Wordy Problems for MathyTeachers

December 2012 Wordy Problems for MathyTeachers 1st Issue Buffalo State College 1

Preface When looking over articles that were submitted to our journal we had one thing in mind: How can you implement this into the classroom? This is really important to us as future educators, because we want to challenge our students on many different levels. Each of these articles can be added as a supplemental resource to enhance a classroom lesson and engage the students in critical higher level thinking skills. Since there are multiple methods of exploration, it is very possible for one classroom to have a variety of methods to solve a single problem and yet arrive at the exact same solution. In the same way of thinking, identical problems could then be expanded in various ways to be explored further. This will portray the creativity and the problem solving skills of your students. These problems are not your usual textbook problems. All of the articles within this journal involve a game aspect or a real life application that students come into contact with in their everyday life. They also might respond more effectively to one of the following problems since it may transmit to their lives; this can help students realize that they will be able to use math and their mathematical reasoning outside the classroom. Students may not have even realized that these aspects of their lives involve math until they are enthralled and engaged in discussion with their classmates about each problem. We believe this journal is a valuable collection of problems involving critical thinking skills which we, as future teachers, believe is an important part of societies everyday life. We hope that these problems will bring out the creativeness of your students and strengthen your students mathematical thinking. Enjoy! Alexandria Langer Lindsay Hojnowski Stephanie Myers Managing Head Editor Editor Editor 2

TABLE OF CONTENTS GOAT S MILK... 4 DAN HOUSE GEOMETRY ON A CHESS BOARD... 9 ZACH TRUNZO WHICH ONE OF THESE IS NOT LIKE THE OTHERS?... 17 PATRICIA DELLA PENNA EXPLORING THE CELL GAME... 21 HEATHER HYLA SKIP THE STEP... 27 CHRISTINE SZAFRAN 3

GOAT S MILK Dan House Buffalo State College Some questions are stated in such a way where it depends on the responder to interpret them. A few might respond one way, while others may think differently and arrive with different results. One particular example that comes to mind is the Tethered Goat. The problem consists of a goat tied to a shed by a rope, and finding the area of grass that the goat can graze. The problem though is that the shed s dimensions are given, but not the shape of the shed. Suppose that the owner uses the goat for its milk and wants to know which area is bigger thus producing more milk. In this article I will discuss and analyze the shed as a rectangle and a parallelogram. First, we begin with the original question: A goat is tethered by a 6 metre rope to the outside corner of a shed measuring 4 metres by 5 metres in a grassy field. What area of grass can the goat graze? When we think of a shed, we think of rectangular shed with length 5 meters and width 4 meters. It is often very useful to draw a diagram. We start by drawing a picture to help determine the area we need to find. 4

Next, one must think of the goat walking with the rope taut, and determine how far the goat could travel in any direction. Since the shed has length 5 m, the goat can be parallel with the shed but extend farther by 1 m because the rope is 6 m long. Using that, the goat can keep going around the corner in a circular motion, with radius 1 m, until it reaches the wall. We want to mark this on diagram to keep partitioning the area into familiar shapes. Likewise, since the width is 4 m, the goat can walk back around and be parallel with the width but extended 2 m past the corner. Again, the goat can keep going past the corner in a circular motion, with radius 2 m this time, until the wall is reached. We need to mark this as well. Finally, the last piece is if the goat starts out parallel to the length, walks in a circular motion, with radius 6 m, and stops when it is parallel to the width. We draw this on our diagram as well and now our diagram should be partitioned as shown below: Now that our diagram is finished, we can begin finding the area. Looking at the diagram, we can now see that each area the goat is able to graze in is part of a circle. All we need to use in determining the area will be the formula for area of a circle:! =!!! 5

We need to modify the formula, however, since we do not have full circles. Notice that each area is one-quarter of a full circle. Therefore, for finding the area of each partition, we must multiply the area formula by ¼. Once doing so, the calculations are simple; we substitute the radius in for each, and sum all of the partitioned areas. However, we do not need to calculate the area of the quarter circle with radius 6 m three times. We can just multiply the area by three.! =!!!!! =!! π(6)! = 9π m! =!!!!! =!! π(2)! = π m! =!!!!! =!! π(1)! =!! π m Finally, we add our areas together and we arrive with a final area of!!"! m. However, we need to consider if the shed was a parallelogram and not a rectangle. For finding the area given a parallelogram shed, we will follow the same process and figure out which will be better for the goat s owner. First, we must draw a diagram to get a better representation. Also, we must decide the angles of our parallelogram. In the diagram, I decided on the obtuse angle being 120 and the acute as 60, and the rope being attached at the obtuse angle.! Similarly, we can mark the diagram with the farthest the goat can go on the rope. The diagram will have similar features except for the shed itself as shown on the next page: 6

All that s left to do is to determine the partitioned areas. We can use the same formula for the quarter circles as we did previously. However, since we are using 60 and 120, we need to determine the fraction of a circle that each angle makes up and multiply the corresponding area by it for the other remaining partitions. For example, 120 /360 is 1/3, and 60 /360 is 1/6.! =!!! =!! π(6)! = 6π! =!!! =!! π(1)! =! π!! =!!! =!! π(2)! =!! π Finally, we add our partitioned areas together and we have a total area of!!!!. This area is less than the area if we used a rectangular shed. So we should assume that the parallelogram would have less of an area. However, by simply moving where the rope is attached from the obtuse to the acute angle, we actually do generate a larger area! 7

By making this simple change, and following the process of marking the goats maximum radius, partitioning the total area into smaller ones, and simple calculations, we arrive to the result of the goat being able to graze an area of!"#!, which is larger than the rectangular shed and ultimately produce more goat milk! A teacher could use this paper and problem in their classroom to develop their students problem solving process. They could pose the original problem (rectangular shed), introduce the geometrical properties of rectangles and parallelograms, and work through the problem with the shed as a parallelogram with the acute angle being 45. Next you could ask them if there exists another shed that will produce greater area for the goat to eat. Finally, pose them the question about which shed would be the most practical even if the area isn t maximized. This will kick-start their problem solving process and help them to understand that there might be another way to interpret a problem. We have discovered that by having a parallelogram shed and positioning the goat properly, we can increase the milk production! Simply by thinking differently about a problem, we can get different results. In that sense, if you cannot find a solution to a problem, try a different approach. It could lead to the solution that you were looking for.! 8

GEOMETRY ON A CHESS BOARD Zach Trunzo Buffalo State College Geometry can be found in everything that you look at. Walls, tiles, pencils, etc. all have geometry. Many things can be looked at and can be picked apart to see multiple shapes of various sizes. One item that took an interest to me was a chess board. A chess board is a large square that contains an 8x8 grid. One can look at a chess board and notice there are a bunch of squares, but how many rectangles are in a chess board? It may surprise you to learn how many different rectangles there are on a chess board, so let s see how many there actually are. First let s start with the squares on a chess board, since squares are in fact a special case of rectangles. Well, the smallest squares we can have on a chess board are 1x1. Since we can fit eight different squares into one column and across all eight rows, we know that there are 64 different 1x1 squares on a chess board (since 8x8=64). We will use this same logic for the other rectangles and squares throughout the article. We will look at how many squares/rectangles can fit into one column and then how many rows they can fit into. We will take these two numbers and it will tell us how many of that sized squares/rectangles there are in the chess board after we multiply the two numbers together. 9

For 2x2 squares, we can fit 7 different squares into two consecutive columns across the board 7 times. This tells us that there are 7 2 2x2 squares, or 49 total. Some sample 2x2 squares on this pictured chess board would be (a1, a2, b1, b2), (a2, a3, b2, b3), (a3, a4, b3, b4), and (a4, a5, b4, b5). For 3x3 squares, we can fit 6 different squares into three consecutive columns across the board 6 times, giving us 36. For 4x4 squares, 5 different squares can fit into four consecutive columns across the board 5 times, giving us 25. For 5x5 squares, 4 different squares can fit into five consecutive columns across the board 4 times, giving us 16. For 6x6 squares, 3 different squares can fit into six consecutive columns across the board 3 times, giving us 9. For 7x7 squares, we can fit 2 different squares into seven consecutive columns across the board 2 times, giving us 4. When we get to our last type of square, 8x8, we know that this is the size of the chess board itself. Therefore, the outline of the chessboard is a square itself and makes it the only 8x8 square. As the size of the square grows, there is a pattern we can see. For 1x1 squares there are 8 2 total, for 2x2 squares there are 7 2 total, for 3x3 squares there are 6 2 total, etc. For every unit the square grows in a side length, the previous base is subtracted from the previous squared term. We can make an expression to find the total number of squares. If n is the side length of a square on a chess board, then (9-n) 2 will give us the total number of different nxn sized squares on the chess board. For example, n=3 plugged into the expression means we are finding the total number of different 3x3 squares on the chess board. This table is how many of each sized square can be found on a chess board. 10

1x1(n=1) 2x2 3x3 4x4 5x5 6x6 7x7 8x8 (9-n) 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 # of squares 64 49 36 25 16 9 4 1 The sum of the bottom row will give us the grand total of all of the different squares on a chess board. This gives us 204 different squares total on the chess board. This gives us a good start on the total number of rectangles in the chess board, since a square is a special type of rectangle. We have only scratched the surface of the total rectangles though, so let s keep going. Let s start looking at the vertical rectangles (meaning the rectangles that have a larger height than its width). The first set of rectangles we ll look at are rectangles that are 1 unit wide; this means rectangles that are 1x2, 1x3, 1x4, 1x5, 1x6, 1x7, and 1x8. For 1x2 rectangles, we can fit 7 different rectangles into each column across the board 8 times, giving us 56. For example, on this chess board, we can make the following seven 1x2 rectangles in column a: (a1, a2), (a2, a3), (a3, a4), (a4, a5), (a5, a6), (a6, a7), and (a7, a8). We can also follow this pattern for columns b-h, giving us 56. For 1x3 rectangles, we can fit 6 different rectangles into each column across the board 8 times, giving us 48. For 1x4 rectangles, we can fit 5 different rectangles into each column across the board 8 times, giving us 40. For 1x5 rectangles, we can fit 4 different rectangles into each column across the board 8 times, giving us 32. For 1x6 rectangles, we can fit 3 different rectangles into each column across the board 8 times, giving us 24. 11

For 1x7 rectangles, we can fit 2 different rectangles into each column across the board 8 times, giving us 16. For 1x8 rectangles, we can fit 1 rectangle into each column across the board 8 times, giving us 8. So when we add all of the one unit wide vertical rectangles together, we get 224. Now let s look at our next set of vertical rectangles, vertical rectangles that are 2 units wide; this means rectangles that are 2x3, 2x4, 2x5, 2x6, 2x7, and 2x8. For 2x3 rectangles, we can fit 6 different rectangles into each pair of consecutive columns and across the board 7 times, giving us 42. For 2x4 rectangles, we can fit 5 different rectangles into each pair of consecutive columns across the board 7 times, giving us 35. For 2x5 rectangles, we can fit 4 different rectangles into each pair of consecutive columns across the board 7 times, giving us 28. For 2x6 rectangles, we can fit 3 different rectangles into each pair of consecutive columns across the board 7 times, giving us 21. For 2x7 rectangles, we can fit 2 different rectangles into each pair of consecutive columns across the board 7 times, giving us 14. For 2x8 rectangles, we can fit 1 rectangle into each pair of consecutive columns across the board 7 times, giving us 7. When we add together all of our two unit wide vertical rectangles, we get 147. Now we will look at our 3 unit wide vertical rectangles; the rectangles that are 3x4, 3x5, 3x6, 3x7, and 3x8. For 3x4 rectangles, we can fit 5 different rectangles into each set of three consecutive columns across the board 6 times, giving us 30. For 3x5 rectangles, we can fit 4 different rectangles into each set of three consecutive columns across the board 6 times, giving us 24. For 3x6 rectangles, we can fit 3 different rectangles into each set of three consecutive columns across the board 6 times, giving us 18. For 3x7 rectangles, we can fit 2 different rectangles into each set of three consecutive 12

columns across the board 6 times, giving us 12. For 3x8 rectangles, we can fit 1 rectangle into each set of three consecutive columns across the board 6 times, giving us 6. When we add up the total amount of three unit wide vertical rectangles, we have 90. Now let s look at our 4 unit wide vertical rectangles, which include rectangles that are 4x5, 4x6, 4x7, and 4x8. For 4x5 rectangles, we can fit 4 different rectangles into each set of four consecutive columns across the board 5 times, giving us 20. For 4x6 rectangles, we can fit 3 different rectangles into each set of four consecutive columns across the board 5 times, giving us 15. For 4x7 rectangles, we can fit 2 different rectangles into each set of four consecutive columns across the board 5 times, giving us 10. For 4x8 rectangles, we can fit 1 rectangle into each set of four consecutive columns across the board 5 times, giving us 5. When we total up all of our four unit wide vertical rectangles, we have 50. Now onto our 5 unit wide vertical rectangles, which include 5x6, 5x7, and 5x8. For 5x6 rectangles, we can fit 3 different rectangles into each set of five consecutive columns across the board 4 times, giving us 12. For 5x7 rectangles, we can fit 2 different rectangles into each set of five consecutive columns across the board 4 times, giving us 8. For 5x8 rectangles, we can fit 1 rectangle into each set of five consecutive columns across the board 4 times, giving us 4. When we total our five unit wide vertical rectangles, we have 24. Now let s look at our 6 unit wide vertical rectangles, which include 6x7 and 6x8. For 6x7 rectangles, we can fit 2 different rectangles into each set of six consecutive columns across the board 3 times, giving us 6. For 6x8 rectangles, we can fit 1 rectangle 13

into each set of six consecutive columns across the board 3 times, giving us 3. When we add up our six unit wide vertical rectangles, we have 9 total. Our last vertical rectangle, seven units wide, only comes in one variety, 7x8. For 7x8 rectangles, we can fit 1 rectangle into each set of seven consecutive columns across the board 2 times. This gives us a total of 2 seven unit wide vertical rectangles on the chess board. Let s recap how many total vertical rectangles there are on a chess board. # of vertical rectangles 1 unit wide 2 units wide 3 units wide 4 units wide 5 units wide 6 units wide 7 units wide 224 147 90 50 24 9 2 The only type of rectangle left on a chess board is a horizontal rectangle (meaning the rectangles that have a larger width than its height). After we have found the amount of total vertical rectangles, finding the total number of horizontal rectangles is quite easy. There are the same amount of vertical rectangles as there are horizontal rectangles. The explanation is simple. Let s imagine that our chess board is the first quadrant of the y coordinate plane. If you graph the line y=x and reflect any of the vertical rectangles upon y=x that line, you will get a unique horizontal rectangle. If you repeat this for every vertical rectangle, you can get every possible horizontal rectangle. I have graphed two examples of this idea. The solid color is the original vertical rectangle and the dotted rectangle is the corresponding horizontal rectangle. Because every vertical rectangle has a corresponding x 14

unique horizontal rectangle, we know that the amount of vertical rectangles is the same as the amount of horizontal rectangles. Now that we have all of the types of rectangles covered, let s see how many we have ended up with (in the following table, V stands for Vertical and H stands for Horizontal): # of Rectangles Squares 204 1 unit wide V Rectangles 224 1 unit tall H Rectangles 224 2 unit wide V Rectangles 147 2 unit tall H Rectangles 147 3 unit wide V Rectangles 90 3 unit tall H Rectangles 90 4 unit wide V Rectangles 50 4 unit tall H Rectangles 50 5 unit wide V Rectangles 24 5 unit tall H Rectangles 24 6 unit wide V Rectangles 9 6 unit tall H Rectangles 9 7 unit wide V Rectangles 2 7 unit tall H Rectangles 2 TOTAL 1296 After we count our results, we see that there are 1296 rectangles on a standard 8x8 chess board! It is pretty remarkable that a simple 8x8 grid contains so many different rectangles. Just think about all of the other places we can find rectangles in our daily lives. How many rectangles can be found in a 15ft x 15ft room with 1ft x 1ft tiles? How many rectangles can be found on a 1280 pixel x 1024 pixel computer screen (Hint: this might take you a really, really, really long time)? We could even use this in a classroom setting with students. This problem will use their geometry skills, along with their logic skills, and will leave them looking at various geometric patterns in our daily lives. An 15

activity like this would be great for an extra credit assignment; it would really expand the student s thinking. Geometry is all around us, whether you know it or not. Take a minute to take a step back, some time, and just think about where it is around you 16

WHICH ONE OF THESE IS NOT LIKE THE OTHERS? Patricia Della Penna Buffalo State College In the classroom, as educators, we must find a way to have students think in an analytical way in which they explore logical reasoning. The use of the following question allows students to not only explore that reasoning but to adapt to change within it to see how the solution changes. We have ten coins, one of which is counterfeit and weighs a different amount (let s assume we know the coin is lighter than its legitimate counterparts). What is the minimum amount of times we need to weigh coins to assure we know which of the coins is counterfeit? We can place three coins on each side of the scale and place the other 4 coins to the side. If these groups weigh the same amount, they do not include the counterfeit. At this point, we weigh the remaining four by splitting them into two coins on each side of the scale. Whichever grouping of two is lighter can then be split into two groups of one to determine the counterfeit. If the original two groups of three are uneven, we can take the lighter three and weigh two of them. If they are even, it is the third coin that is counterfeit. The solution to the problem is three. 17

What if the situation changes and instead of a standard scale, we use a three way scale? On each of the three scales, we place three coins so there is only one remaining coin placed to the side. If they are all even, the counterfeit is the last coin that has not been weighed. However, if one of the groupings of coins is different from the other two, it contains the counterfeit. From here, we may split the pile so each of the three coins is on its own scale, finding which one is the counterfeit. Using a three way scale changes our answer to only having to weigh the coins twice. Let s continue this exploration by contemplating the situation of a four way scale. If we begin by weighing eight coins, two on each scale. If these groupings are even in weight, we know the counterfeit is one of the remaining two. In this case, we can take a second weighing to determine which of the two is counterfeit by placing the two coins on their own scales and on the other two scales, placing one coin from the original weighing. If one of the first groupings weighs a different amount from those on the other three scales, we can split the group of two as to weigh them individually on the four scales to identify the counterfeit. In the case of five scales, the number of times we need to weigh continues to be two in that we can weigh the coins individually in two sets, one on each scale. We can also put two coins on each scale to identify the group of two that weighs a different amount. We can then split the groups into individual coins on each scale, including the two that were of the uneven weighing. 18

Six scales requires us to weigh coins twice as well, first weighing six individually and then if they are all even, removing four of those six and replacing them with the remaining four. A similar solution is true for eight scales. This does not change until we have nine scales in which we can determine the counterfeit by only having to weigh coins once, placing one coin on each of the nine scales. If they are uneven, we can determine the counterfeit as the one that is uneven is the counterfeit coin. If the originally weighed coins are of equal weight, we know it is the remaining tenth coin that is the counterfeit. In conclusion for n coins being weighed, we can identify the counterfeit by weighing coins only one time when we have n-1 way scales. Another way to alter the problem for students is to have them recognize and identify a pattern based on the number of coins. Beginning with the trivial case of one coin, this requires us to weigh coins zero times, as we know one is counterfeit so this must be the fake coin. If there are two coins, we only need to weigh once (again, assuming we know the counterfeit coin is lighter). If there are three coins, we only need to weigh once again, as we can compare two, and if they are even, we know the third is the fake. Once we reach four coins, our scale is now more than one less than our number of coins so we are required to weigh twice. First we can compare two of them, then the remaining two. With five coins, we can compare groupings of two coins, and then split the lighter of the groupings, or know the fifth is the counterfeit if they are even. With six coins, we also weigh groups of two, if they are even we weigh the remaining two to identify the fake coin. If we begin with seven coins we can weigh two groups of three. If they are even, the counterfeit is the last coin, but if they are uneven, we can weigh two 19

coins of the lighter grouping to identify the counterfeit. We can discover the counterfeit of eight coins and nine coins the same way. It isn t until ten coins, that we are required to weigh three times. This is a question that will let students not only think but it s also an opportunity to bring in a physical aspect to the classroom. Students can use or even create scales to test their theories and to show their peers if they found a more efficient way of identifying the counterfeit coin. This is a question that could also be combined with a science class to create a project, perhaps during a unit in which they are covering mass in comparison to size of an object. At the end of their discovery process, students could be asked to even write an article describing their findings! 20

EXPLORING THE CELL GAME Heather Hyla Buffalo State College In this article I will be discussing the attributes for the Cell Game and my analysis of the game. This can be used in any math class. If you need a free day where you catch up students whom have been absent, you could have your students play the cell game. You can also check to see if they have mastered the game. The main reason for playing the game would be so all of your students are actively engaged in a hands-on activity. This would even be a great activity to do right before a holiday break if one does not wish to start a new topic. Not saying that this is just a way to waste time, however, this game also builds students knowledge and abilities on problem solving strategies. After playing this game a few times students will start to see when they have won and when they have lost. They will start to break the problem down and try to see if they can always win. The directions for the game are: Player 1 and Player 2 take turns placing a checker on one empty cell or two adjacent empty cells. The winner is the player who occupies the last empty cell. Who wins the game if the game is played correctly? The Cell game is a two player game that has 10 cells that a player can put their checker down in. Each turn, players can place checkers in either one cell or two adjacent cells. Since this is a two player game the players alternate turns. The last person to make a play wins. 21

After playing the Cell game several times I had started to notice what Player 1 could do to guarantee a win for themselves. The player who gets the board down to 4 remaining cells will win if they play their checkers correctly. This will be demonstrated below. My analysis is in favor of Player 1; however Player1 or Player 2 could win the game. To see this better we will start with the winning combinations. Player 1 wins the game if they place the last checker in the last cell or the last two adjacent cells. From here we will be able to see the patterns. If you see:3 or 4 cells left you will lose, 5 or 6 cells left you will win, 7 or 8 cells you will lose the game, and if you see 9 or 10 cells you will win by following my strategy. From this analysis one can see that it should be favorable for player 1 to win the game. Another thing I noticed about the Cell game is that when a player takes the center cell, it is easier to see the patterns. If you make the board symmetric, you can see who will win a lot faster. One other key point is that when placing the 5 th or 6 th checker to knock the board down to 4 remaining cells you should never leave 4 adjacent cells; if you do they will lose. In the diagram below I will show some winning moves and some losing moves. The last winning move is highlighted in yellow. Key: Player 1: Red Player 2: Blue (Top diagram shows open cells, and bottom diagram shows the players move) 1 Cell remaining: Win Player 1 O1 O1 O4 O5 O3 O3 O6 O2 O2 O1 O1 O4 O5 O3 O3 O6 O7 O2 O2 22

2 Cells remaining: Win Player 1 O1 O1 O4 O4 O3 O3 O2 O2 O1 O1 O4 O4 O3 O3 O5 O5 O2 O2 3 Cells remaining: Lose Player 1 O1 O1 O4 O4 O3 O2 O2 O1 O1 O4 O4 O3 O6 O5 O5 O2 O2 4 Cells remaining: Lose Player 1 O1 O1 O3 O3 O2 O2 O1 O1 O3 O3 O5 O4 O4 O6 O2 O2 5 Cells remaining: Win Player 1 O3 O1 O1 O2 O2 O5 O3 O4 O4 O1 O1 O6 O2 O2 O7 6 Cells remaining: Win Player 1 O1 O1 O2 O2 O1 O1 O4 O4 O3 O3 O5 O5 O2 O2 7 Cells remaining: Lose Player 1 O1 O1 O2 O1 O1 O4 O4 O3 O3 O5 O5 O6 O2 23

O1 8 Cells remaining: Lose Player 1 O1 O1 O1 O3 O3 O5 O4 O4 O6 O2 O2 O1 9 Cells remaining: Win Player 1 O2 O1 O3 O8 O7 O5 O5 O6 O9 O4 O2 O1 10 Cells remaining: Win Player 1Win O1 O3 O8 O7 O5 O5 O6 O9 O4 O2 The winning combinations to getting the game down to four cells are as follows: O1 O3 O5 O5 O4 O2 These are the first 5 plays. I will now discuss the remaining plays. No matter how the next play is made, Player 1 will win the game. This is because Player 1 had knocked the board down to four remaining spots. From here, Player 2 can take two adjacent cells. This would leave Player 1 to take the last two adjacent cells. Another option is that Player 2 could take one cell in one grouping (two adjacent cells), which will allow Player 1 to place their checker in the other grouping. This would force Player 2 to take one of the remaining single cells. Player 1 is left with the last cell; therefore they win the game. The diagram on the next page shows the moves of Player 1 and Player 2, when Player 2 plays 1 checker in 1 cell. Player 1 should not place a checker in a cell within the grouping. 24

O1 O3 O7 O5 O5 O6 O4 O2 Another way player 1 could win is if they get the board to have only 4 single, empty cells remaining. O1 O3 O4 O4 O5 O2 This would be a win for player 2 because they would alternate turns leaving player 2 with taking the last cell. The only way that one is not guaranteed a win is if they leave the board with any combination of four adjacent cells. O1 O4 O3 O5 O5 O2 This is the only time the Player 1 is not guaranteed to win with the four cells remaining. If Player 2 places two checkers right in the center they win the game. O1 O4 O3 O7 O6 O6 O8 O5 O5 O2 Player 1 should never place their chips as follows to ensure a win for themselves. If Player 1 takes the two center adjacent cells they win. O1 O3 O3 O4 O4 O2 O1 O3 O3 O5 O5 O4 O4 O2 O1 O6 O3 O3 O5 O5 O4 O4 O7 O2 If Player 2 wanted to ensure a win they should have left the board as follows: O1 O3 O3 O4 O2 O1 O5 O3 O3 O4 O6 O6 O2 O1 O5 O3 O3 O7 O7 O4 O6 O6 O2 25

After reading this article, this you should be able to win every time on a board of size 10. The key to the game is thinking ahead and planning your moves carefully. Players should know how to place their checkers so they are guaranteed to win. Also, one should look for patterns when playing as outlined above. They will be able to tell who is going to win once the board reaches four remaining cells. This also works for a board of size 8. If you are not convinced try it out. like: Using the board above, we can see that player 1 will win when the board looks 02 O4 O4 O1 O1 03 O5 O5 By using the same method as above we know that the player who eliminates all but four cells will win. The only way a player will lose is if the board has four adjacent cells. 02 O2 O1 O1 This will cause player 2 to lose because player 1 playing optimally will play as follows: 02 O2 O3 O3 O1 O1 02 O2 O4 O3 O3 O5 O1 O1 26

SKIP THE STEP Christine Szafran Buffalo State College Skip the step is a thinking exercise in which the individual focuses on figuring out the different number of ways it is possible to get to the top of a set of stairs of a given length, if he or she can climb one or two stairs per step. The person may ascend the stairs two at a time or one at a time, giving rise to numerous different ways to walk up the stairs. The goal of this experiment is to come up with a pattern to predict the number of methods the person can use to climb the stairs. The best course of action to investigate this is to simply work out the different methods. The investigation is as follows. Number of stairs Note: A way in which the person can climb the stairs will be denoted as: Number of stairs Methods of ascent Number of methods 3 (2, 1) (1, 2) (1, 1, 1) 3 In this example, there are 3 stairs, and in the first method the person took the first step over 2 stairs and the second over 1. The second method uses two steps as well, the first over 1 stair, the second over 2. The third method uses three steps over 1 stair each. This gives the number of methods as 3. Methods of ascent Number of methods 1 (1) 1 2 (2) (1,1) 2 3 (2,1)(1,2)(1,1,1) 3 4 (2,2)(2,1,1)(1,2,1)(1,1,2)(1,1,1,1) 5 5 (2,2,1)(2,1,2)(1,2,2)(2,1,1,1)(1,2,1,1)(1,1,2,1)(1,1,1,2)(1,1,1,1) 8 6 (2,2,2)(2,2,1,1)(2,1,2,1)(2,1,1,2)(1,2,2,1)(1,2,1,2)(1,1,2,2) (2,1,1,1,1)(1,2,1,1,1)(1,1,2,1,1)(1,1,1,2,1)(1,1,1,1,2)(1,1,1,1,1,1) 7 (2,2,2,1)(2,2,1,2)(2,1,2,2)(1,2,2,2)(2,2,1,1,1)(2,1,2,1,1)(2,1,1,2,1) (2,1,1,1,2)(1,2,1,1,2)(1,1,2,1,2)(1,1,1,2,2)(1,2,2,1,1)(1,1,2,2,1) (1,2,1,2,1)(2,1,1,1,1,1)(1,2,1,1,1,1)(1,1,2,1,1,1)(1,1,1,2,1,1) (1,1,1,1,2,1)(1,1,1,1,1,2)(1,1,1,1,1,1,1) n 13 21 F n 27

Looking at the pattern, one may deduce that the number of methods the person may climb the stairs follow the rule; Number of methods of climbing n stairs = F n-1 Where F n is the n th term of the Fibonacci sequence (1,2,3,5,8,13,21,34, ) Note: The Fibonacci sequence here is started at the second 1 to fix the index of the function. This, in and of itself, is a good way to get a class to look at patterns and to also see what real life examples the Fibonacci sequence has. To take this exercise one step further, it may be of worth to analyze what happens when the person may take more than one step at a time. For instance, the following analysis is over the same situation, but instead of being able to take the stairs 1 or 2 at a time, the person may take them 1, 2, or 3 at a time. Number of stairs Methods of ascent # of methods # of methods n- # of methods (n-1) # of methods (n-1) - # of methods (n-2) 1 (1) 1 - - 2 (2)(1,1) 2 1-3 (3)(1,2)(2,1)(1,1,1) 4 2 1 4 (3,1)(1,3)(2,2)(2,1,1)(1,2,1)(1,1,2)(1,1,1,1) 7 3 1 5 (3,2)(2,3)(3,1,1)(1,3,1)(1,1,3)(2,2,1)(2,1,2)(1,2,2) (2,1,1,1)(1,2,1,1)(1,1,2,1)(1,1,1,2)(1,1,1,1) 6 (3,3)(3,2,1)(3,1,2)(2,1,3)(2,3,1)(1,3,2)(1,2,3)(2,2,2) (2,2,1,1)(2,1,2,1)(2,1,1,2)(1,2,2,1)(1,2,1,2) (1,1,2,2)(2,1,1,1,1)(1,2,1,1,1)(1,1,2,1,1)(1,1,1,2,1) (1,1,1,1,2)(1,1,1,1,1,1) 7 (3,3,1)(3,1,3)(1,3,3)(3,2,2)(2,3,2)(2,2,3)(3,2,1,1)(3,1,2,1) (3,1,1,2)(2,3,1,1)(2,1,3,1)(2,1,1,3)(1,3,2,1)(1,3,1,2) (1,2,3,1)(1,2,1,3)(1,1,3,2)(1,1,2,3)(3,1,1,1,1)(1,3,1,1,1) (1,1,3,1,1)(1,1,1,3,1)(1,1,1,1,3)(2,2,2,1)(2,2,1,2)(2,1,2,2) (1,2,2,2)(2,2,1,1,1)(2,1,2,1,1)(2,1,1,2,1)(2,1,1,1,2) (1,2,2,1,1)(1,2,1,2,1)(1,2,1,1,2)(1,1,2,2,1)(1,1,2,1,2) (1,1,1,2,2)(2,1,1,1,1,1)(1,2,1,1,1,1)(1,1,2,1,1,1) (1,1,1,2,1,1)(1,1,1,1,2,1)(1,1,1,1,1,2)(1,1,1,1,1,1,1) n 13 6 2 24 11 4 44 20 7 T n 28

This pattern follows the three-step Fibonacci sequence, also called the Tribonacci sequence. The sequence is defined by: T n : an = an 1 + an 2 + an 3 A conjecture may be made that the numbers of methods may follow a Fibonacci + X pattern, in which the number of previous terms added to form each new term equals the number of steps able to be taken. For example, if a person may take a step covering four stairs, the sequence may follow: F 4 n : an = an 1 + an 2 + an 3 + an 4 Looking at patterns such as this can fit into a few different classes at the high school level. In fact, one of the Standards for Mathematical Practice from the Common Core Learning Standards for New York State suggests that students should be able to look for and make use of structure and pattern. It may be overly exhausting to have students pursue the Tribonacci sequence or those of higher order themselves. If the methods are worked out beforehand and given to the class, then the students can investigate the pattern. 29