Math 202 Test Problem Solving, Sets, and Whole Numbers 19 September, 2008

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1 Math 202 Test Problem Solving, Sets, and Whole Numbers 19 September, 2008 Ten questions, each worth the same amount. Complete six of your choice. I will only grade the first six I see. Make sure your name is on the top of each page you return. Explain your reasoning for each problem whenever appropriate; that helps me give partial credit. Perform scratch work on scratch paper; keep your explanations clean. Make final answers obvious by boxing or circling them. nd remember to read and answer the entire question. 1 Problem solving: searching This problem is from G. Pólya s How to Solve It. mong Grandfather s papers a bill was found: 72 turkeys $ 67.9 The first and last digits of the number that obviously represented the total prive of those fowls are replaced here by blanks, for they have faded and are now illegible. What are the two faded digits and what was the price of one turkey? Explain your reasoning. Hint: The total is less than $500 and greater than $300. The price per turkey is not a whole number of dollars, but it is a whole number of cents. Structure your explanation according to Pólya s principles: understanding the problem, devising a plan, carrying out the plan, and looking back. 1.1 Solutions Understanding the problem We need a total number T such that T/72 is an integer number of pennies. From the hint, we know the first digit is either 3 or 4. nd the total (in pennies) must be a multiple of 72, an even number, so we can limit the final digit to one of 0, 2, 4, 6, or Devising a plan We could form a list. There are 10 possible combinations to try. In each line, we list the first and last digit, the total price, and the price per turkey. ny lines where the price per turkey is an integer number of pennies are possible solutions. 1

2 1.1.3 Carry out the plan First Last Total Total/ The only entry that produces an integer number of pennies is /72 = 5.11, so the price must have been $ The price per turkey is $ Looking back We can check the result easily, = In the longer expansion of T/72, some of the patterns look familiar. The decimal expansions eventually repeat a constant digit. This is the effect of dividing by 3 or 9, and both are factors of lternate method Looking back suggests another method for solving the problem. Let f and d be the first and last digits. Then the number of pennies in the total is P = 10000f + d = 1000(10f + 6) + d We need this number to be divisible by 8 and by 9. Note that 1000 is divisible by 8 already, so we need only that d be divisible by 8. We could form a smaller, five entry list to find d. Or we could note that (790 + d)/2 = d/2 must be a multiple of four, thus 95 + d/2 must be a multiple of four (the 300 portion already is divisible by 4), and d = 2. Now we have the number of pennies as P = 10000f and we must find an f such that P is divisible by 9. n integer is divisible by 9 if the sum of its digits are divisible by 9 (we will discuss this later in class). So f = f + 24 must be divisible by 9. This is true only when f = 3. Thus f = 3, d = 2, and the final price again is $ with the cost per turkey being $ Problem solving: finding and following dependencies You are watching the line form in front of a bank teller before the teller opens. The line forms and is served in order of arrival. 2

3 ill arrives 10 minutes after Joan. Robert held the door open for Morgan and thus arrived after Morgan. Chris rushed in the door shortly before Morgan and Robert arrived, and no one else arrived between Robert, Morgan, and Chris. Chris thankful to see only one person in line. ssuming that these are all the people who arrived, what is the order in which the people are served? Explain how you arrive at your answer. Structure your explanation according to Pólya s principles: understanding the problem, devising a plan, carrying out the plan, and looking back. 2.1 Solution Understanding the problem The problem is to place the five people ill, Joan, Robert, Morgan, and Chris in order of arrival. The arrivals are described in five statements; the third bullet point contains two statements. Most of the statements directly relate relative arrival times between people. Two are somewhat implicit: No one arrived between Robert, Morgan, and Chris. Only one person arrived before Chris. Chris appears in both these rules Devising a plan We try placing people in order by following the dependencies in the statements. Not all statements are explicit, and we will keep those statements in mind throughout. ecause Chris appears in the more implicit rules, we start by placing Chris. Others will be placed relative to Chris Carrying out the plan The first person to place is Chris. We know Chris arrived before Morgan and Robert and behind one unspecified person. The unspecified person must be one of the remaining two people, either ill or Joan. ut we know Joan was in line when ill arrived. ecause no one was served before the teller opened, Joan must have been first. From the last and first rules, we know that the line begins with Joan, and Chris. The rules relating Chris, Morgan, and Robert place Morgan and Robert after Chris. ecause Robert held the door open, Morgan was first. nd we know that Chris, Morgan, and Robert arrived without anyone else between them, so now we have 3

4 Joan, Chris, Morgan, and Robert. Only one person remains, ill. He arrived after Chris, because Chris saw only one person waiting. nd no one arrived between Chris, Morgan, and Robert, so ill must be last. The final arrival order is Joan, Chris, Morgan, Robert, and finally ill Looking back We can verify each statement in the problem against the ordering. The key here was keeping the implicit rules in mind and trying to make them concrete quickly. 3 Problem solving: patterns What is the units digit (last digit) of 3 25? Using the pattern found for 3 k, find the units digit of 9 25, 27 25, and 81 25? Remember that 9 = 3 2, 27 = 3 3, and 81 = 3 4. Now find the units digit of Given the units digit of 3 25 and 5 25, what is the units digit of 15 25? Explain your reasoning throughout. Structure your explanation according to Pólya s principles: understanding the problem, devising a plan, carrying out the plan, and looking back. 3.1 Solution Understanding the initial problem The multi-part problem has three components: finding the units digit of 3 25, 9 25 = 3 50, = 3 75, and = ; finding the units digit of 5 25 ; and finding the units digit of ecause 15 = 3 5, = , so the last portion will be answered by the two earlier pieces Devising a plan Remembering that we need track only the last digit, we can skip breaking this into multiple sub-problems. We simply make a table of units digits for powers of 3 and powers of 5. Then we find where 25, 50, 75, and 100 would be in the powers of 3, and where 25 would be in the powers of 5. Then we can multiply the units digits of 3 25 and 5 25 to obtain the units digit of

5 3.1.3 Carrying out the plan First we consider powers of 3: k units digit of 3 k The units digit of powers of 3 repeats every 4 entries. We can write 25 = , so 3 25 will have a units digit of 3. ecause 9 25 = 3 50 and 50 = , the units digit here is 9. nd = 3 75 with 75 = , so the units digit is 7. Finally = and 100 = , so the units digit is 1. Summarizing, 3 25 has a units digit of 3, 9 25 has a units digit of 9, has a units digit of 7, and has a units digit of 1.. The powers of five are much less complicated: k units digit of 5 k The units digit of 5 k is 5 when k 1, so 5 25 has a units digit of 5.. Now = , so we first multiply their units digits. That leaves 15, with its units digit of 5. So has units digit Looking back The result is difficult to check without modular arithmetic and a computer or calculator. ut we can at least check that makes sense by considering only the units digit. We already discussed that we need only the units digit, and we demonstrated that 5 k has a units digit of 5 for any k 1. So the result of 5 matches. nd notice that the powers of powers of three (3 25, 9 25, etc.) walked along the table. Note that powers multiply, so we multiply the expansion for 24 = by an integer k to see that k ( ) = (6k) 4 + k. So for 9 = 3 2, we look at the k = 2 line, 27 = 3 3 leads to the k = 3 line, and so on. 5

6 4 Set theory: definitions and relations nswer clearly: Write out the set {2x + 1 x J, 3 < x 2} by listing all its elements appropriately. Write the following set succinctly in set-builder notation: {7, 10, 13, 16, 19, 22}. Given two sets and, when is? Given two sets and, when is? What is a proper subset? What are two ways to write an empty set symbolically? Fill in each with the most appropriate relation (,,, =,, or no relation at all): 1 { } {x/3 x is a whole number less than five} for all sets 6 {1, 2, 3, 4, 5, 6} {x x is an odd number divisible by two} {6} {1, 2, 3, 4, 5, 6} {1, 2, 4, 8,..., 64} {1, 3, 9, 27,..., 729} Under what circumstances are each of the following true for sets and? = \ = for any set 4.1 Solutions nswers: { 3, 1, 1, 3, 5} {7 + 3x x J, 0 x 5} when all elements of are in. Logically, x x. when all elements of are in. Logically, x x. proper subset, here denoted, is one where but there is some x such that x /. Two expressions: {}, Relations: 1 {x/3 x is a whole number less than five}, because 3/3 = 1 and 3 < 5. { } for all sets. No relation at all. The set { } is not empty. 6

7 6 {1, 2, 3, 4, 5, 6} = {x x is an odd number divisible by two} {6} {1, 2, 3, 4, 5, 6} {1, 2, 4, 8,..., 64} {1, 3, 9, 27,..., 729} Conditions: = when. On the left-hand side, = {x x x }. We can write the right-hand side as = {x x }. So we need x to always be true when x is true, or x x. This is the subset relation. \ = when =. The left is \ = {x x x / }, and the right is just = {x x }. If the result has any entries, then it must have elements x such that x and x /. This is impossible, so the result must be empty. for any set when =. Here is a different way to think about the problem. If for any, then 1 and 2 for two specific sets. Then we must have ( 1 2 ). We could have picked any two sets as 1 and 2, including disjoint sets where 1 2 =. Thus, or =. 5 Set theory: operations and relations Define the result of the following operations using set-builder notation: \ ( \ ) C Draw two Venn diagrams for each of the following illustrating different ways and can be related (e.g. one is a subset of another, all sets are distinct, or other possibilities): \ ( \ C) ( \ C) 5.1 Solution = {x x or x } \ = {x x and x / } ( \ ) C = {x (x and x / ) or x C} 7

8 Two examples of : Two examples of \ : \ \ = Two examples of ( \ C) ( \ C): ( \ C) ( \ C) ( \ C) ( \ C) C C 6 Cardinalities and one-to-one mappings Let and be finite sets. Explain why. Explain why. If =, how are and related? Venn diagrams may be useful. Find a one-to-one correspondence between the following pairs of sets. The correspondence can be a listing if the sets are finite or an equation relating elements if the sets are infinite. = {1, 2, 4, 8,..., 64} and = {1, 3, 9, 27,..., 729} = {red, orange, yellow, green, blue, indigo, violet}, the traditional colors of the rainbow (roygbiv), and be the days of the week. 8

9 = {x x is an even positive integer} and = {x x is an odd positive integer} 6.1 Solution because removes elements and must be no larger than. because adds elements and must be no smaller than. If =, then =. If there were some element in but not in, then >. The same argument applies to elements in but not in. Thus there are no elements that are in one set but not the other, and the sets are equal. (Venn diagrams...) = {1, 2, 4, 8,..., 64} = {2 k k J, 0 j 6} and = {1, 3, 9, 27,..., 729} = {3 k k J, 0 j 6}. So a one-to-one correpondence maps 2 k to (and from) 3 k. One such mapping is red Monday, orange Tuesday, yellow Wednesday, green Thursday, blue Friday, indigo Saturday, and violet Sunday. Interestingly enough, Isaac Newton insisted on including violet in the standard rainbow colors so there would be seven colors. The mapping f(a) = a + 1 is a one-to-one mapping from to. This mapping has f 1 (b) = b 1 as its inverse mapping from to. 7 Properties of operations Is the union operation closed over all sets? Is subtraction closed over whole numbers? If either is not, provide an example. Define the commutative and associative properties for addition on whole numbers, and union of sets. Illustrate both properties for whole numbers with number lines. Define the distributive property for multiplication over addition for whole numbers, and 9

10 unions over intersections for sets. Illustrate the distributive property for sets with Venn diagrams. 7.1 Solutions The union is closed over sets, but subtraction is not closed over whole numbers. The subtraction 1 3 implies there is a number k such that 3 + k = 1. That number k cannot be a whole number. For addition of whole numbers, the commutative property states that a+b = b+a. The associative property states that (a + b) + c = a + (b + c). For the union of sets, the commutative property states that =. The associative property states that ( ) C = ( C). 8 Illustrating multiplication Illustrate the following properties using sketched areas or volumes: the commutative property, the associative property, the multiplicative identity, and the distributive property of multiplication over addition. Remember that a factor of an integer M is another integer d that cleanly divides into M. So 8 is a factor of 40 because 5 8 = 40. Using boxes of area 16, find all the factors of 16. Which property allows you to construct only some of the boxes? Using a box of area 15 and one of area one, illustrate 16 = Draw a similar illustration for 16 = q 7 + r where q and r are positive integers, and r is the smallest positive integer for which there is a q. You have just illustrated division with remainders. 8.1 Solutions (to draw...) 10