Detailed Solutions to Problem Solving Exercises

Transcription

1 1. A Guessing Game First of all, we should note that the objective is to determine the best strategy in order to maximize our average score per game. We are not trying to beat our classmates, although that might be fun. So we should imagine that we are playing the game alone, or, blindfolded. There are 36 equally likely rolls for the two dice. Of these 36 possibilities, 11 of them contain a 1. The probability of getting a 1 is therefore 11 / 36, and the probability of not getting a 1 is 25 / 36. When a 1 isn t rolled, the average score increase on a roll is 8. If a 1 is rolled, the larger my score is (before the 1 was rolled) the greater my loss. The expected (i.e., average) increase (E) in my score, if I decide to stay standing, is equal to the expected increase (C) in my score when a 1 isn t rolled, minus the expected decrease (D) in my score when a 1 is rolled. Taking each of these two possibilities The expected increase (C) in my score when a 1 isn t rolled, is the average increase when a 1 isn t rolled (8) times the probability of not rolling a 1 ( 25 / 36 ). Therefore C = 8 ( 25 / 36 ). The expected decrease (D) in my score when a 1 is rolled, is whatever my score happens to be (x) just before the 1 was rolled times the probability of rolling a 1 (which is 11 / 36 ). Therefore D = x ( 11 / 36 ). In summary, E = C D E = 8 ( 25 / 36 ) x ( 11 / 36 ) How is this formula useful? Well before my first roll, x is zero, so the expected increase in my score is 8 ( 25 / 36 ) At some later point in the game, if I happen to have a score of 10 (and decide to stay standing), then the expected increase in my score is 8 ( 25 / 36 ) 10 ( 11 / 36 ) = 2.5, which means on average we gain 2.5 points if we stay standing for the next roll. If instead I happen to have a score of 50, then the expected increase in my score is 9.72, which means on average we lose 9.72 points. We are interested in determining the value of x such that the expected increase of our score is zero, which is the boundary between a positive expected increase (i.e., a gain) and a negative expected increase (i.e., a loss). Putting zero in for E in the above equation gives us x Therefore, the best mathematical strategy is to remain standing whenever our score is 18 or below, and sit down once our score exceeds 18. 1

2 2. Monty s Choice This problem received national attention when it appeared in Parade Magazine in 1990, in Marilyn vos Savant s column, Ask Marilyn. Peter Taylor s book includes a wonderful (and outrageous) account of the various letters that it prompted. After having a fair bit of debate, it s nice to have the students do a simulation of the game, as follows. In groups of three, one is the contestant, one is Monty, and one is the tosser of a die and a coin. The tosser throws the die and coin so that Monty, but not the contestant, can see the result. This roll of the die determines which door has the car; a 1 or 2 represents door A, a 3 or 4 represents door B, and a 5 or 6 represents door C. The contestant then chooses a door. Monty looks at the die to see if the contestant guessed the door with the car behind it, or not. If the contestant guessed correctly, then Monty looks at the coin heads means Monty opens the higher of the two remaining doors, and tails means he opens the lower of the two remaining doors. (With the simulation game, opening door B simply means that Monty announces to the contestant, Door B has a goat behind it. ) If the contestant didn t guess the door with the car, then Monty ignores the coin and simply opens the remaining door with the goat. Once Monty has shown a door with a goat behind it to the contest, Monty says to the contestant that he may now either switch to the other remaining door, or stick to the door he originally chose. Each group should do a simulation at least 10 times with the switching strategy and an equal number of times with the sticking strategy. Once all of the data has been collected for the whole class, it should become apparent that the switching strategy is best, yielding around a ⅔ probability of winning. Of course now the discussion can turn to trying to explain why this is true. The explanation for the sticking strategy is easier. In this case, the probability of success is clearly not affected by which door Monty opens; therefore the probability of winning is ⅓. The explanation for the switching strategy is trickier. One way to look at it is that the contestant wins every time with the switching if, and only if, he initially didn t guess the correct door. Why is this so? Well, if the contestant chose the wrong door, then Monty will open the other wrong door, and then switching to the third, remaining door is always a winner. In other words, the probability of winning with a switching strategy is the same as the probability that the initial choice of doors was wrong, which is clearly a probability of ⅔. 2

3 3. Two Ants A. There are many possible paths, but only one shortest path. If the ant just walks along edges, the distance would be 9 feet. The shortest path, however, is best seen if we unfold the box and place it on a flat surface. There are many possible ways to unfold the box, depending on which edges we choose to cut. The drawing here shows one way to unfold and lay the box flat, such that the shortest path becomes obvious. We now simply connect the two dots (the diametrically opposite corners of the box) to get the shortest path, which has a length of approximately 6.40 feet. Visualizing this in three dimensions, this path takes the ant (which started at the top front right corner of the box) diagonally across the top and left side of the box. Of course, the shortest path could also have been diagonally across the right side and bottom of the box, which would have been easily seen if we had cut the box differently. B. Following the above method, we first cut selected edges of the box, unfold it, and lay it out flat such that the male ant is seen closest to the top front edge of the box. The question then before us is: where is the best place to put the back face such that the shortest path becomes evident? The above drawing shows the four possibilities for the placement of the back face each placement of the back face yielding a different path by connecting the two ants with a straight line. The path that most people first think of takes the ant straight across the top face and has a length of 7 feet, or 84 inches. Next, consider the path that resulting from when the back face is in the lowest position, and the path where the back face is two squares from the top. Both of these paths have a length of 81.4 inches (calculated using the Pythagorean Theorem). The shortest path of all, which has a length of exactly 80 inches, or 6⅔ feet, comes from placing the back face at the top-most square in the drawing. This path takes the ant across all but one of the faces of the box. C. Given that the distance between two points is the shortest possible path between them, the two furthest points are the centers of the square faces, for a total length of exactly 7 ft. 3

4 D. I begin with the hypothesis that in order to be as far as possible from the male ant, the female must be on the opposite square face fairly close to the corner that is diametrically opposite from the male ant. There are several paths possible that go from the male to the female, but clearly, in this case, any path that crosses three of the rectangular faces cannot be the shortest path. That leaves us with four possible paths that might be the shortest. With the drawing here, the back square wall is shown in the four possible different foldouts, but the female ant is shown in each case in the same location. Connecting the female to the male with a straight line shows the four different paths. If, for the moment, I only consider paths B and C, then I can see that as long as the female is somewhere along the diagonal line of the back square (the diagonal that passes through the right bottom back corner of the box) then paths B and C will have the same lengths. Likewise, at any point along this diagonal, paths A and D will have the same lengths. Now, imagining the female ant moving along this diagonal line, let s consider the lengths of paths A and B only. As it moves along this diagonal, the length of one of these two paths increases while the length of the other path decreases. The objective is to be at the location along this diagonal where the minimum path is the greatest. This location is where the lengths of these two paths (A and B) are equal. In fact, from this particular location, the lengths of all four of the paths (A, B, C and D) are equal. If the female ant strays from this location, in any direction, then the length of at least one of the paths will decrease, and therefore the minimum path length is also decreased. Our goal therefore is to find exactly where this location is such that the lengths of path and A and B are the same. I assign x to the distance that the female is from the right back edge, and y to the distance that the female is up from the bottom back edge (i.e., the floor). (Note that with each back square in the drawing these edges have been rotated.) The lengths are therefore: Path A: (4 x) 2 + (y+5) 2 Path B: (7 x) 2 + (2 y) 2 Setting these equations equal, as well as setting x equal to y (since the location must be along the square s diagonal), yields: x = y = 3 / 5, which means the female ant is located near the right bottom back corner of the box, 0.6 ft up from the floor and 0.6 ft away from the right wall. All four paths (A, B, C and D from the male ant to the female at this location) have a length of ft. 4

5 4. Cut Plane We will first solve a simpler version of the problem, and then see what insight it gives to the originally stated problem. By reducing the number of planes, we can fairly easily visualize that 0 planes yield 1 region; 1 plane yields 2 regions; 2 planes yield 4 regions; and 3 planes yield 8 regions. One might be inclined at this point to see a pattern and conclude that the next step is 4 planes yield 16 regions, and therefore 10 planes will yield 1024 (2 10 ) regions. However, on closer inspection, we start with the 8 regions that were produced by 3 planes and see that the 4 th plane cuts through all but one of these 8 regions, thereby showing that 4 planes divide space into 15 regions. It is even more difficult to picture how many regions are produced by 5 planes. Certainly, trying to picture how many regions are created by 10 planes would be impossible. The major insight, is to simplify the number of dimensions of the original problem. Therefore we ask: how many regions on a plane are produced by a certain number of lines? (Again where all of the lines intersect with one another, but never three lines meeting at the same point.) This is quite manageable, and, as long as we are careful, we can progress quite quickly up to nine lines. It usually doesn t take long before the students notice the pattern that generates this 2-D table. The differences in the regions column keeps going up by one. And once that has been discovered, the students will likely discover two things: 3-D planes regions D lines regions The differences in the 3-D table (1, 2, 4, 7, etc.) appear to come from the 2-D table. 2. If we reduce from two dimensions to one dimension (i.e., How many regions on a line are produced by a certain number of points? ), then the resulting regions column is simply the counting numbers (1, 2, 3, 4 ). Quite nicely, we see that the differences in the 2-D table are given by the 1-D table, which seems to support our proposition stated in #1, above. 5

6 Of course, we can go ahead and fill out the 3-D right away, but we are left with a bit of doubt as to whether our proposition stated in #1 (i.e., the differences in the 3-D table come from the 2-D table) is definitely true. How can we demonstrate that this is true? Here is one possible approach Let s start with the 2-D case of three lines (which divide the plane into 7 regions). The question is: How many additional regions are created by the fourth line? To answer this question, we start by looking at the drawing on the right. We see that the new 4 th line intersects the other lines at 3 points. Each of the 4 segments that the 1-D line is broken into corresponds to a new region on the 2-D plane. In other words, our above 2-D question ( How many additional regions are created by adding a 4 th line? ), can now be translated into this 1-D question: How many regions are produced by the 3 points of intersection on this new (4 th ) line? The answer to both of these questions is 4, which can be found in the 1-D table, given below. And in the same way, we can fill in the next place of the 3-D table by transforming this 3-D question: How many additional regions are created by the 5 th plane?, into this 2-D question: How many regions are produced by the 4 lines of intersection on this new (5 th ) plane? The answer to both questions (which is 11) can be found on the 2-D table. We therefore know that 5 planes divide space into = 26 regions. Continued on the next page 6

7 Now, we can fill out our tables, even past three dimensions. 1-D points regions D lines regions D planes regions D hyperplanes regions Lastly, we leave with an interesting thought: what would all of this be in projective geometry? Answer: The number of regions would be found one row higher. For example, in 3-D projective space, 7 planes divide space into 42 regions. 5. Angles in a Star Polygon A. 180 B. 540 C. No, the results would be the same. One explanation is that each angle of the star polygon can be considered an inscribed angle, with each of these angles subtending an arc of the circle. With part a, each arc gets subtended once, and with part b, each arc is subtended three times and it doesn t matter if the points are evenly spaced or not. D. 180 E F. S = 180 (N 2X) 7

8 6. Factors and Perfect Numbers A. The number 9,489,150,000 has a prime factorization of To produce a factor of the number, we can choose anywhere from 0 to 4 two s, 0 to 5 three s, 0 to 5 five s, 0 or 1 eleven, and 0 or 1 seventy-one. This is now the same kind of problem as the typical wardrobe problem (e.g., How many different outfits are possible if you have 2 pairs of shoes, 5 pants, and 4 shirts to choose from? there are 40 possible outfits). In the case of determining the number of factors of this number, we have 5 choices of two s, 6 choices of three s, 6 choices of five s, 2 choices of eleven s, and 2 choices of seventy-one s. Therefore, this number has 720 factors. B. For x 12 1 (x 6 +1)(x 6 1) (x 6 +1)(x 3 +1)(x 3 1) Then we recognize perfect cubes to get: [(x 2 +1)(x 4 x 2 +1)][(x+1)(x 2 x+1)][(x 1)(x 2 +x+1)] The binomial factors are: (x+1), (x 1), (x 2 +1), (x 2 1), (x 3 +1), (x 3 1), (x 4 1), (x 6 +1), (x 6 1), (x 8 1) Some possible laws are: If n is odd then (x+1) is a factor of (x n + 1). For any n (x 1) is a factor of (x n 1). If i is a factor of n then (x i 1) is a factor of (x n 1). For x 15 1, we know that (x 1), (x 3 1), and (x 5 1) must all be factors. We can start by using polynomial division to divide (x 3 1) into (x 15 1) to get: x 12 + x 9 + x 6 + x 3 + 1, and then factoring the (x 3 1), we finally get: x 15 1 (x 1)(x 2 +x+1)(x 12 + x 9 + x 6 + x 3 + 1). But why isn t (x 5 1) in our final answer? To investigate this question, we start by dividing (x 5 1) into (x 15 1), or, better yet, we factor (x 15 1) using perfect cubes to get: x 15 1 (x 5 1)(x 10 +x 5 +1), and then dividing (x 5 1) by (x 1), we get: x 15 1 (x 1)(x 4 +x 3 +x 2 +x+1)(x 10 +x 5 +1). But above, we said that: x 15 1 (x 1)(x 2 +x+1)(x 12 + x 9 + x 6 + x 3 + 1). Why don t these two different approaches end up with the same result? It must be that neither has been completely 8

9 factored. Of the four different polynomial factors, we can be confident that (x 2 +x+1) is prime, so it must divide evenly into (i.e., be a factor of) either (x 4 +x 3 +x 2 +x+1) or (x 10 +x 5 +1). It ends up that (x 2 +x+1) doesn t go evenly into (x 4 +x 3 +x 2 +x+1), but it does divide into (x 10 +x 5 +1), for an unexpected outcome of (x 8 x 7 +x 5 x 4 +x 3 x+1). With some satisfaction we note that (x 4 +x 3 +x 2 +x+1) also divides evenly (x 12 +x 9 +x 6 +x 3 +1) for a result of (x 8 x 7 +x 5 x 4 +x 3 x+1). We have finally completely factored (x 15 1) to (x 1)(x 2 +x+1)(x 4 +x 3 +x 2 +x+1)(x 8 x 7 +x 5 x 4 +x 3 x+1). C. Following some of the above ideas, we get: ( )(2 30 1) ( )( )(2 15 1) [( )( )][(2 5 +1)( )][(2 5 1)( )] This would be a good point to evaluate the parentheses: [(1025)( )][(33)(993)][(31)(1057)], and then breaking these down further, we get: [(5 2 41)( )][(3 11)(3 331)][(31)(7 151)] Note that with a few of the larger numbers, we only need to try dividing the number by all of the prime numbers up to the square root of the number. With , we d have to try dividing by all of the prime numbers up to , which is Dividing the class into groups, and giving them a table of prime numbers, quickly leads to the discovery that it s divisible by 13, leaving us with a quotient of 80581, and that number is divisible by 61, leaving a quotient of None of the primes up to 1321 divide evenly into 1321, so it must be prime. Our final answer is:

10 D. After some experimentation students are often led to listing the factors of the perfect number in pairs. For example, using the 28, 496 and 8128, we get: Given that the number of rows is n, the last two numbers (which multiply to become the perfect number) are 2 n-1 and 2 n 1, and so the perfect number, P, is given by P = (2 n-1 )(2 n 1). However, we must note that if n = 2, 3, 5, 7 then we get the perfect numbers P = 6, 28, 496, 8128, but if n = 4, 6, 8 and others, then this formula won t generate a perfect number. Why not? Well, in that case, we will get extra factors that aren t included in this pattern, and therefore the sum of the proper factors will be greater than the number itself, so the number will be abundant instead of perfect. It then follows that the last number of the second column, namely (2 n 1), needs to be a prime number. Prime numbers of this form are known as Mersenne primes. So the question now becomes: what determines whether (2 n 1) is prime or not? Well, for one thing, n itself must be prime (we know this through one of the laws that we discovered on the previous problem). But other than that we need to check to see if it is prime. Of all of the prime values for n between 10 and 60, only n = 13, 17, 19, 31 produce values for (2 n 1) that are prime, which means The 5 th perfect number is: 33,550,336 The 6 th perfect number is: 8,589,869,056 The 7 th perfect number is: 137,438,691,328 The 8 th perfect number is: 2,305,843,008,139,952,128 (For more about perfect numbers, see Appendix B from my book, Making Math Meaningful A Middle School Math Curriculum for Teachers and Parents.) 10

11 7. Cell Phone Decisions This is a good exercise for students to experience a practical application of Cartesian geometry, so I would encourage the students to graph all three equations and see the results. The x-axis ought to be the number of cell phone minutes and the y-axis ought to be the total cost for the particular plan. The end result is that the break-even point between the first two plans is 494 minutes per month, and the break-even point between the last two plans is 1000 minutes per month. 8. Power Series and Darts. A. The given series is 1 + x + x 2 + x x n. We know that (x+1)(x 1) = x 2 1 and that (x 2 +x+1)(x 1) = x 3 1 and that (x 3 + x 2 + x + 1)(x 1) = x 4 1. Therefore (1 + x + x 2 + x x n )(x 1) = x n+1 1, from which we can see that the original identity is true. B. Given 0 < x < 1, the above identity becomes: 0 i x i = 1 1 x and if x = ⅓, then we get 1 + ⅓ + (⅓) 2 + (⅓) 3 + = 3 / 2 11

12 C. The given series is 1 + 2x + 3x 2 + 4x (n+1)x n. 1 + x + x 2 + x x n = xn+1 1 x 1 x + x 2 + x x n x( xn 1 ) x 1 = xn+1 x x 1 x 2 + x x n = xn+1 x 2 x 1 x x n = xn+1 x 3 x 1 etc., until x n = xn+1 x n x 1 Now, adding up all of the above equations, we get: 1 + 2x + 3x 2 + 4x (n+1)x n = (n+1)xn+1 (1+x+x 2 + +x n ) x 1 Which eventually simplifies to: 1 + 2x + 3x 2 + 4x (n+1)x n = (n+1)xn+2 (n+2)x n (x 1) 2 D. Given 0 < x < 1, the above identity becomes: 0 i (i+1)x i = 1 (1 x) 2 Although, not at first obvious, x is equal to ⅓ in the example. Plugging in ⅓ tells us that the infinite series equals 9 / 4. E or 8.33% or 7.64% ( 12) % % % 6. ( 12) % 12 F. This is a weighted average problem. 15 (0.2) + 32(0.3) + 24(0.5)

13 G. Imagine that someone playing darts hit the target 70% of the time on the 1 st throw, 20% of the time on the 2 nd throw, and 10% of the time on the 3 rd throw (thereby never needing more than three throws). The expected number of throws for that person to hit the target would then be: 1(0.7) + 2(0.2) + 3(0.1) 1.4 throws. This is, of course, the same principle (weighted average) that we used to solve the above bus problem. But with the given dartboard problem, the probability of hitting the target is 1 / 12 ( 8.33%) for each throw. Therefore, there is a chance (however small) that we won t score until after several million attempts. This problem, therefore, combines both weighted average and infinite series. i is the number of throws needed to score, and p(i) is the probability that we score on exactly that number of throws (and not earlier). The expected number of throws needed to score is: Now we can use the formula from part D above, which says: (i+1)x i = i 0 1 (1 x) 2 but can also be written as (i)x i i-1 = Using this to solve 1 12 i 1 i 1 (i)( i 12) 11 i-1 gives a result of (1 x) 2 Therefore, we expect to hit the target, on average, in 12 throws, which may be the answer we thought of intuitively at the start. The Dartboard Principle states: The expected (average) number of attempts needed to achieve a desired outcome is the reciprocal of the probability of getting the desired outcome in one attempt. 13

14 9. Baseball Cards. Let s start by asking a few simpler questions: Question: What would be the expected (average) number of flips of a coin in order to get both possibilities (heads and tails)? Solution: After one flip you have one of the two possibilities heads or tails. Since the chance of getting the other possibility on the next flip is ½, the expected number of flips needed to get this other possibility will be 2. (This is the Dartboard Principle, from the above Dartboard problem.) Therefore, the expected total number of flips is 3. Question: What would be the expected (average) number of roles of a die in order to get all six possible roles? Solution: After one die has been rolled, the expected number of rolls needed to get a different result is 6 / 5, because the probability of getting it immediately is 5 / 6 (again, this is the Dartboard Principle). Now we have achieved two different results in / 5 rolls. The expected number of rolls needed to get a third different result is 6 / 4, (because there is a 4 / 6 probability of getting a different result in the next roll). Continuing this pattern, we see that the expected number of rolls to get all six results is: 6 6 / / / / / / i i 1 Question: After having collected the 70 th different baseball card (out of 200) what would be the expected (average) number of purchases needed in order to get another different card? Solution: Again, using the Dartboard Principle, the probability that the next purchase will result in a different card is 130 / 200. Therefore the expected number of purchases needed in order to get a different card is 200 / 130. So now we return to the question at hand: What is the expected number of days needed to collect all 200 cards? Using what we have learned from the above questions, we get: 200 / / / / i i 1 i i 1 Unfortunately, the series Σ 1 slowly diverges and has no easy formula i associated with it. However, with some technological assistance, we find 200 that i 1 1 i , and therefore our final answer is that we expect it to take about 1176 days to collect all 200 cards

15 10. Building Trains A. One effective strategy is to start with shorter trains and build up. There is 1 way to build a train of length 0, 1 way to build a train of length 1, 2 ways to build a train of length 2, etc. Let x n be the number of possible trains having a length of n. The sequence (starting with n=0) is then x n = 1,1,2,3,5,8,13 We hopefully notice that the value of any x n is simply the sum of the two previous terms (x n = x n-1 + x n-2 ). This is the Fibonacci sequence. Continuing in this way we get achieve a result of 233 possible trains having a length of 12. B. Any train starts with an A (length of 2), or with a C (length of 1). If it starts with an A, then the rest of the train has a length of n-2. Therefore the number of possible trains having a length of n that start with an A is the same as the number of possible trains having a length of n-2. Likewise, if the train of length n starts with a C, then the rest of the train has a length of n-1. Therefore the number of possible trains having a length of n that start with a C is the same as the number of possible trains having a length of n-1. This allows us to conclude that the number of trains having a total length of n is equal to the number of trains of length n-1 (if it begins with a C) plus the number of trains of length n-2 (if it begins with an A). C. 1a) x n = 2x n and x n = 4 2 n 1 or x n = 2 n+2 1 (Done using the same method as shown with the problem below.) b) x n = 5x n and x n = 7 5n 3 (see proof below) 4 Since x n = 5x n x 1 = 5x x 2 = 5x (5x 0 + 3) + 3 = 5 2 x x 3 = 5x (5 2 x ) + 3 = 5 3 x and so on and by noticing the pattern, we get: x n = 5 n x n n = 5 n x 0 + 3(5 n n ) Now we simplify inside the parentheses by using the formula a n + a n-1 + a n a 2 + a + 1 = an+1 1 a 1 which gives us: x n = 5 n x 0 + 3( 5n ) simplifying to 7 5 n

16 C. 1c) x n = 3x n-1 4 ; x n = 5 3 n + 2 (same method as above.) 2) x n = (ax 0 x 0 + b)a n b a 1 (Proof shown below.) Given x n = ax n-1 + b x 1 = ax 0 + b x 2 = ax 1 + b a(ax 0 + b) + b a 2 x 0 + ab + b x 3 = ax 2 + b = a(a 2 x 0 + ab + b) + b a 3 x 0 + a 2 b + ab + b x n = a n x 0 + a n-1 b + a n-2 b+ + ab+ 1) = a n x 0 + b(a n-1 + a n a + 1) x n = a n x 0 + b( an 1 ) which simplifies to a 1 x n = (ax 0 x 0 + b)a n b a 1 D. 1) As with the A-C train problem (see above), start with shorter trains and build up. There are 2731 trains of length 12. 2) The sequence (starting with a train length of n=0) is: x n = 1,1,3,5,11,21,43,85,171,341,683,1365,2731 3) There are a few different patterns that we could notice. One method is to double the previous term then alternately add or subtract one. Simpler, however, is the following recursive formula: x n = x n x n-2 4) For odd values of n we have the sub-sequence of t m = 1,5,21,85,341,1365 (if m = n-1 2, then t m = x n ) The recursive formula is t m = 4t m-1 + 1, which then yields (by using the above general general formula) the general formula: t m = 4 4m 1 which simplifies to t 3 m = 22m Now we translate this into terms of x n by letting m = ½(n-1), which results in the general formula: x n = 2n+1 1 (for odd n values) 3 Using a similar process for even values of n we find that x n = 2n (for even n values) 3 We can then combine the two formulas, and account for the plus and minus, with this general formula: x n = 2n+1 + (-1) n 3 (for all n values) 16

17 E. 1a) x n = 1, 4, 6, 14, 26, 54 b) x n = 1, 3, 5, 11, 21, 43 c) x n = 1, 2, 4, 8, 16, 32 d) x n = 1, 1, 3, 5, 11, 21 e) x n = 1, 0, 2, 2, 6, 10 f) x n = 1, -1, 1, -1, 1, -1 g) x n = 1, -2, 0, -4, -4, -12 2) At any given value for i, the ratio is the same. For example, if we choose the sequences given above from problems a, d, e, and choose i to be 4, then we get a ratio of (26-11):(11-6), which simplifies to 3:1. For any value of i the ratio will be 3:1. One way to interpret this is to say that the sequence given by d is 3 times further from a than it is from e. 3) The sequences from c and f are geometric progressions. c: x n = 1,2,4,8,16,32 x n = 2 n f: x n = 1,-1,1,-1,1,-1 x n = (-1) n 4) Using the first three terms of the sequence (1, r, r 2 ) and plugging it into the recursive formula x n = x n-1 + 2x n-2 gives us r 2 = r + 2. Solving this equation yields the values r = 2 or 1. This means that the only sequences that start with 1, are geometric, and satisfy the given recursive formula must have the first two terms as 1,2 or 1,-1. 5a) = 82.3 b) ( 2 / 5 )(68) + ( 3 / 5 )(90) = 81.2 c) ( 2 / 7 )(23) + ( 5 / 7 )(37) = $33/hr 6) x n = ⅔ 2 n + ⅓ ( 1) n (which is the same as problem D4 above.) F. 1) r 2 = r + 1. Solving this (with the quadratic formula) gives us: r = which is Φ, the Golden Ratio, or 2 r = which we will call Φ^ ( phi hat ). 2 1) The ratio is (Φ 1):(1 Φ^ ). Since Φ 1 = Φ^, and 1 Φ^ = Φ, we can say that the ratio of the distances is Φ^ :Φ. 17

18 3) If the ratio of the distances had been 2:5, then the weights would have been 2 / 7 and 5 / 7. The 7 comes from adding 2 and 5. So, with a ratio of Φ^ :Φ we add Φ^ + Φ, which simplifies to 5, which leaves us with weights of Φ^ / 5 and Φ/ 5. Now, the two geometric sequences we are working with are: x n = 1, Φ, Φ 2, Φ 3 given by x n = Φ n, and x n = 1, Φ^, Φ^ 2, Φ^ 3 given by x n = Φ^ n. Since the Fibonacci sequence sits between these two sequences, with a distance ratio of Φ^ : Φ, we can use the method of weighted average (as with the test average problems from #E5, above) to determine a general formula for the Fobonacci sequence. x n = Φ 5 Φn Φ^5 Φ^ n which simplifies to: x n = 1 5 (Φn+1 Φ^ n+1 ) Binet s Formula!! Isn t it amazing that this formula generates Fibonacci numbers? 11. Hitting Ten After some experimentation, it may be best to make some tables, where each table determines the probability of hitting N. We hope that we can build up to N = 10, and in the process see some pattern. With each table, we flip the coin as many times as needed, and state whether N was hit, or not. (Recall that heads is worth 1 and tails is worth 2. With the below tables S = success, F = failure ). N=1 1 S 2 F P 1 = 1 2 N=2 11 S 12 F 21 S 22 S P 2 = 3 4 There are a few interesting patterns to notice. The top half of any given table is exactly the same as the entire previous table. (Can you explain why this is so?) The bottom half of each 18 N=3 111 S 112 F 121 S 122 S 211 S 212 S 221 F 222 F P 3 = 5 8 N= S 1112 F 1121 S 1122 S 1211 S 1212 S 1221 F 1222 F 2111 S 2112 S 2121 F 2122 F 2211 S 2212 S 2221 S 2222 S P 4 = 11 16

19 table has the same number of successes as the top half, except it s always one more or less, alternating. For example, with the N=3 table, the top half has 3 S s, and the bottom half has one less (2 S s), and with the N=4 table, the top half has 5 S s, and the bottom half has one more (6 S s). Focusing now on the probabilities, the denominators are clearly powers of two. Each numerator is equal to twice the previous numerator, plus or minus 1 (which is the same sequence that appears in Building Trains, in the above problem). But the pattern that is perhaps most useful, is that each probability is just a slight adjustment from the previous table s probability. We simply take the previous probability and add or subtract 1 over the next power of two. For example, P 3 = 3 1, and P = So the question now 8 16 becomes: how can we find a general formula for P N? Well, if we start from the beginning, we see that: P 0 = 1 P 3 = P 1 = P 4 = P 2 = P N = N We can now recognize this as the general sum of a geometric series in the form 1 + x + x 2 + x x n, where x = ½. The general formula for this series is: Now plugging in x = ½ gives us P N = ⅔[1 ( ½) N+1 ] n i 0 x i = xn+1 1 x 1 This formula shows that as N becomes large, P N approaches ⅔. Since our goal is to determine the probability of hitting 10, we simply put 10 in for N and get 683 / 1024, or 66.70% very close to ⅔ indeed. 19

20 12. Lines and Points By looking at the initial example that starts with 4 lines drawn on the page, we can begin to analyze the problem and see some patterns. Starting with 4 lines. The number of points of intersection is 4 C 2 = 6. Each line has 3 points on it. Each point has 2 lines through it, and on these two lines there are a total of = 5 points. From each point, there is 6 5 = 1 point to which a new line can be drawn. The total number of new lines that can be drawn is = 3. Starting with 5 lines. The number of points of intersection is 5 C 2 = 10. Each line has 4 points on it. Each point has 2 lines through it, and on these two lines there are a total of = 7 points. From each point, there are 10 7 = 3 points to which a new line can be drawn. The total number of new lines that can be drawn is = 15. Starting with n lines. The number of points of intersection is n C 2. Each line has n-1 points on it. Each point has 2 lines through it, and on these two lines there are a total of 2(n 1) 1 = 2n 3 points. From each point, there are n C 2 (2n 3) points to which a new line can be drawn. The total number of new lines is: ½[ nc 2 ( n C 2 2n + 3)]. Here is a table of the results: n # of starting lines nc 2 # pts intersection # new lines

21 13. Four Sons Of course, there are different methods for solving this problem. One possibility is recognizing that we are working with a sequence of the form x n = a x n-1 + b. In this case x n is how much money the old man has after giving money to the n th son. We can then see that: x n = ¾(x n-1 4) = ¾ x n-1 3 which then gives us: x 1 = ¾x 0 3 x 2 = 9 16 x x 3 = 27 x x 4 = 81 x , which is also 64 x 4 = 81x where x 0 is the original number of coins, and x 4 is the number of coins after giving coins to the fourth son. So now the question is: what is the smallest positive integer that can be put into x 0 such that it will yield an answer for x 4 that is also a positive integer? Writing a simple computer program can show that the first integral answer to this Diophantine equation is x 0 = 244 (and x 4 = 69). 21

22 14. Secret Santa This problem is certainly very interesting, and there are several different paths to solving it. One approach is to start with just 1 person, create a table that builds up one person at a time, and then look for patterns. With the below table, P is the probability that no person draws his own name. For 1 person, we get P = 0. For 2 people, we get P = 1 2 = 0.5 For 3 people, we get P = For 4 people, we get P = 9 24 = For 5 people, we get P = For 6 people, we get P = For 7 people, we get P = For 8 people, we get P = We notice a few things. Firstly, the answer converges very quickly. In fact, quite surprisingly, once we get past 5 people, the probability (that nobody draws his own name) remains unchanged to three significant digits. The denominators of P are n!, which might be expected because it is the number of ways of arranging n people. The pattern with the numerators is more tricky. If we call the n th numerator h n, then it turns out that: h n = n h n-1 + (-1) n and most interestingly, that P n = 1 1 1! + 1 2! 1 3! + 1 4! 1 5! 1 n! This problem is most famously known as the Hat-Check problem. Euler solved the problem and recognized that the solution (approximately ) approaches 1 / e, as the number of people (n) approaches infinity. Peter Taylor refers to this same problem as the game of snap. He offers a nice solution and a very thorough explanation at 22

23 15. Prime Factorization a) The prime factorization of 109,350 is Given that a square number must have all even exponents in its prime factorization, we can see that we need one more 2 and one more 3. Therefore, the answer is 6. b) First some background if a number has the prime factorization , then we only need to look at the exponents of the 2 and the 5 in order to conclude that the number ends in 6 zeroes. Likewise if a number s prime factorization is , then we know that that number must end in 7 zeroes. Now, to address the question at hand let N = 4273! We know that the number of 2 s in the prime factorization of N must be greater than the number of 5 s (i.e., the exponent of the 2 must be greater than the exponent of the 5). Therefore, we simply need to determine the number of 5 s in the prime factorization of 4273! We can systematically do this by asking ourselves the following questions: How many numbers between 1 and 4273 are Divisible by 5? Answer: 854. Divisible by 25 (which is 5 2 )? Answer: 170. Divisible by 125 (which is 5 3 )? Answer: 34. Divisible by 625 (which is 5 4 )? Answer: 6. Divisible by 3125 (which is 5 5 )? Answer: 1. With a little bit of thought, we can now conclude that the number of zeroes in N must be equal to the sum of the above answers, which is c) Let N = 9,489,150,000. N s prime factorization is To produce a factor of N, we can choose anywhere from 0 to 4 two s, 0 to 5 three s, 0 to 5 five s, 0 or 1 eleven, and 0 or 1 seventy-one. This is now the same kind of problem as the typical wardrobe problem (e.g., How many different outfits are possible if you have 2 pairs of shoes, 5 pants, and 4 shirts to choose from? there are 40 possible outfits). In the case of determining the number of factors of N, we have 5 choices of two s, 6 choices of three s, 6 choices of five s, 2 choices of eleven s, and 2 choices of seventy-one s. Therefore, N has 720 factors. 23

24 For some of the below problems, the following theorem is helpful: If a number Q has exactly n factors, where n is a prime number, then Q s prime factorization must be in the form Q = k (n 1), where k is a prime number. d) There are many possible solutions, including 12 or 20. e) There are two possible variations for this: either k 9, where k is a prime number (e.g., 2 9 = 512), or k h 4, where k and h are prime numbers (e.g., = 891). f) Since n is a prime number (see above theorem), our answer must be k 6, where k is a prime number (e.g., 2 6 = 64). g) There are two possible variations for this: either k 8, where k is a prime number (e.g., 5 8 = 390,625), or k 2 h 2, where k and h are prime numbers (e.g., = 100). h) Since n is a prime number (see above theorem), our answer must be k 12, where k is a prime number, such as 2 12 = i) There are two possible variations for this: either k 14, where k is a prime number (e.g., 3 14 = 4,782,969), or k 2 h 4, where k and h are prime numbers (e.g., = 400). 24