per debate? 1</p><p>How many debates until a champion is determined? 116</p><p>LOOK BACK</p><p>The championship team is determined after 116 debates. Can we verify this answer another way? Suppose we had used the twin strategies of looking for a pattern in a simpler case. Take a moment and draw diagrams that shows how many matches must be played to determine the winner when there are 3, 4, 5, and 6 players.</p><p>We observe that the required number of debates is 1 fewer than the number of teams competing. Indeed, 117 teams will require 116 debates to determine a winner.</p><h4 id="heading4">USE REASONING: ELIMINATE Impossible Situations</h4><p>Example 5 involves the combined strategies of elimination and reasoning, as well as the familiar strategy of making a table. The table furnishes us with an organized method for eliminating incorrect situations. Selecting the clues that lead to elimination is very important.</p><p>EXAMPLE 5</p><p>Five students-Leon, Sarah, Russo, Sue, and Sharon-participate in a debate tournament in which each team must have at least one affirmative debater and one negative debater. Two of the students attend Harp College, and three attend Sloan. Three are affirmative debaters, and two are negative. Leon and Sue attend the same college. Russo and Sharon attend different schools. Sarah and Russo are on the same side. Sue and Sharon are on opposite sides. A negative debater from Harp College was selected outstanding debater. Who was that person?</p><p>SOLUTION</p><p>UNDERSTAND THE PROBLEM</p><p>Two of the five students attend Harp, and three attend Sloan. We are given facts about Leon and Sue, Russo and Sharon, Sarah and Russo, and Sue and Sharon. We also know that a negative debater from Harp College was named outstanding debater.</p><p>PRINT PAGE</p><h6 id="page26">26</h6><p>We are to find out which of the five debaters received this honor.</p><p>DEVISE A PLAN</p><p>We will construct a table (Table 1-2) and reason from given facts to answer “yes” or “no” for each student for each of the categories affirmative, negative, Harp College, and Sloan College.</p><p>BEGIN TABLE:</p><p>TABLE 1-2</p><p>Category: Affirmative</p><p>Leon: No</p><p>Sarah: Yes</p><p>Russo: Yes</p><p>Sue: BLANK SPACE</p><p>Sharon: BLANK SPACE</p><p>Category: Negative</p><p>Leon: Yes</p><p>Sarah: No</p><p>Russo: No</p><p>Sue: BLANK SPACE</p><p>Sharon: BLANK SPACE</p><p>Category: Harp College</p><p>Leon: No</p><p>Sarah: Yes</p><p>Russo: BLANK SPACE</p><p>Sue: No</p><p>Sharon: BLANK SPACE</p><p>Category: Sloan College</p><p>Leon: Yes</p><p>Sarah: No</p><p>Russo: BLANK SPACE</p><p>Sue: Yes</p><p>Sharon: BLANK SPACE</p><p>END TABLE.</p><p>CARRY OUT THE PLAN</p><p>Leon and Sue attend the same college. Russo and Sharon attend different colleges. What does this tell you? (Leon and Sue, along with either Russo or Sharon attend Sloan.) Sarah must attend Harp. Why? Sarah and Russo are affirmative debaters along with either Sue or Sharon. Leon is a negative debater. Why? The other negative debater is either Sue or Sharon. Sue is not from Harp College. Sharon can be from Harp College. Therefore, the outstanding debater is Sharon.</p><p>If you do not yet feel like an expert problem solver, do not be overly worried. We will revisit these same ideas throughout the text and use many more strategies in the days ahead. What you should understand now is that no single strategy is a magic wand that you can use for all problems, but that all strategies provide ideas that you can use in devising your own plans to solve problems.</p><p>BEGIN TEXT BOX:</p><p>Just for Fun</p><p>IMAGE</p><p>What is the least number of pitches that a pitcher can throw in the course of pitching a full-length baseball game?</p><p>END TEXT BOX.</p><h4 id="heading5">EXERCISE SET 3</h4><p>R 1. If it takes 12 minutes to cut a log into 3 pieces, how long would it take to cut the log into 4 pieces? (HINT: Draw a picture.)</p><p>2. Charles and Gary earn the same amount of money, although one works 5 days more than the other. If Charles earns $56 a day and Gary earns $96 a day, how many days does each work? (HINT: Try guess, test, and revise on this one.)</p><p>3. Without adding all 30 numbers, find the sum of the first 30 odd counting numbers. (HINT: Look at simpler cases, and look for a pattern.)</p><p>4. Jack goes to the well to get some water. He has a 3-liter pail and a 5-liter pail, and he wishes to return with exactly 1 liter of water. How can he do this? (HINT: Work backward!)</p><p>5. Jack’s 5-liter pail in Exercise 4 springs a leak, and he must carry a 3-liter pail and a 7-liter pail to the well. How can he return with 1 liter of water in this case?</p><p>In each of Exercises 6 through 22, record the strategies that you try as you look for a solution, even if you are not successful.</p><p>PRINT PAGE</p><h6 id="page27">27</h6><p>6. Forty-four players enter a single-elimination tennis tournament on the Fourth of July. How many matches must be played to determine a winner?</p><p>7. Each member of the Wheeler family gives a gift at Christmas to all other members of the family. Find the total number of gifts given by the ten members of the family.</p><p>8. Arrange the first 6 counting numbers in the 6 circles so that the sum of each side (3 circles) of the triangle totals (a) 9 and (b) 10. (c) Are there other possible common sums?</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here are 6 circles. One on the top, two in the middle, and three in the bottom, that make up a triangle together.</p><p>END PRODUCER’S NOTE.</p><p>T 9. Eighty-six baseball teams participate in a double-elimination tournament. (A team continues to compete until it has lost two games.) How many games must be played to determine a champion</p><p>(a) if the champion does not lose a game?</p><p>(b) if the champion loses one game?</p><p>10. How can one use a 3 minute timer and a 5 minute timer to measure 7 minutes?</p><p>11. Draw a straight line across the face of a clock so that the sums of the numbers on each side of the line are the same.</p><p>12. What is the largest sum of money-all in coins, with no silver dollars-that I could have in my pocket without being able to give change for a dollar, half dollar, quarter, dime, or nickel?</p><p>13. You have $1 in change. You have at least one of each coin less than a half dollar. You do not have a half dollar. What is the smallest number of coins you can have?</p><p>14. Jodi traded dollar bills for dimes and quarters and received the same number of each. What is the smallest number of dollars she could have had?</p><p>15. How many cubes are there in this figure? (Assume that all hidden cubes are contained in the figure.)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here is a structure made of cubes. The structure is three cubes high with one layer of cubes on the top tier, two layers of cubes on the second tier and three layers of cubes on the bottom tier.</p><p>END PRODUCER’S NOTE.</p><p>16. Twenty-five coins total $1. If one of the coins is a quarter, what are the other coins?</p><p>17. Can you determine which of ten dimes is counterfeit (lighter than others) with only three weigh-ings on a balance scale? How?</p><p>18. These 4 steps are made of cubes. How many cubes are needed to make 20 steps?</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here are 10 cubes that make up 4 steps. One cube on the top, two on the next step down, three cubes on the next step and 4 cubes on the bottom step.</p><p>END PRODUCER’S NOTE.</p><p>19. How can you obtain 6 gallons of cider from a tank if you have only a 4-gal. container and a 9-gal. container?</p><p>C 20. If a bird and a half lays an egg and a half in a day and a half, how many eggs will 33 birds lay in 11 days?</p><p>21. If 3 cats catch 3 rats in 100 minutes, how many cats will catch 100 rats in 100 minutes?</p><p>22. If 3 hens lay 3 eggs in 3 days, how many eggs will 300 hens lay in 300 days?</p><p>23. Amo, Terry, Aaron, Mike, and Chris are starters on the Windsor Forest basketball team. Two shoot with their left hands, and three shoot with their right hands. Two are over 6 ft. tall, and three are under 6 ft. Arno and Aaron shoot with the same hand; Mike and Chris use different hands to shoot. Terry and Chris are in the same height range, but Aaron and Mike are in different height ranges. The player who plays center is over 6 ft. tall and is left-handed. Who is he? (HINT: Try using reasoning to eliminate impossible cases on this one. Completing the following table might be helpful.)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here is a blank table with the column headings of: Arno, Terry, Aaron, Mike, Chris and the row headings of: Left Hand, Right Hand, Taller than 6 ft, Shorter than 6 ft.</p><p>END PRODUCER’S NOTE.</p><p>24. Kanisha and Byron are playing a game called NIM. Each of them has a large pile of pennies, and the game pile starts empty. In turn, they move 1, 2, or 3 pennies from their pile to the game pile. The player who brings the total in the game pile to 24 wins the game. Byron plays first. Devise a strategy so that Kanisha can be sure of winning. (HINT: Since the goal is clearly stated, try working backward to find the best strategy for Kanisha to use. If there are 21, 22, or 23 pennies on Byron’s last move, he can win. Thus Kanisha should arrange for Byron to have a 20 on his last move; clearly she can win if he adds 1, 2, or 3 to 20. Similarly, she should arrange for Byron to play with a game pile of 16 on his next to last move. Describe the rest of the strategy.)</p><p>PRINT PAGE</p><h6 id="page28">28</h6><p>25. Three soccer teams-the Cats, the Dogs, and the Ants-play a three-game tournament. We have the following information: Each team played two games. The Cats won two games; the Dogs tied one game. The Dogs scored a total of two goals; the Ants scored three goals. One goal was scored against the Cats, four against the Dogs, and seven against the Ants. Find the scores of all three games.</p><p>26. It takes 8 days for a truck to cross the desert. The truck can carry gasoline for only 5 days. What is the smallest number of trucks to start the trip in order for one to get across and the others to return?</p><p>27. Cut a 3-by-8 rectangle into two pieces with one cut, and form a 2-by-12 rectangle with the pieces.</p><p>28. Labels were incorrectly placed on each of the three boxes. Explain how it is possible to determine the correct labels by removing just one ball from one box.</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here are three boxes with the following contents:</p><p>Box 1: 2 red balls</p><p>Box 2: 1 red ball 1 black ball</p><p>Box 3: 2 black balls</p><p>END PRODUCER’S NOTE.</p><p>29. A standard 8-by-8 checkerboard is shown below. Is it possible to cover this checkerboard with dominos (one domino covers two blocks) so that the two corners labeled with * are left uncovered? Explain. (HINT: Experiment with a miniature 2-by-2 checkerboard, a 3-by-3 checkerboard, and a 4-by-4 checkerboard until you understand the pattern.)</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here is a checkerboard. The squares that have * in them are in opposite corners.</p><p>END PRODUCER’S NOTE.</p><p>30. Thirty points are marked on a circle, and straight line segments are drawn to connect each point with all the other points. How many line segments are drawn?</p><p>REVIEW EXERCISES</p><p>31. (a) Conjecture a pattern for this sequence, and complete the next three terms: 1, 2, 4, BLANK SPACE, BLANK SPACE, BLANK SPACE.</p><p>(b) Conjecture a different pattern for this same sequence and fill in the next three terms.</p><p>32. Jane has 20 coins consisting of nickels and dimes, totaling $1.20. How many of each does she have?</p><p>PCR Excursion</p><p>33. The sequence of numbers 1, 3, 5, 7, 9, …. is called the sequence of odd counting numbers. The sums 1, <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn></mrow></math></span>, and <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>5</mn></mrow></math></span> are the first three partial sums of the odd counting numbers and are represented geometrically below:</p><p>1 one dot</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn></mrow></math></span> four dots, three dots connected with a line</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>5</mn></mrow></math></span> nine dots, three dots and five dots connected with lines</p><p>A. (a) Represent the next 3 of these partial sums geometrically.</p><p>(b) Because the dots in <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn></mrow></math></span> form a square with 2 dots on a side,</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn><mo>+</mo><msup><mn>2</mn><mrow><mn>2</mn></mrow></msup></mrow></math></span></p><p>Because the dots in <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>5</mn></mrow></math></span> form a square with 3 dots on a side,</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>5</mn><mo>=</mo></mrow></math></span> BLANK SPACE</p><p>Describe the next 3 of these partial sums in a similar way.</p><p>(c) Use the pattern you observed in (b) to find the sum of the first 8 odd counting numbers; the first 100 odd counting numbers.</p><p>(d) Write the sum of odd counting numbers that is equal to<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mi>n</mi><mrow><mn>2</mn></mrow></msup></mrow></math></span>.</p><p>B. Look at the following rectangular arrays of dots:</p><p>2 by 3 (i) (6 dots)</p><p>3 by 4 (ii) (12 dots)</p><p>4 by 5 (iii) (20 dots)</p><p>(a) Observe that because rectangle (i) has 2 rows and 3 columns, it has <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>⋅</mi><mn>3</mn></mrow></math></span> dots. Alternatively, looking at the dots on diagonal lines, we see that it has <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn></mrow></math></span> dots. We can conclude from this that</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>2</mn><mi>⋅</mi><mn>3</mn></mrow></math></span></p><p>Because rectangle (ii) has 3 rows and 4 columns, it has <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>3</mn><mi>⋅</mi><mn>4</mn></mrow></math></span> dots. Alternatively, the dots on diagonal lines indicate that it has <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn></mrow></math></span> dots. What can we conclude from rectangle (ii)? What similar relationship can we find in rectangle (iii)?</p><p>(b) Draw a rectangle of dots with 5 rows and 6 columns. What can we learn? How about a rectangle with 6 rows and 7 columns?</p><p>(c) Using the pattern you observed above, what fact can you learn about this sum?</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>4</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>6</mn><mo>+</mo><mn>7</mn><mo>+</mo><mn>7</mn><mo>+</mo><mn>6</mn><mo>+</mo><mi>…</mi><mo>+</mo><mn>1</mn></mrow></math></span></p><p>(d) Write a sum equal to<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>9</mn><mi>⋅</mi><mn>10</mn></mrow></math></span>.</p><p>(e) Write a sum equal to<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mfenced open="(" close=")"><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow></mfenced></mrow></math></span>.</p><p>(f) Because <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn></mrow></math></span> is equal to both <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mfenced open="(" close=")"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn></mrow></mfenced></mrow></math></span> and<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>4</mn><mi>⋅</mi><mn>3</mn></mrow></math></span>, we can conclude that<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mfenced open="(" close=")"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn></mrow></mfenced><mo>+</mo><mn>4</mn><mi>⋅</mi><mn>3</mn></mrow></math></span>. Dividing by 2 yields</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>=</mo><mfrac><mrow><mn>4</mn><mi>⋅</mi><mn>3</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></math></span></p><p>Find a similar pattern for the sums <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>4</mn></mrow></math></span> and<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>4</mn><mo>+</mo><mn>5</mn></mrow></math></span>.</p><p>(g) Use the reasoning of step (f) to determine a formula for the sum<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>4</mn><mo>+</mo><mi>…</mi><mo>+</mo><mi>n</mi></mrow></math></span>.</p><p>PRINT PAGE</p><h6 id="page29">29</h6><h3 id="heading6">SOLUTION TO INTRODUCTORY PROBLEM</h3><p>UNDERSTANDING THE PROBLEM. First, let’s restate the problem. Three times the following procedure is executed:</p><p>(a) The remaining pile is divided into three equal piles with one extra coconut.</p><p>(b) One of the three piles is hidden, while the other two equal piles are combined and left.</p><p>Finally, the remnant after these three incursions is divided into three equal piles, with one coconut remaining.</p><p>DEVISING A PLAN. Work the problem backward. Notice that the sum of the last three piles<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>+</mo><mn>1</mn></mrow></math></span> must be an even number because it is the sum of two equal previous piles. This situation must be true for the total number of coconuts at all times except at the beginning. This fact allows us to rule out possibilities as we look at various cases.</p><p>CARRYING OUT THE PLAN. Let <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi></mrow></math></span> be the number of coconuts received by each sailor at the end. Now <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi></mrow></math></span> cannot be even, because <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>+</mo><mi>n</mi><mo>+</mo><mi>n</mi><mo>+</mo><mn>1</mn></mrow></math></span> would be odd, which is a contradiction. If<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>1</mn></mrow></math></span>, then</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>1</mn><mo>+</mo><mn>1</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>4</mn></mrow></math></span> (Carl has divided into piles of 2.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>2</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>7</mn></mrow></math></span> (odd)</p><p>which is a contradiction. If<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>3</mn></mrow></math></span>, then</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>3</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>3</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>10</mn></mrow></math></span> (Carl has divided into piles of 5.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>5</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>16</mn></mrow></math></span> (Bob has divided into piles of 8.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>8</mn><mo>+</mo><mn>8</mn><mo>+</mo><mn>8</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>25</mn></mrow></math></span> (odd) which is a contradiction. If<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>5</mn></mrow></math></span>,</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>5</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>16</mn></mrow></math></span> (Carl has divided into piles of 8.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>8</mn><mtext></mtext><mo>+</mo><mtext></mtext><mn>8</mn><mtext></mtext><mo>+</mo><mtext></mtext><mn>8</mn><mtext></mtext><mo>+</mo><mtext></mtext><mn>1</mn><mtext></mtext><mo>=</mo><mtext></mtext><mn>25</mn></mrow></math></span> (odd)</p><p>which is a contradiction. If<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>7</mn></mrow></math></span>,</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>7</mn><mo>+</mo><mn>7</mn><mo>+</mo><mn>7</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>22</mn></mrow></math></span> (Carl has divided into piles of 11.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>11</mn><mo>+</mo><mn>11</mn><mo>+</mo><mn>11</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>34</mn></mrow></math></span> (Bob has divided into piles of 17.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>17</mn><mo>+</mo><mn>17</mn><mo>+</mo><mn>17</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>52</mn></mrow></math></span> (Mark has divided into piles of 26.)</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>26</mn><mo>+</mo><mn>26</mn><mo>+</mo><mn>26</mn><mo>+</mo><mn>1</mn><mo>=</mo><mn>79</mn></mrow></math></span></p><p>The least possible number of coconuts in the pile is 79.</p><p>LOOKING BACK. Mark divides the pile into 3 stacks of 26, hides his 26, and gives the extra coconut to the monkey. This leaves 52 coconuts in the pile. Bob divides the pile into 3 stacks of 17, hides his 17, and gives the extra coconut to the monkey. This leaves 34 coconuts in the pile. Carl divides the pile into stacks of 11, hides his 11, and gives the extra to the monkey. This leaves 22 coconuts in the pile. In the morning they divide the pile, giving each sailor 7 coconuts and the monkey 1.</p><p>SUMMARY AND REVIEW</p><p>NUMBER SEQUENCES</p><p>1. An arithmetic sequence is characterized by a common difference between terms.</p><p>2. In a geometric sequence, a term divided by the preceding term gives a common ratio.</p><p>PROBLEM SOLVING</p><p>Polya’s four steps:</p><p>1. Understand the problem.</p><p>2. Devise a plan.</p><p>3. Carry out the plan.</p><p>4. Look back.</p><p>Understanding a problem:</p><p>1. Identify what you are trying to find (the unknowns).</p><p>2. Summarize the information that is available to you (the data).</p><p>3. Don’t impose conditions that do not exist.</p><p>4. Strip the problem of irrelevant details.</p><p>Strategies for problem solving:</p><p>1. Make a chart or a table.</p><p>2. Draw a picture or a diagram.</p><p>3. Guess, test, and revise.</p><p>4. Look for a pattern.</p><p>5. Try a simpler version of the problem.</p><p>6. Restate the problem.</p><p>7. Eliminate impossible situations.</p><p>8. Use reasoning.</p><p>9. Work backward.</p><p>PRINT PAGE</p><h6 id="page30">30</h6><h3 id="heading7">CHAPTER TEST</h3><p>1. Study two rows at a time of the accompanying triangle to determine patterns of regularity. Note that the sum of two consecutive numbers in a row gives the number between them in the next row. In the fourth row, for example, the sum of 1 and 3 equals 4(circled) in the fifth row, and the sum of 3 and 3 is 6 (circled). Complete the sixth and seventh rows of the array.</p><p>BEGIN PRODUCER’S NOTE:</p><p>Shown here is a triangle made up of numbers in the following order from top to bottom:</p><p>1</p><p>1 1</p><p>1 2 1</p><p>1 3 3 1</p><p>1 4 6 4 1</p><p>Numbers 4 and 6 are circled on the bottom row.</p><p>END PRODUCER’S NOTE.</p><p>2. A farmer was asked by a passing stranger how many chickens and how many goats he had in his farmyard. The farmer answered that his animals had 90 legs and 62 eyes. How many of each animal did the farmer have?</p><p>3. A small boy determines to increase his physical stamina. Each morning he does pushups, increasing the number completed by 6 each day. After 5 days, he has done a total of 100 pushups. How many did he do each day?</p><p>4. A developer wants to place lots around each of four sides of a square block. If each of the four sides can be divided into seven lots (including corner lots), how many lots are available?</p><p>5. One of a group of nine dimes is counterfeit and lighter than the others. Explain how you could identify the counterfeit dime with only two weigh-ings on a balance scale.</p><p>6. What is the 36th term of the arithmetic sequence 17, 20, 23, 26, …?</p><p>7. Determine the next three entries in each sequence.</p><p>(a) 2, 7, 12, 17, ….</p><p>(b) 1, 5, 1, 1, 5, 1, 1, 1, 5, ….</p><p>(c) A, C, D, E, G, H, I, K, L, M, ….</p><p>8. Three times Ralph’s weight added to 54 kg is equal to 300 kg. How much does Ralph weigh?</p><p>9. Can you determine a pattern?</p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>=</mo><mn>3</mn></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>4</mn><mo>+</mo><mn>5</mn><mo>+</mo><mn>6</mn><mo>=</mo><mn>7</mn><mo>+</mo><mn>8</mn></mrow></math></span></p><p><span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>9</mn><mo>+</mo><mn>10</mn><mo>+</mo><mn>11</mn><mo>+</mo><mn>12</mn><mo>=</mo><mn>13</mn><mo>+</mo><mn>14</mn><mo>+</mo><mn>15</mn></mrow></math></span></p><p>Find the next two lines.</p><p>10. How can one measure 2 quarts from a faucet with a 6-quart container and a 10-quart container?</p><p>11. If 2 carpenters can build one deck in 1 day, how many days will it take 4 carpenters to build 8 decks?</p><p>12. Yolanda leaves Atlanta early in the morning for a very important presentation in Houston. She usually drives at a speed of 60 miles per hour. One hour later, Rolf discovers that Yolanda has forgotten her file needed for this presentation. He immediately starts after her at a rate of 75 miles per hour. How many hours will it take for Rolf to catch Yolanda?</p><p>13. The expression for the <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi></mrow></math></span>th term of a sequence is<span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>3</mn><mi>n</mi><mo>−</mo><mn>1</mn></mrow></math></span>. Write the first five terms of this sequence.</p><p>14. Consider this sequence: 2, 6, 18, 54, ….</p><p>(a) Find the next three terms.</p><p>(b) Find the tenth term.</p><p>(c) Write an expression for the <span class="mathmlEquation"><math xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi></mrow></math></span>th term.</p><p>15. Admission to a certain movie was $6 for adults and $3 for children. If 300 tickets were sold for a total of $1320, how many adult tickets were sold? How many children’s tickets were sold?</p><p>END OF FILE.</p></body></html>