Idea Transcript
Math 211 – Assignment 1 Solutions
1) Section 1.1 #2
Understanding the problem: • 5 people are entered in a racquetball tournament. • The 5 people play each other player exactly once. • If they only play each other exactly once, then Person A playing Person B is the same as Person B playing Person A (AB=BA). • Lets call each of the players A B C D and E, respectively. • We need to find out how many games were played between these five players (given the restrictions above). Devising a plan: Strategy one – Picture/graph Let a point represent each of the players A-‐E and a line between them represent a played game. By drawing as many lines as I possibly can between the points (without allowing any duplicates – this would mean the players played each other more than once), then the total number of lines will give the total number of games played. Strategy two – Create a table List the players in the rows and columns of a table. Let each cell of the table be named by its corresponding combination of row and column player. Select from these possible pairings those that meet the problem’s criteria, namly: • Not those on the diagonals, since these represent a player playing themselves. • Only those pairings on one side of the diagonal since the table is symmetric. • Carrying out the plan: Strategy one – Picture/graph A B Counting the number of lines in the graph shows that there are 10 games played in the tournament of C E 5 players. ! ! ! !" " D
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Strategy two – Create a table
A
B
C
D
E
A
AA
AB
AC
AD
AE
B
BA
BB
BC
BD
BE
C
CA
CB
CC
CD
CE
D
DA
DB
DC
DD
DE
E
EA
EB
EC
ED
EE
A
B
C
D
E
A
AA
AB
AC
AD
AE
B
BA
BB
BC
BD
BE
C
CA
CB
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There are duplicate pairings on either side of the diagonal, so only one side of pairings will be counted.
Diagonals have players playing themselves and cannot be counted.
CE
D
DA
DB
DC
DD
DE
E
EA
EB
EC
ED
EE
The possible unique pairings are those shaded, of which there are 10.
Looking back: The graph strategy suggests another strategy – namely a formula. We can see from the graph that at each point (player) there can be a maximum of four lines extending to another point (since there are only 4 other points to connect to). The number of lines extending out of the points can be found by the number of points multiplied by the number of lines extending from it, or 5×4=20. This calculation, however, counts each line exactly
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twice – once from point x to y, and once from point y to x. In order to only count each line once (since a line represents a match and it is the number of matches that we want to count), we simply have to divide our doubled count by 2, or (5×4)÷2=20÷2=10. It seems reasonable to expect that for any number of players, this pattern would be the same, namely that each of n people could be paired with n-‐1 people, but this would count duplicates, so we would need to divide the doubled count by 2. This gives in general the formula (n×(n−1))÷2. The table strategy shows a pattern that we know something about and can use to verify the formula above, namely it shows that the number of matches is also equal to 1+2+3+4, or the sum of n−1 consecutive numbers (note that is it not the sum of n consecutive numbers or else in this case that would mean the sum 1+2+3+4+5, which is 5 too many). The sum of the first n consecutive numbers was determined by Gauss to be given by the formula (n×(n+1))÷2. Since we only want to find the sum to the n−1 term, we can substitute n−1 for n everywhere it occurs in the formula: n(n + 1) (n − 1)((n − 1) + 1) (n − 1)(n − 1 + 1) (n − 1)n n(n − 1) = = = = 2 2 2 2 2 So we see that the adjustment to Gauss’ formula gives the same formula we determined from the graph. We can also note that this problem is very similar to the handshake problem, and could therefore be solved using any of the other strategies we saw in class (provided we account for duplicates). -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 2) Section 1.1 #8 Understanding the question: • There are less than 10 coins whose value adds to 42 cents. • The coins may be any of pennies, nickels, dimes, or quarters. • We know that pennies are valued at 1 cent, nickels at 5 cents, dimes at 10 cents, and quarters at 25 cents. • Given their values, we know that we can have either 1 or 0 quarters (2 or more quarters is more than 42 cents), we can have 0, 1, 2, 3, or 4 dimes (more than this will give more than 42 cents), we can have 0 – 8 nickels, and we can have 0-‐9 pennies (if we only consider number of coins possible). • Using these observations, we want to determine how many combinations of less than 10 coins will equal 42 cents. Devising a plan: Strategy 1 – A case based table
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Complete a table listing the number of each coin combination adding to 42 cents. We can start with the cases of either having 1 or 0 quarters. If we have one quarter, then we can have either 1 or 0 dimes. If we 1 dime, then we can have either have 1 or 0 nickels. If we have 1 nickel, then we must have 2 pennies. In a similar fashion, we can work through the different cases. Strategy 2 – Algebra/guess and check combo We can describe a combination of quarters, dimes, nickels, and pennies by the equation a(.25) + b(.10) + c(.05) + d(.01) = P where a, b, c, d represent the numbers of quarters, dimes, nickels, and pennies respectively and P represents the value of the collection. Therefore we can substitute .42 for P and “guess and check” for a, b, c, and d. Carrying out the plan Strategy one – A case based table Quarters Dimes Nickels Pennies
1
1
1
2
1
1
0
7
1
0
3
2
1
0
2
7
More than 10 coins
1
0
2
7
More than 10 coins
0
4
0
2
0
3
2
2
0
3
1
7
0
2
4
2
0
1
6
2
More than 10 coins
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Continuing in this fashion (i.e. replacing nickels with pennies, dines with nickels) requires more than 9 coins, so we have uncovered all possible combinations of coins equaling 42 cents. There are 7. Strategy two – Algebra/guess and check combo Using this strategy I want to substitute values for a, b, c, d in the equation a(.25) + b(.10) + c(.05) + d(.01) = .42 Since I want to have less than 10 coins total, the sum of a, b, c, and d needs to be less than 10 (since a, b, c, d represent numbers of quarters, dimes, nickels, and pennies, respectively). Once I started this strategy, however, I realized it was not especially helpful in finding the solution to the problem (the number of combinations equaling 42 cents). To solve the problem, I need to be able to verify in some way that I have in fact discovered every possible combination meeting the requirements of the problem. In fact, as soon as I start to add order to my process of guessing, this strategy becomes similar to the first. Looking back This problem requires that we think systematically about possibilities and organize and record our thinking (strategy one). This strategy can vary, however, in how it is implemented. For instance, rather than starting with the cases of having either one or no quarters, we could start with the cases of having either 2 pennies, or 7 pennies. Note that we must, in every case, have either 2 or 7 pennies since this is the only way to reach 42 from multiples of 5 (nickels, dimes, and quarters all have values that are multiples of 5) with fewer than 10 coins. We can then build the remaining cases from this starting case in a similar fashion as the first strategy. This problem also lends itself to extensions involving higher values and greater variation in denomination. For instance, we could add one dollar bills to the mix and find all the combinations of less than 13 coins/bills having a value of $1.27. 3) Section 1.1 #11 Understanding the problem: • Given only a 4 gallon container and a 9 gallon container, how can we measure 6 gallons? • A student solution for measuring 1 gallon is presented as an illustration.
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We want to explain the student illustration to help make sense of the 6 gallon problem. The student solution illustrates each step of the process of getting to 1 gallon. The process of getting 1 gallon took four steps: o Measure 9 gallons o Fill the 4 gallon container from the 9 gallons o Empty the 4 gallons. This leaves 5 gallons in the 9 gallon container. o Again fill the 4 gallon container from the 5 gallons of water. o Now there is 1 gallon of water in the 9 gallon container.
Devising a plan Strategy one – Guess and check We want to come up with a similar procedure for reaching 6 gallons. We can just try some different fillings and emptyings of the 4 and 9 gallon container, recording each step along the way. Strategy two – A little algebra Since we are emptying and filling only multiples of 4 and 9 gallons, we can set up an equation to represent the problem. We can use the equation 4a + 9b = 6 . In this case, the a and b represent the number of times the 4 and 9 gallon containers are filled and/or emptied. The values of a and b are either positive (adding water) or negative (pouring out water) whole numbers. Since we only have one equation and two unknowns, we cannot solve this equation directly. We will have to employ some level of guess and check using this strategy as well. Carrying out the plan Strategy one – Guess and check After some trial and error, here is one approach that works: • Fill 9 gallon container • Fill the 4 gallon container from the 9 gallons. This leaves 5 gallons of water in the 9 gallon container. • Empty the 4 gallon container. • Fill the 4 gallon container once again from the 5 gallons in the 9 gallon container. • Empty the 4 gallon container. • Pour the 1 gallon of water from the 9 gallon container to the 4 gallon container. • Fill the 9 gallon container. • Fill the 4 gallon container from the water in the 9 gallon container.
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•
Since there was already one gallon of water in the 4 gallon container, only 3 gallons of water are added from the 9 gallons of water in the 9 gallon container. This leaves 6 gallons of water in the 9 gallon container.
• Strategy two – A little algebra In strategy one, we fill the 9 gallon container 2 times, and empty the 4 gallon container 3 times. This can be represented by letting a=-‐3 (the 4 gallon container is emptied 3 times, so a is negative) and b=2 (the 9 gallon container is filled 2 times, so b is positive). Substituting these values into our equation we can verify that this actually give us 6 4a + 9b = 6 gallons: 4(−3) + 9(2) = −12 + 18 = 6 Yea! It does. Looking back This problem is easily extended to the question of how many different amounts of water can be measured using only a 4 and 9 gallon container. The first strategy is best suited to this type of extension since we can follow step by step what is happening. The second strategy only describes the number of times the containers are filled an emptied. Moreover, the filling and emptying of one container is cancelled out (+1-‐1=0) in the formula, making it difficult to work backwards from values of a and b to describe the physical actions of filling and emptying. It actually turns out that we can measure every whole gallon measure of water between 1 and 13 gallons. (I shall leave showing this as a bonus problem worth 3 points). Another extension problem then is the question of which other container sizes will measure every whole gallon of water between 1 and the sum of the container sizes. For instance, we cannot measure 1 gallon of water from a 4 gallon container and a 6 gallon container, but we can measure every amount if we have a 2 gallon container and a 5 gallon container. It turns out that the two containers must be relatively prime, or not share a common factor (don’t worry – we will be learning about all this stuff in another few weeks!). -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 4) Section 1.1 #18 Understanding the problem • Sue Ellen and Angela both saved $51 • They both added money to their piggy banks each week. • Sue Ellen added $1 to her piggy bank each week
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Angela added $2 to her piggy bank each week Sue Ellen started with $7 more in her piggy bank than Angela Suppose Angela started with $X. Then Sue Ellen would have started with $(X+7) How much did Sue Ellen have in her piggy bank when they started?
Devising a plan: Strategy one: Work backwards We can determine how much Sue Ellen started with if we work backwards from Sue Ellen and Angela having $51, subtracting $2 from Angela’s total and $1 from Sue Ellen’s total each week (since working forwards this is how much they added each week) and stopping when Sue Ellen’s adjusted total minus $7 equals Angela’s adjusted total (since at the starting point Sue Ellen had $7 more than Angela). Basically what we are doing is the opposite or inverse of each of the steps that would have taken place moving forward. So, rather than adding we are subtracting, and rather than starting with Sue Ellen having $X+7, we are ending with Sue Ellen having $X-‐7. Strategy two: Use algebra Using what we know from working backwards, we can set up an equation to solve for the number of weeks it takes for the two girls to save for their family trip. Once we know how many weeks it took, we can subtract that many dollars (since she added one dollar per week) from the total of $51 to know how much she started with. The formula is similar to the process of working backwards described in the first strategy. Let a= the number of weeks. Then we want to set up an equation with one side representing Sue Ellen’s savings and the other side representing Angela’s savings as it changes going back in time. Sue Ellen’s savings : 51 − 1a − 7 Starting from $51, we subtract $1 per week since this is how much was added each week, and $7 one time since this is how much more Sue Ellen started with than Angela. Angela’s savings: 51 − 2a Starting from $51, we subtract $2 per week since this is how much was added each week. Setting the two sides equal gives an equation whose solution is the number of weeks it took for their savings to be equal. Formula: 51 − 1a − 7 = 51 − 2a
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Carrying out the plan: Strategy one:
Sue Ellen
Angela
Final week
51
51
-‐1 wk
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50-‐7= 43
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-‐2 wk
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49-‐7= 42
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-‐3 wk
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48-‐7= 41
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-‐4 wk
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47-‐7= 40
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-‐5 wk
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46-‐7= 39
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-‐6 wk
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45-‐7= 38
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44-‐7= 37
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-‐7 wk
Sue Ellen starts with $44 which is $7 more than the $37 that Angela starts with.
Strategy two: Since our equation has only one unknown, we can solve it using standard algebra techniques: 51 − 1a − 7 = 51 − 2a
51 − 1a − 7 + (−51) = 51 − 2a + (−51) −1a − 7 = −2a −1a − 7 + (2a) = −2a + (2a)
a−7=0 a − 7 + (7) = 0 + (7)
a=7
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This shows that it took 7 weeks for the two girls to save a total of 51 dollars from their respective starting amounts. Since Sue Ellen added $1 per week, this means that she added a total of $7 over the 7 weeks, and therefore she started with $51-‐$7=$44. Looking back One observations that can be made looking at the first strategy that can help, or at least speed along, our solving of problems similar to this one is that that by adding $2 each week and reaching $51, Angela will always have an odd value in her piggy bank. Looking at the table from strategy one, we can see that this is the case – we always have values that end in 1,3, 5, 7, or 9. Thus, when looking for when to stop working backwards, we need to stop at a point where Sue Ellen’s savings ends in 0, 2, 4, 6 or 8, since otherwise subtracting 7 will result in a value ending in an even number. (recall that an odd plus/minus an odd equals an even, an even plus/minus an odd equals an odd, and an even plus/minus an even equals an even). In general we can think of this problem as having 4 parts to it: how much each girl started with in their savings, how much each girl added to their savings each week, how many weeks the girls saved, and how much savings the girls ended with. Extension problems based on this problem could therefore vary which parts where unknown. For the current problem, it is unknown how many weeks the girls saved, and how much the girls started with (although we did know a relationship between their starting values). An extension problem, therefore might ask how much each girl would have to save each week to reach $100 if we know that Angela started with $30 more than Sue Ellen, but that Sue Ellen saved twice as much as Angela (Solve this extension problem in full detail in two ways for an additional 3 bonus points). -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 5) Section 1.1 #25 Understanding the problem • Millie counted 30 masts on 8 ships at the tall ship exhibit. • Each ship has either 3 masts or 4 masts. • All eight ships cannot have 4 masts since this would give more than 30 masts total. • All eight ships must have more than 3 masts since this would give fewer than 30 masts total. • Want to find how many of the ships have 4 masts. Devising a plan: Strategy one – Draw a picture
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We can draw the eight ships, and since we know that each ship must have at least 3 masts, we can start by drawing 3 masts on each of the 8 ships. We can then add a fourth mast to each of the ships until the total number of masts reaches 30. The number of ships that have a fourth mast added gives the solution to the problem. Strategy two – Use number sense We can reason through this problem fairly easily just thinking about the numbers. We know that there has to be at least 3 masts on each of the eight ships. This gives us 24 masts. We know that there are 30 masts in total, so we have 6 masts to place. Since none of the ships have more than 4 masts, we can only place one extra mast per ship… we can wait for the punchline until carrying out the plan. Strategy three – A system of two equations Since there are 8 ships total, the number of ships with 3 masts plus the number of ships with 4 masts must equal 8. If we let x = the number of ships with 3 masts, and y = the number of ships with 4 masts, then x + y = 8. We also know that a total of 30 masts were counted. So if there are x ships with 3 masts and y ships with 4 masts, then 3x + 4y =30. Since we have two equations in two unknowns, we can solve for each of our variables x and y using either the technique of substitution or the technique of elimination. In answering the problem, we only need to solve for y (The number of ships with 4 masts). Carrying out the plan: Strategy one: Draw a picture Start by drawing 8 ships with 3 masts each. There 24 masts on the 8 ships. We need to add one mast to each ship until we count 30 masts altogether.
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25 26 27 28 29 30 So there are 6 ships with 4 masts. Strategy 2 – Use number sense As stated above… We can reason through this problem fairly easily just thinking about the numbers. We know that there has to be at least 3 masts on each of the eight ships. This gives us 24 masts. We know that there are 30 masts in total, so we have 6 masts to place. Since none of the ships have more than 4 masts, we can only place one extra mast per ship… Therefore, the remaining 6 masts must be placed on 6 different ships. Thus, 6 ships will have 4 masts. Strategy 3 – A system of two equations We know that the situation can be modeled by the system of two equations: x+y=8 3x + 4y = 30 Using the technique of substitution, we can solve the first equation for x, substitute this expression for x in the second equation, then solve the equation for the remaining y variable (the number of ships with 4 masts).
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x+y=8 x=8−y 3x + 4y = 30 3(8 − y) + 4y = 30 24 − 3y + 4y = 30
24 + y = 30 24 + y + (−24) = 30 + (−24) y=6 Thus solving our system of equations also results in a solution of 6 boats having 4 masts. Looking back This is a relatively easy question; however, extending it to include ships of 5 masts, for instance, invites thinking about solving systems of more than two equations (either informally or formally). Adding more options for numbers of masts that each boat can have may also lead to questions of how many combinations of ships there might be. For instance, suppose Millie counted a total of 40 masts on 11 ships, and there were ships with 3, 4 or 5 masts. Without any additional information we can ask for all the different combinations of ships Millie could have seen (for now we have two equation in 3 unknowns – we would need one more piece of information relating the number of ships and the number of masts in order to solve for a single solution). If we add an additional piece of information, say that there are twice as many 3 mast ships as there are 4 mast ships, and only one third as many 5 mast ships as there are 3 mast ships, then we can solve for a single solution to how many of each ship type she counted. (Solve both these problems for 3 bonus points). -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 6) Making Connections Section 1.1 #4 The fourth expectation in the Process Standards on Problem Solving is to “monitor and reflect on the process of mathematical problem solving”. Polya’s fourth problem solving step (Looking Back) addresses this expectation quite explicitly by asking us to verify, or check, that our result offers a reasonable solution to the problem posed; by asking us to reflect on our work, thinking about other ways we might demonstrate the same solution; and by asking us to look more deeply at the question, thinking of variations and extensions that may prove interesting. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐
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7) Section 1.2 #28 Understanding the problem: • Consider the given pattern of dots in the shape of a triangle • The sequence given by the number of dots in each successive term of the sequence are called the “triangular numbers” • Want to write the next three numbers in the sequence • Want to determine how many dots are in the 100 th figure • Want to determine a general formula for the number of dots in the nth figure. • By counting the sum of the levels of the triangle, we can see that the number of dots in any term is equal to the sum of numbers up to and including that term. Ex. Term 4 1 +2 =10 +3 +4 • We can also orient the pattern so it is more “square”: • Now we can see the summands of the previous example in terms of columns rather than rows. • The more square orientation suggests a possible array structure. • If we duplicate and flip our square pattern (the red dots), then line it up with the original square pattern (the black dots), then we create a rectangular array whose total dots are easy to count – the total is just the product of the number of dots along the horizontal and the number of dots along the vertical. So the total is 5×4=20. Recall though that by construction, this array is just two copies of the original figure we are trying to count.
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Devising a plan Given what we know now about the pattern, it seems like we can find the next three numbers in the sequence simply by adding the next 3 consecutive numbers. So, since we know that the 4th term is 10, the 5th term can be found by taking the sum 1+2+3+4+5 or 10+5, the 6th term can be found by taking the sum 1+2+3+4+5+6 or 10+5+6, and the 7th term can be found by taking the sum 1+2+3+4+5+6+7 or 10+5+6+7. To find the 100th term, we could in theory find the sum of the first 100 numbers, but this would take a long time. In this case it would be better to use a formula. We can use the square array to come up with a general formula for the number of dots in the nth term of the sequence. As shown by the example of the 4th term, constructing a rectangular array from two copies of a “square” triangle gives one dimension that is equal to the value of the term, and another dimension that is one more than the value of the term. If we generalize this result in terms of any term n, this gives the number of dots in the rectangular array as n(n+1). However, the rectangular array represents 2 copies of the original triangle, so the number of dots in just one copy would be equal to one half this n(n + 1) product, or . 2 Carrying out the plan The next three numbers in the sequence are: 5th term: 10+5=15 Dots from 4th term =10 +5 6th term: 10+5+6=15+6=21 7th term: 10+5+6+7=21+7=28 We can verify these results using the formula: n(n + 1) 5(5 + 1) 5(6) 30 = = = = 15 5th term: 2 2 2 2 n(n + 1) 6(6 + 1) 6(7) 42 = = = = 21 6th term: 2 2 2 2
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7th term:
n(n + 1) 7(7 + 1) 7(8) 56 = = = = 28 2 2 2 2
We can use the formula to determine the number of dots in the 100th term: n(n + 1) 100(100 + 1) 100(101) 10100 = = = = 5050 100th term: 2 2 2 2 Looking back: It seems that we have again come across a “handshake problem” problem, and therefore we could use the array technique described in this problem for solving problems like the tournament problem (question 1 on this assignment). Since other techniques for solving problems of this type have already been explored, I will leave the looking back component of this problem at that. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 8) Section 1.2 #29 Understanding the problem • Consider the sequence of dots in the shape of a square. • The sequence given by the number of dots in each successive term of the sequence are called the “square numbers” • Want to write the next three numbers in the sequence • Want to determine how many dots are in the 100 th figure • Want to determine a general formula for the number of dots in the nth figure. • Since the pattern is square, the horizontal and vertical dimensions are equal. • The horizontal dimension (the number of dots lying on the horizontal) is equal the sequence term, i.e. n. • We can count the number of dots as either a sum of n rows (or columns) of n dots, or in terms of a product given by the n x n square array. 5 +5 +5 =25 5 +5 +5 5×5=25 5
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Given the pattern’s array structure, and that the horizontal and vertical dimensions are equal to the term in the sequence, we can calculate the number of dots in any term of the sequence using the formula n×n, or n2. Therefore, we can either find the number of dots in the 5th, 6th, and 7th terms of the sequence by using the formula, or finding the sum of 5 5’s, 6 6’s, and 7 7’s respectively.
5th term: 5+5+5+5+5=5×5=25 6th term: 6+6+6+6+6+6=6×6=36 7th term: 7+7+7+7+7+7+7=7×7=49 For the 100th term, finding the sum of 100 100’s is more than a bit tedious, so we can simply use the formula: 100th term: 100×100=10000 Another way to visual this problem is given by problem 12, so I will leave it till then -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 9) Section 1.2 #30 • Consider the sequence of dots in the shape of a pentagon. • Each term of the sequence is built from the combination of a triangle and a square. Triangle – triangular number Square – square number • We can use what we already know about triangular and square numbers to make sense of the pentagonal numbers and generalize to a formula for the nth figure. • The square component of the figure corresponds exactly with the square number sequence; that is, the number of dots in the square component of the nth figure is equal to n×n or n2. • The triangle component of the nth figure is off by one; that is, the triangular component of the nth pentagonal number is the n-‐1th triangular number. So for the third pentagonal number given in the example above, the triangle component is given by the second triangular number, and not the third.
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The formula for the triangular number in the nth term of the pentagonal sequence can be generated from the formula for the triangular numbers in problem #28, n(n + 1) namely . In order to reduce confusion over a commonly named variable lets 2 k(k + 1) instead write where k refers to the kth term in the sequence of triangular 2 numbers. What we are trying to find is the triangular number that is one term prior to the term of the pentagonal number. Since we are speaking of the nth pentagonal number in the sequence, one less is n-‐1, and so we want to solve for k=n−1.
k(k + 1) (n − 1)((n − 1) + 1) (n − 1)(n − 1 + 1) (n − 1)n = = = 2 2 2 2
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Triangle – triangular number Rectangle – oblong number Re
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So the number of dots in the triangular component of the nth pentagonal number is (n − 1)n given by . 2 Combining this result with what we know about the square component of the pentagonal number gives us the compete formula for the number of dots in the nth (n − 1)n (n − 1)n = n2 + pentagonal number, namely (n × n) + . 2 2 We can also consider the nth pentagonal number in terms of the nth triangular number and a rectangular number (oblong number).
The triangular component corresponds exactly with the triangular number sequence; that is, the number of dots in the triangular component of the nth n(n + 1) pentagonal number is the same as the nth triangular number given by . 2 The horizontal dimension of the rectangular component of the nth pentagonal number is equal to n – there are as many dots drawn horizontally as there are terms in the sequence. The vertical dimension of the rectangular component of the nth pentagonal number is one fewer than n, so (n-‐1) Since the dots in the rectangular component of the pentagonal number are in a rectangular array, we can find the number of dots by finding the product of the
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number of dots in the horizontal and vertical dimensions. This gives us the general formula for the nth rectangular component as n×(n-‐1). Combining these two results (the formula for the triangular and rectangular components of the nth pentagonal number) gives us another formula for the n(n + 1) number of dots in the nth pentagonal number, namely n(n − 1) + . 2 Now we have two formulas that supposedly give the same information (what is the nth pentagonal number). Lets see if they are the same by trying to make the second formula look like the first:
n(n + 1) 2n(n − 1) n(n + 1) 2n(n − 1) + n(n + 1) = + = 2 2 2 2 2 2 2n − 2n + n + n = 2 2 2 2n + n − n = 2 2 2n + (n 2 − n) = 2 2 2n n2 − n = + 2 2 (n − 1)n = n2 + 2 Thus, despite counting the pentagonal numbers in different ways, the resulting formulas are equivalent. Using the first formula, we can now find the 5th, 6th, 7th and 100th pentagonal number:
n(n − 1) +
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(n − 1)n (5 − 1)5 (4)5 20 = 52 + = 25 + = 25 + = 25 + 10 = 35 2 2 2 2 (n − 1)n (6 − 1)6 (5)6 30 = 62 + = 36 + = 36 + = 36 + 15 = 51 6th term: n 2 + 2 2 2 2 (n − 1)n (7 − 1)7 (6)7 42 = 72 + = 49 + = 49 + = 49 + 21 = 70 7th term: n 2 + 2 2 2 2 100th term: (n − 1)n (100 − 1)100 (99)100 9900 n2 + = 100 2 + = 10000 + = 10000 + = 10000 + 4950 = 14950 2 2 2 2 -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 10) Activity Set 1.1 #6 5th term: n 2 +
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Given the following third term in a sequence, we are asked to create 3 sequences that would have this third term and explain how someone might build the next few figures in the sequence. Third term: Sequence I: 1st term 2nd term 3rd term 4th term • Each term in this sequence is given by an (n+1) by (n+1) square array that is missing a square in the upper right hand corner. • We can build each successive term by adding a “L” shape of squares to the outside of the previous figure. • A general formula for the number of squares in the nth term of this sequence is (n+1)(n+1)-‐1. Sequence II: st nd rd th 1 term 2 term 3 term 4 term • Each successive term in the sequence adds a row of 4 squares to the base of the preceding tem
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There is a constant difference of 4 between each the terms, so we know that there will be some “4 times something” component to our formula. Each successive term another 4 squares, so the “something” will be related to n – the sequence term. We can think of building a formula in two ways from this point: either we start with 7 squares for n=1 and add 4 squares to each successive term, or we start with 3 squares, and add 4 squares starting from n=1. The first case is then 7+4(something in n), but we need to make sure that for n=1, the formula gives an output of 7 so the 4 must multiply with something equal to zero. This gives the formula 7+4(n-‐1). In the second case, our formula would be 3+4n since we want the n=1 term to equal 7. Note, however, that the two formulas are the same when simplified.
Sequence III:
1st term 2nd term 3rd term 4th term • Beginning from 3 columns of 4 squares and a single square in the fourth row of a fourth column, we add a single square at each successive term, building a fourth column of 4 squares, then a fifth column and so on, starting from the forth row. • This pattern will give a rectangle in every term that is a multiple of 4. Each multiple of 4 will increase the width of the rectangle by one. Thus, the 4th term is a 4X4 rectangle, the 8 term will be a 4X5 rectangle, the 12th term will be a 4X6 rectangle, and so on. • Adding 1 square at each successive term in the sequence gives a constant difference of 1; thus we can expect a formula to for finding the number of squares the nth term of the sequence to include a “1 times something in n” term. As with the previous sequence, the formula we come up with will be different depending on how we “look” at the pattern. • One formula is given by 12+n. In this case, the additional single square is added starting with the first term, and is added to 12 preexisting squares. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 11) Activity set 1.2 #1 • The problem asks us to find the 6th and 7th figures of a sequence of tile figures. The first tile figure contains a single tile, and each successive figure adds one tile to each the height and width. Thus, starting from one tile, two additional tiles are added to
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each successive figure. So we have, 1, 3, 5, 7, 9, … hum, looks a lot like another sequence we know. o f tiles in the 6th The number o f tiles in the 7th The number term figure are given by the figure a re given by the sum of sum o f t he t iles l ying v ertically the t iles l ying vertically + the + the number of tiles lying number of tiles lying vertically. horizontally. S o 6+5 = 11 So 7 +6 = 1 3
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6th term 7th term The sequence of tile figures gives the sequence of odd numbers. The 10th odd number can be given by the 10th figure in the tile sequence.
We can count the tiles as the number vertically + the number horizontally 10+9 = 19
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To determine the number of tiles in the 20th figure, we can follow the same pattern observed in the 6th, 7th, and 10th figures. The number of tiles in the 20th figure will be the sum of the tiles lying vertically and the number of tiles lying horizontally. The number of tiles lying vertically will be equal to 20, and the number of tiles lying horizontally will be equal to 19. Thus the number of tiles will be equal to 20+19 =39, which is the 20th odd number. The 50th figure is expected to similarly follow the pattern of having 50+49=99 tiles. The pattern suggests that for the nth figure, there will be n+(n-‐1) tiles. Simplifying gives us the more common “odd number” formula of 2n-‐1 – which is expected given that the sequence of tiles represents the sequence of odd numbers.
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-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 12) Writing and Discussion, Section 1.2 #2 • The question asks us to explain with a diagram why n2=(n-‐1)2+(n-‐1)+n 12 22 = 12 +2+1
32 = 22 +3+2
Blue = 2 tiles + 1 tiles Red = 3 tiles + 2 tiles Green = 4 tiles + 3 tiles
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42 = 32 +4+3
As shown in the diagram, each successive square number is build from the previous square number by wrapping a set of “L” shaped tiles around it. The number of tiles in the L shape, as we learned in problem 11, is given by the sum of the number of tiles lying horizontally and the number of tiles lying vertically. In this case, the number of tiles lying horizontally is the same as n. For example, for n=4 , or the 4th square number, we have a 4X4 square that is built from a 3X3 square, a horizontal row of 4 tiles, and a vertical column of 3 tiles (See the picture for 42 above) We can also look at this from a more “numerical” perspective. Consider again the case of 42.
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42 = 4 × 4 =4+4+4+4 = (3 + 1) + (3 + 1) + (3 + 1) + (3 + 1) = 3+ 3+ 3+ 3+1+1+1+1 = (3 + 3 + 3) + 3 + 4 = (3 × 3) + 3 + 4 = 32 + 3 + 4
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The above substitutions further support our claim that n2=(n-‐1)2+(n-‐1)+n
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