The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in

The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 7 or higher.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Fair Game? Five identical balls are numbered 1 to 5 and then placed in a bag. The bag is shaken so the balls are mixed up. A player draws three of the balls from the bag and then determines the sum of the numbers on the balls. The player wins the game if the sum of the numbers on the balls is odd. Determine the probability that a player wins the game. Express your answer in fraction form. Strand Data Management and Probability

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem Problem of the Week Problem C and Solution Fair Game? Five identical balls are numbered 1 to 5 and then placed in a bag. The bag is shaken so the balls are mixed up. A player draws three of the balls from the bag and then determines the sum of the numbers on the balls. The player wins the game if the sum of the numbers on the balls is odd. Determine the probability that a player wins the game. Express your answer in fraction form. Solution In order to determine the probability, we must determine the number of ways to obtain a sum that is odd and divide it by the total number of possible selections of three balls from the bag. We will count all of the possibilities by systematically listing the possible selections. Select 1 and 2 and one higher number: 123, 124, 125; three possibilities. Select 1 and 3 and one higher number: 134, 135; two possibilities. Select 1 and 4 and one higher number: 145; one possibility. Select 2 and 3 and one higher number: 234, 235; two possibilities. Select 2 and 4 and one higher number: 245; one possibility. Select 3 and 4 and one higher number: 345; one possibility. By counting the outcomes from each case, there are 3 + 2 + 1 + 2 + 1 + 1 = 10 possible selections of three balls from the bag. We must now determine how many of these selections have an odd sum. We could take each of the possibilities, determine the sum and then count the number which produce an odd sum. However, we will present a different method which could be useful in other situations. The sum of three numbers is odd in two instances: there are three odd numbers or there is one odd number and two even numbers. Selections with three odd numbers: 135. Selections with one odd number and two even numbers: 124, 234 and 245. The total number of selections where the sum is odd is 1 + 3 = 4. Therefore the probability of winning, by selecting three balls with an odd sum, is 4 10 = 2. A game is considered fair if the 5 probability of winning is 50%. In this game, the winning probability, as a percentage, is 40%. Extending the problem: If nine balls numbered 1 to 9 are placed in the bag and three balls are removed, then the probability that the sum of the numbers on the balls is odd is 10 = 47.6%. Can you verify this? 21

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Can You DIGIT? The three digit number 5A4 is divisible by 4 and the three digit number 37B is divisible by 3. Determine the largest positive difference between 5A4 and 37B. Strands Data Management and Probability, Patterning and Algebra

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Can You DIGIT? Problem The three digit number 5A4 is divisible by 4 and the three digit number 37B is divisible by 3. Determine the largest positive difference between 5A4 and 37B. Solution The largest positive difference will occur when 5A4 is a large as possible and 37B is as small as possible. We are therefore looking for the largest possible value of A and the smallest possible value of B. A number is divisible by 4 if the last two digits of the number are divisible by 4. For 5A4 to be divisible by 4, the only possible values of A are 0, 2, 4, 6, and 8 since 04, 24, 44, 64, and 84, respectively, are the only two digit numbers that end in 4 and are divisible by 4. (04 is technically not a two-digit number but a larger number could end in these two digits.) Since we want 5A4 to be as large as possible, A = 8 and 5A4 = 584. A number is divisible by 3 if the sum of its digits is divisible by 3. For 37B to be divisible by 3, the only possible values of B are 2, 5, and 8. When B = 2 the number is 372. It is divisible by 3 since 3 + 7 + 2 = 12 is divisible by 3. When B = 5 the number is 375. It is divisible by 3 since 3 + 7 + 5 = 15 is divisible by 3. When B = 8 the number is 378. It is divisible by 3 since 3 + 7 + 8 = 18 is divisible by 3. No other three digit numbers of the form 37B will be divisible by 3 since no other values of B, other than 2, 5, and 8, give a digit sum that is divisible by 3. Since we want 37B to be as small as possible, B = 2 and 37B = 372. The largest positive difference occurs when A = 8 and B = 2 so that 5A4 37B = 584 372 = 212. the largest positive difference is 212.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Decathlon s On Willy Makit is participating at the local level of a decathlon in which he competes in 10 events. In each event the maximum possible score is 10 points. To advance to the regional level of the decathlon a competitor must earn a minimum total score of 75 points. On the first five events, Willy had an average score of 6.8. On the next three events his average score was 7.2. What must Willy average on his final two events in order to have a total score of exactly 75 points? Strands Data Management and Probability, Number Sense and Numeration

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Decathlon s On Problem Willy Makit is participating at the local level of a decathlon in which he competes in 10 events. In each event the maximum possible score is 10 points. To advance to the regional level of the decathlon a competitor must earn a minimum total score of 75 points. On the first five events, Willy had an average score of 6.8. On the next three events his average score was 7.2. What must Willy average on his final two events in order to have a total score of exactly 75 points? Solution To determine an average, we add the numbers together and divide the total by the number of numbers. Total Average = Number of Numbers So to determine the total we would multiply the average by the number of numbers. Total = Average Number of Numbers Total Score for First 5 Events = 6.8 5 = 34.0 Total Score for Next 3 Events = 7.2 3 = 21.6 Total Score to Move On = 75.0 Total Score Needed for Last Two Events = 75.0 34.0 21.6 = 19.4 Average Score for Final 2 Events = 19.4 2 = 9.7 Willy needs to average 9.7, near perfect scores, on his last two events to have a total score of exactly 75 points in order to advance to the regional round of the decathlon. The pressure is definitely on!

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C What is the Deal Anyway? Three playing cards are placed in a row, from left to right. One card is a club ( ), one card is a diamond ( ), and one card is a heart ( ). The number on each card is different. One card is a four, one card is a five and one card is an eight. Using the following clues, determine the exact order of the cards, from left to right, including the suit and number. 1. The club is somewhere to the right of the heart. 2. The 5 is somewhere to the left of the heart. 3. The 8 is somewhere to the right of the 4. Strand Data Management and Probability

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution What is the Deal Anyway? Problem Three playing cards are placed in a row, from left to right. One card is a club ( ), one card is a diamond ( ), and one card is a heart ( ). The number on each card is different. One card is a four, one card is a five and one card is an eight. Using the following clues, determine the exact order of the cards, from left to right, including the suit and number. 1. The club is somewhere to the right of the heart. 2. The 5 is somewhere to the left of the heart. 3. The 8 is somewhere to the right of the 4. Solution There are six ways to order the suits: (,, ), (,, ), (,, ), (,, ), (,, ), and (,, ). The first clue tells us that the club is somewhere to the right of the heart. We can eliminate the first three from the above list since the club is to the left of the heart in each case. There are now only three ways to order the suits: (,, ), (,, ), and (,, ). The second clue says that the 5 is somewhere to the left of the heart. This means that the heart cannot be the leftmost card. This eliminates the last two possibilities from the above list. The only possibility is (,, ) and the 5 must be the five of diamonds giving us (5,, ). The third clue tells us that the 8 is somewhere to the right of the 4. With only two spots to decide, we can conclude that the 8 must be in the rightmost (third) spot and the last two cards are the 4 of hearts and the 8 of clubs. This gives us (5, 4, 8 ). the cards are dealt in the following order: 5 of diamonds, 4 of hearts, and 8 of clubs.

Problem of the Week Problem C Fractions to the Max In diagram 1 below, four-fifths of a circle is shaded dark grey. In diagram 2, one-sixth of a circle is shaded dark grey. In diagram 3, three-halves of a circle is shaded dark grey. Diagram 1 Diagram 2 Diagram 3 The three fractions illustrated have been formed using the positive integers 1, 2, 3, 4, 5 and 6 exactly once. When the three fractions are added together, the sum is 4 5 + 1 6 + 3 2 = 24 30 + 5 30 + 45 30 = 74 30 = 37 15 = 2 7 15 Suppose that six different numbers are selected from the set {1, 2, 3, 4, 5, 6, 7}. These six numbers are then used to form three fractions. Each of the three numerators and three denominators must contain a different digit from the selection. The three fractions are then added together. Determine the largest possible sum that can be obtained. Strands Data Management and Probability, Number Sense and Numeration

Problem Problem of the Week Problem C and Solution Fractions to the Max Suppose that six different numbers are selected from the set {1, 2, 3, 4, 5, 6, 7}. These six numbers are then used to form three fractions. Each of the three numerators and three denominators must contain a different digit from the selection. The three fractions are then added together. Determine the largest possible sum that can be obtained. Solution Solution 1 We can start by observing that to get a fraction with the highest value we need a 7 in the numerator. The choice of denominators is possibly obvious as well. 7 = 7, 7 = 3.5, 7. = 2.3, 1 2 3 7 = 1.75, 7 = 1.4 and 7. = 1.2. 7 is the largest fraction and any numerator other than 7 will 4 5 6 1 produce a lower value. Now we have five numbers left to place: {2,3,4,5,6}. Of these remaining numbers, since 6 is the largest it should go in the numerator. Then 6 = 3, 2 6 = 2, 6 = 1.5 and 6 = 1.2. 6 is the largest fraction and any numerator other than 6 will 3 4 5 2 produce a lower value. Now we have three numbers left to place: {3,4,5}. Of these remaining numbers, since 5 is the largest it should go in the numerator. Then 5. = 1.7 3 and 5 = 1.25. 5 is the largest fraction and any numerator other than 5 will produce a lower 4 3 value. We can now determine the largest possible sum. Largest Possible Sum = 7 1 + 6 2 + 5 3 = 42 6 + 18 6 + 10 6 = 70 6 the largest possible sum is 35 3 or 11 2 3. = 35 3 If you look closely, you will see that the largest number and the smallest are together in one fraction, the second largest and the second smallest are together in another fraction, and the third fraction uses the third largest and the third smallest numbers.

Solution 2 The largest fractions will be created by putting the three smallest numbers, 1, 2, and 3, in the denominators and then placing the numbers 4, 5, 6, and 7 in the numerators. We will do this in every possible way, determine the sums and choose the largest. There are 24 different possible sums in which 1, 2, and 3 are the denominators and 4, 5, 6, 7 are the numerators. Only one sum will be worked out showing all of the steps. The other sums would be determined in a similar fashion. 4 1 + 5 2 + 6 3 = 24 6 + 15 6 + 12 6 = 51 6 Using 4,5,6 Using 4,5,7 Using 4,6,7 Using 5,6,7 in the Numerators in the Numerators in the Numerators in the Numerators 4 1 + 5 2 + 6 3 = 51 6 4 1 + 6 2 + 5 3 = 52 6 5 1 + 4 2 + 6 3 = 54 6 5 1 + 6 2 + 4 3 = 56 6 6 1 + 4 2 + 5 3 = 58 6 6 1 + 5 2 + 4 3 = 59 6 4 1 + 5 2 + 7 3 = 53 6 4 1 + 7 2 + 5 3 = 55 6 5 1 + 4 2 + 7 3 = 56 6 5 1 + 7 2 + 4 3 = 59 6 7 1 + 4 2 + 5 3 = 64 6 7 1 + 5 2 + 4 3 = 65 6 the largest possible sum is 7 1 + 6 2 + 5 3 = 70 6 or 11 2 3. Algebraic Aside 4 1 + 6 2 + 7 3 = 56 6 4 1 + 7 2 + 6 3 = 57 6 6 1 + 4 2 + 7 3 = 62 6 6 1 + 7 2 + 4 3 = 65 6 7 1 + 4 2 + 6 3 = 66 6 7 1 + 6 2 + 4 3 = 68 6 5 1 + 6 2 + 7 3 = 62 6 5 1 + 7 2 + 6 3 = 63 6 6 1 + 5 2 + 7 3 = 65 6 6 1 + 7 2 + 5 3 = 67 6 7 1 + 5 2 + 6 3 = 69 6 7 1 + 6 2 + 5 3 = 70 6 Above we stated that the largest fractions will be created by putting the three smallest numbers, 1, 2, and 3 in the denominators. Is there a way to justify why the 5, 6 and 7 are used in the numerator and where they are? We want to maximize the expression a 1 + b 2 + c 3. We can express this with a common denominator of 6 as follows: 6a 6 + 3b 2 + 2c 6a + 3b + 2c =. 3 6 If we are selecting values from 4, 5, 6 and 7, the largest value will be created when a = 7, b = 6 and c = 5. The largest sum is 7 1 + 6 2 + 5 3.

Problem of the Week Problem C Pane - Filled Problem A contractor is building new houses. Each house has a large window with width 2 feet and height 6 feet to be made up of six identical 1 foot by 2 feet glass panes. The contractor wants each house to look slightly different. The window size is fixed, but he plans on arranging the glass panes in the windows so that no two windows have the same configuration. Two possible arrangements of the glass panes are given below. Assuming that the glass panes cannot be cut, how many different arrangements can be made? Strands Data Management and Probability, Measurement

Problem of the Week Problem C and Solution Pane - Filled Problem Problem A contractor is building new houses. Each house has a large window with width 2 feet and height 6 feet to be made up of six identical 1 foot by 2 feet glass panes. The contractor wants each house to look slightly different. The window size is fixed, but he plans on arranging the glass panes in the windows so that no two windows have the same configuration. Two possible arrangements of the glass panes are given below. Assuming that the glass panes cannot be cut, how many different arrangements can be made? Solution Let s consider the ways that the glass panes can be arranged. First, notice that there must always be an even number of window panes that have a vertical orientation (standing on end). All glass panes are horizontal (and none are vertical) This can only be done one way. Four glass panes are horizontal and two are vertical There could be 0, 1, 2, 3 or 4 horizontal panes below the two vertical panes. So there are 5 ways that four glass panes are horizontal and two are vertical. Two glass panes are horizontal and four are vertical We need to consider sub cases: Case 1: There are no horizontal panes between the vertical panes. There could be 0, 1 or 2 horizontal panes below the four vertical panes. So there are 3 ways that two glass panes are horizontal, four are vertical and there are no horizontal panes between the vertical panes. Case 2: There is one horizontal pane between the vertical panes. There could be 0 or 1 horizontal panes below the bottom two vertical panes. So there are 2 ways that two glass panes are horizontal, four are vertical and there is one horizontal pane between the vertical panes. Case 3: There are two horizontal panes between the vertical panes. There cannot be any horizontal panes below the bottom two vertical panes. So there is 1 way that two glass panes are horizontal, four are vertical and there are two horizontal panes between the vertical panes. All glass panes are vertical This can only be done one way. Therefore, the total number of different configurations of the glass panes is 1 + 5 + (3 + 2 + 1) + 1 = 13. So the contractor can build 13 houses before he has to start duplicating window patterns.

Problem of the Week Problem C Reaching in for Cash A local radio station, PROBLEM 3.14, is running a game in the area mall. All proceeds from the game benefit a local charity. A drum contains envelopes which contain a card inside. Printed on the card is the amount of money you win by selecting that particular envelope. You are able to win $5, $10, $20, $50 or $100. The total value of all of the cards that say You win $5 is $500; the total value of all of the cards that say You win $10 is $500; the total value of all of the cards that say You win $20 is $500; the total value of all of the cards that say You win $50 is $500; and the total value of all of the cards that say You win $100 is $500. Once you select your envelope, your prize is revealed and you are given the appropriate amount of money. The card is then placed back in the envelope and returned to the drum. Since the envelope containing the card is returned to the drum after each play, the chances of winning any particular amount remains the same. It costs $15 to play the game. If you randomly choose one of the envelopes from the drum, what is the probability that you win more than you paid to participate in the game? PROBLEM 3.14PrizeDrum Strands Data Management and Probability, Number Sense and Numeration

Problem of the Week Problem C and Solution Reaching in for Cash PROBLEM 3.14PrizeDrum Problem A local radio station, PROBLEM 3.14, is running a game in the area mall. All proceeds from the game benefit a local charity. A drum contains envelopes which contain a card inside. Printed on the card is the amount of money you win by selecting that particular envelope. You are able to win $5, $10, $20, $50 or $100. The total value of all of the cards that say You win $5 is $500; the total value of all of the cards that say You win $10 is $500; the total value of all of the cards that say You win $20 is $500; the total value of all of the cards that say You win $50 is $500; and the total value of all of the cards that say You win $100 is $500. Once you select your envelope, your prize is revealed and you are given the appropriate amount of money. The card is then placed back in the envelope and returned to the drum. Since the envelope containing the card is returned to the drum after each play, the chances of winning any particular amount remains the same. It costs $15 to play the game. If you randomly choose one of the envelopes from the drum, what is the probability that you win more than you paid to participate in the game? Solution In order to determine the probability, we need to know the total number of envelopes and the total number of envelopes containing each of the specific prize amounts. These numbers are determined in the following chart. Value of Prize Total Value of Number of Envelopes in the Envelope Prizes at this Level Containing this Prize Value $5 $500 $500 $5 = 100 $10 $500 $500 $10 = 50 $20 $500 $500 $20 = 25 $50 $500 $500 $50 = 10 $100 $500 $500 $100 = 5 The total number of envelopes is 100 + 50 + 25 + 10 + 5 = 190. The total number of envelopes containing a prize greater than $15 is 25 + 10 + 5 = 40. The probability of winning an amount greater than what was paid to play is given in the following calculation: number of envelopes containing more than $15 total number of envelopes = 40 190 = 4 19. The probability of winning one of the top three prizes is 4 19. You will not win very often but remember the profits of this game are given to a local charity.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Stack Em Up Three cubes with side lengths 1 m, 2 m and 3 m are stacked on top of each other as shown. Determine the total surface area of the stack, including the bottom. Strands Geometry and Spatial Sense, Measurement, Number Sense and Numeration

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Stack Em Up Problem Three cubes with side lengths 1 m, 2 m and 3 m are stacked on top of each other as shown. Determine the total surface area of the stack, including the bottom. Solution To determine the areas we will primarily use Area = length width. Each cube has four exposed square sides so the total area of all the sides is 4 (1 1)+4 (2 2)+4 (3 3) = 4 (1)+4 (4)+4 (9) = 4+16+36 = 56 m 2. To determine the exposed top area of each of the cubes look down on the tower and see a cross-section like the one below. This exposed area is exactly the same as the side area of one face of the largest cube. Therefore, the top exposed area is 3 3 = 9 m 2. The top area and the bottom area are the same. Therefore, the bottom area is 9 m 2. The total surface area is 56 + 9 + 9 = 74 m 2. Extension: Three cubes with side lengths x, y and z are stacked on top of each other in a similar manner to the original problem such that 0 < x < y < z. Show that the total surface area of the stack, including the bottom, is 6z 2 + 4y 2 + 4x 2.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C What s Your Angle Anyway I? In ABD, C is on AB such that AC = CB = CD and BCD = 70. Determine the measure of ADB. If you want to solve a more general version of this problem, consider solving this week s Problem of the Week Problem D, What s Your Angle Anyway II?. If you want to further extend this problem, consider solving this week s Problem of the Week Problem E, What s Your Angle Anyway III. Strand Geometry and Spatial Sense

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution What s Your Angle Anyway I? Problem In ABD, C is on AB such that AC = CB = CD and BCD = 70. Determine the measure of ADB. Solution Solution 1 Since ACB is a straight line, ACD + DCB = 180 but DCB = 70 so ACD = 110. In ACD, since AC = CD, ACD is isosceles and CAD = CDA = x. The angles in a triangle sum to 180 so in ACD CAD + CDA + ACD = 180 x + x + 110 = 180 2x = 70 x = 35 Similarly, in BCD, since CB = CD, BCD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in CBD CBD + CDB + BCD = 180 y + y + 70 = 180 2y = 110 y = 55 Then ADB = CDA + CDB = x + y = 35 + 55 = 90. the measure of ADB is 90. See solution 2 for a more general approach to the solution of this problem.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING It turns out that it is not necessary to find the angles in the problem. Solution 2 Here is a second solution to the problem. In CAD, since CA = CD, CAD is isosceles and CAD = CDA = x. In CBD, since CB = CD, CBD is isosceles and CBD = CDB = y. The angles in a triangle sum to 180 so in ABD BAD + ADB + ABD = 180 x + (x + y ) + y = 180 2x + 2y = 180 x + y = 90 But ADB = ADC + CDB = x + y = 90. the measure of ADB is 90.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Irregular Area y x The dots on the diagram are one unit apart, horizontally and vertically. Determine the area of the figure. Strands Geometry and Spatial Sense, Measurement

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem Problem of the Week Problem C and Solution Irregular Area The dots on the diagram are one unit apart, horizontally and vertically. Determine the area of the figure. y Solution On the diagram, draw two horizontal lines and two vertical lines to form a rectangle such that the vertices of the irregular area are on the sides of the rectangle. Label the four vertices of the rectangle and the four vertices of the irregular area as shown on the diagram. To find the area of the irregular shape, determine the area of the rectangle and subtract the area of the four triangles that are not part of the area of the irregular shape. Since the dots are one unit apart horizontally and vertically we can determine the various lengths: AB = 5, BC = 4, AC = 9 CD = 4, DE = 3, CE = 7 EF = 7, F G = 2, EG = 9 GH = 2, HA = 5, GA = 7 To find the area of the rectangle, multiply the length AC by the width CE obtaining 9 7 = 63 units 2. x Each of the corners of the rectangles has a right angle. So each of the four triangles is right angled and we can use the lengths of the two sides that meet at the right angle in the calculation of the area of the triangle. Using Area = base height 2, we calculate the areas: Area ABH = Area BCD = Area DEF = AB HA 2 BC CD 2 DE EF 2 Area F GH = F G GH 2 = 5 5 2 = 4 4 2 = 3 7 2 = 2 2 2 = 25 2 = 12.5 units2 = 16 2 = 8 units2 = 21 = 10.5 units2 2 = 4 2 = 2 units2 Therefore the area of the irregular shape is 63 12.5 8 10.5 2 = 63 33 = 30 units 2.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Be Mine Valentine A valentine is constructed by pasting two white semi-circles, each with radius 3 cm, and a white triangle onto a 12 cm square sheet of red paper as shown below. (The dashed line and the right angle symbols will not actually be on the finished card.). You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. Strands Geometry and Spatial Sense, Measurement

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Be Mine Valentine Problem A valentine is constructed by pasting two white semi-circles, each with radius 3 cm, and a white triangle onto a 12 cm square sheet of red paper as shown below. (The dashed line and the right angle symbols will not actually be on the finished card.). You are going to write your valentine a message in red ink on the white region of the card. Determine the total amount of area available for your special valentine greeting. Solution Place the given information on the diagram. 12 cm 3 cm 12 cm The total area for writing the message is the area of the two semi-circles plus the area of the white triangle. Since there are two semi-circles of radius 3 cm, the total area is the same as the area of a full circle of radius 3 cm. The area of the two semi-circles is πr 2 = π(3) 2 = 9π cm 2. The height of the triangle is the length of the square minus the radius of the semi-circle. Therefore the height of the triangle is 12 3 = 9 cm. The base of the triangle is 12 cm, the width of the square. The area of the triangle is 1 2 base height = 1 2 (12)(9) = 54 cm2. The total area for writing the message is (9π + 54) cm 2. This area is approximately 82.3 cm 2. Happy Valentine s Day.

Problem of the Week Problem C A Griddy Performance Three of the vertices of square ABCD are located at A(0,3), B(4,0), and C(7,4). Determine the area of square ABCD. Strands Geometry and Spatial Sense, Measurement, Number Sense and Numeration

Area ABCD = Area of Large Square 4 Area of One Triangle = Length Width 4 (Base Height 2) = 7 7 4 (4 3 2) = 49 4 6 = 49 24 = 25 units 2 the area of the square is 25 units2. Problem of the Week Problem C and Solution A Griddy Performance Three of the vertices of square ABCD are located at A(0,3), B(4,0), and C(7,4). Determine the area of square ABCD. Solution Solution 1 Problem In this solution we will determine the area of ABCD without using the Pythagorean Theorem We will first determine the coordinates of the fourth vertex D. To do so, observe that to get from A to B, you would go down 3 units and right 4 units. To get from B to C, you move 3 units to the right and then 4 units up. Continuing the pattern, go up 3 units and left 4 units you get to D(3,7). Continuing, as a check, go left 3 units and down 4 units, and you arrive back at A. The coordinates of D are (3, 7). Draw a box with horizontal and vertical sides so that each vertex of the square ABCD is on one of the sides of the box. This creates a large square with sides of length 7 containing four identical triangles and square ABCD. Each of the triangles has a base 4 units long and height 3 units long.

= 3 2 + 4 2 AB 2 = OA 2 + OB 2 Using the Pythagorean Theorem, we can find AB 2 which is AB AB, the area of the square. Let the origin be O(0,0). Then OAB forms a right triangle. OA, the distance from the origin to point A on the y-axis, is 3 units. OB, the distance from the Solution 2 In this solution we will determine the area of ABCD using the Pythagorean Theorem. Since ABCD is a square, it is only necessary to find the length of one side. We can determine the area by squaring the length of the side. the area of the square is 25 units 2. = 25 = 9 + 16 origin to point B on the x-axis, is 4 units.

Problem of the Week Problem C Check the Perimeter A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, a median is drawn from A meeting BC at M. The perimeter of ABC is 24. The perimeter of ABM is 18. The perimeter of ACM is 16. Determine the length of the median AM. A B M C Strands Geometry and Spatial Sense, Patterning and Algebra

Problem of the Week Problem C and Solution Check the Perimeter Problem A median is a line segment drawn from the vertex of a triangle to the midpoint of the opposite side. In ABC, a median is drawn from A meeting BC at M. The perimeter of ABC is 24. The perimeter of ABM is 18. The perimeter of ACM is 16. Determine the length of the median AM. B Solution Solution 1 Let AB = t, BM = p, MC = q, CA = r, and AM = m. The perimeter of ABM = t + p + m = 18. The perimeter of ACM = q + r + m = 16. The perimeter of ABC = t + p + q + r = 24. Consider the perimeter of ABM plus the perimeter of ACM. This is equal to (t + p + m) + (q + r + m) = t + p + m + q + r + m = (t + p + q + r) + 2 m. So, if we add the perimeter of ABM to the perimeter of ACM, we will obtain the perimeter of ABC plus two times the length of the median, m. In other words, since the perimeter of ABM is 18, the perimeter of ACM is 16 and the perimeter of ABC is 24, we find that 18 + 16 = 24 + 2 m. It follows that 34 = 24 + 2 m and 2 m must be 10. Therefore m, the length of the median, is 5. Something to think about: In the solution we never used the fact that AM is a median and that BM = MC. In other words, we never used the fact that p = q. This means that there could be other triangles that satisfy the conditions of the problem without AM being the median. Indeed there are! Try creating a few different triangles with AM = 5 that satisfy all the conditions of the problem except the condition that AM is a median. It turns out that there is only one triangle that satisfies all the conditions of the problem so AM is a median. You will need some high school mathematics to be able to find the precise dimensions. t p m M A q r C

A t m r Solution 2 B p M q C In this solution, we take a more algebraic approach to solving the problem. It involves more formal equation solving. Let AB = t, BM = p, MC = q, CA = r, and AM = m. The perimeter of ABM = t + p + m = 18 or t + p = 18 m. (1) The perimeter of ACM = q + r + m = 16 or q + r = 16 m. (2) The perimeter of ABC = t + p + q + r = 24. (3) We will now combine the results. t + p = 18 m (1) q + r = 16 m (2) So t + p + q + r = 18 m + 16 m, by combining the two results. But the perimeter is t + p + q + r = 24. 24 = 18 m + 16 m 24 = 34 2m 24 34 = 34 34 2m 10 = 2m 10 = 2m 2 2 5 = m The length of the median AM = m = 5.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Wired A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2. Each part is bent to form a square. Determine the ratio of the area of the larger square to the smaller square. Strands Measurement, Patterning and Algebra

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Wired Problem A piece of wire 60 cm in length is to be cut into two parts in the ratio 3 : 2. Each part is bent to form a square. Determine the ratio of the area of the larger square to the smaller square. Solution Let the length of the longer piece of wire be 3x cm and the length of the shorter piece of wire be 2x cm. Then 3x + 2x = 60 or 5x = 60 and x = 12 follows. Then the longer piece of wire is 3x = 3(12) = 36 cm and the smaller piece of wire is 2x = 2(12) = 24 cm. These two lengths correspond to the perimeters of the respective squares. Each of the wires is bent to form a square. The length of each side of the square is the perimeter of the square divided by 4. Therefore the side length of the larger square is 36 4 = 9 cm and the side length of the smaller square is 24 4 = 6 cm. The area of a square is calculated by squaring the side length. The area of the larger square is 9 2 = 81 cm 2 and the area of the smaller square is 6 2 = 36 cm 2. The ratio of the area of the larger square to the area of the smaller square is 81 : 36. This ratio can be simplified by dividing each term by 9. The ratio in simplified form can then be written as 9 : 4. Therefore the ratio of the area of the larger square to the area of the smaller square is 9 : 4. An observation: The ratio of the area of the larger square to the area of the smaller square is 9 : 4 = 3 2 : 2 2. Is it a coincidence that the ratio of the area of the larger square to the area of the smaller square is equal to the squares of each term in the given ratio? Also notice that the ratio of the area of the larger square to the area of the smaller square is equal to the ratio of the square of the perimeter of the larger square to the square of the perimeter of the smaller square. In this case, the perimeter of the larger square is 36 cm and the perimeter of the smaller square is 24 cm. Then 36 2 : 24 2 = 1296 : 576 = 1296 : 576 = 9 : 4. 144 144 It is left to the solver to see if these two results are true in general.

Problem of the Week Problem C Fund Raising - One Step at a Time A temporary set of stairs has been constructed to get dignitaries on and off a stage at a fundraising event. The steps are of equal depth and equal height. The entire structure is 120 cm high, 120 cm from front to back and 100 cm wide. The stairs are to be carpeted and the two sides are to be painted. One of the two sides that is to be painted is shaded in the diagram. The back and bottom of the structure will not be seen and will not be carpeted or painted. Determine the total area to be carpeted and the total area to be painted. 120 cm 100 cm 120 cm Strand Measurement

120 cm 100 cm Problem of the Week Problem C and Solution Fund Raising - One Step at a Time 120 cm Problem A temporary set of stairs has been constructed to get dignitaries on and off a stage at a fundraising event. The steps are of equal depth and equal height. The entire structure is 120 cm high, 120 cm from front to back and 100 cm wide. The stairs are to be carpeted and the two sides are to be painted. One of the two sides that is to be painted is shaded in the diagram. The back and bottom of the structure will not be seen and will not be carpeted or painted. Determine the total area to be carpeted and the total area to be painted. Solution In order to solve both parts of the problem, the depth and height of each individual step must be calculated. The entire structure is 120 cm from front to back and four steps cover the entire depth. Therefore, each step is 120 cm 4 = 30 cm wide and high. The area to carpet is made up of 8 identical rectangles, each 30 cm wide and 100 cm long. Using the formula Area = Length W idth, the area of one rectangle is 30 100 = 3 000 cm 2. The area of all surfaces to be carpeted is 8 3 000 = 24 000 cm 2. There are many different ways to find the area of the side pieces. One solution would be to break the figure into four equal width rectangles, one four steps high, one three steps high, one two steps high and the final rectangle one step high. The area of one side would then be 30 (4 30) + 30 (3 30) + 30 (2 30) + 30 (1 30) = 30 120 + 30 90 + 30 60 + 30 30 = 3 600 + 2 700 + 1 800 + 900 = 9 000 cm 2. The total area of the two sides to be painted is 2 9 000 = 18 000 cm 2. A second method to calculate the area of one side involves breaking the figure into triangles. Draw a diagonal from the top left corner to the bottom right corner. This diagonal line would hit the bottom corner of each step as shown in the diagram. The larger triangle has a base and height of 120 cm. Each of the four smaller triangles has a base and a height of 30 cm, the width and height of each step. Using the formula Area = Base Height 2, the area = 120 120 2 + 4 30 30 2 = 7 200 + 1 800 = 9 000 cm 2. The total area of the two sides to be painted is then 2 9 000 = 18 000 cm 2. the total area to carpet is 24 000 cm 2 and the total area to paint is 18 000 cm 2. 120 cm 120 cm 120 cm 120 cm 100 cm 100 cm

Problem of the Week Problem C All Things Being Equal Square BCDE and ACD have equal areas. Square BCDE has sides of length 12 cm. AD intersects BE at F. Determine the area of quadrilateral BCDF. Strands Measurement, Number Sense and Numeration, Patterning and Algebra

Problem of the Week Problem C and Solution All Things Being Equal Problem Square BCDE and ACD have equal areas. Square BCDE has sides of length 12 cm. AD intersects BE at F. Determine the area of quadrilateral BCDF. Solution The area of square BCDE = 12 12 = 144 cm 2. The area of ACD equals the area of square BCDE. Therefore area ACD = 144 cm 2. The area of a triangle is calculated using the formula base height 2. It follows that: Area ACD = (CD) (AC) 2 144 = 12 AC 2 144 = 6 AC 24 cm = AC But AC = AB + BC so 24 = AB + 12 and it follows that AB = 12 cm. Area ACD = Area Square BCDE Area ABF + Area Quad. BCDF = Area DEF + Area Quad. BCDF Since the area of quadrilateral BCDF is common to both sides of the equation, it follows that the area of ABF equals the area of DEF. Area ABF = Area DEF (AB) (BF ) 2 = (DE) (EF ) 2 12 BF 2 = 12 F E 2 6 BF = 6 F E BF = F E But BF + F E = BE = CD = 12 so BF = F E = 6 cm. Then the area of DEF = DE F E 2 = 12 6 2 = 36 cm 2. The area of quadrilateral BCDF = area of square BCDE area DEF = 144 36 = 108 cm 2 Therefore the area of quadrilateral BCDF is 108 cm 2.

Problem of the Week Problem C Tunnel Vision A train 1000 metres long travels through a 3000 metre tunnel. Thirty seconds pass from the time the last car has just completely entered the tunnel until the time when the front of the engine emerges from the other end. Determine the speed of the train, in kilometres per hour. Strands Measurement, Number Sense and Numeration

Problem Problem of the Week Problem C and Solution Tunnel Vision A train 1000 metres long travels through a 3000 metre tunnel. Thirty seconds pass from the time the last car has just completely entered the tunnel until the time when the front of the engine emerges from the other end. Determine the speed of the train, in kilometres per hour. Solution A diagram to represent the problem will make it very easy to visualize. 1000 m train 3000 m tunnel At the time the entire train is just inside the tunnel, there is 3000 1000 = 2000 metres left to travel until the front of the engine emerges from the other end. The engine has to travel 2000 metres in 30 seconds. We can calculate the speed of the train by dividing the distance travelled by the time required to travel the distance. The speed of the train is 2000 m 30 seconds = 200 3 m/s. Now our task is to convert from m/s to km/h. We will do this in two steps: first convert metres to kilometres and then convert seconds to hours. (1) 200 m 3 s = 200 m 3 s 1 km 1000 m 200 km = 3000 s = 1 km 15 s (2) 1 km 15 s = 1 km 15 s 60 s 60 min 1 min 1 h = 3600 km 15 h = 240 km 1 h The train is travelling at a speed of 240 km/h.

Problem of the Week Problem C Pool Anyone? A couple has a rectangular backyard measuring 24 m by 18 m. They would like to install a circular pool. They have submitted a plan in which they have divided up their backyard into nine identical rectangles (as shown). The circumference of the pool would pass through the four vertices of the middle rectangle and the pool would completely cover the middle rectangle. The distance across the centre of the pool equals the length of the diagonal of the middle rectangle. They have submitted their plan to the municipality in order to obtain a permit for the construction of their pool. A city by-law states that a new pool in a backyard may not cover more than one-sixth of the area of the backyard. Will their plan be approved? You may find the following useful: The Pythagorean Theorem states, In a right angle triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides. In the right angle triangle shown, c is the hypotenuse and c 2 = a 2 + b 2 Strands Geometry and Spatial Sense, Measurement, Patterning and Algebra

A couple has a rectangular backyard measuring 24 m by 18 m. They would like to install a circular pool. They have submitted a plan in which they have divided up their backyard into nine identical rectangles. The circumference of the pool would pass through the four vertices of the middle rectangle and the pool would completely cover the middle rectangle. The distance across the centre of the pool equals the length of the diagonal of the middle rectangle. They have submitted their plan to the municipality in order to obtain a permit for the construction of their pool. A city by-law states that a new pool in a backyard may not cover more than one-sixth of the area of the backyard. Will their plan be approved? The diameter of the pool is 10 m so it follows that the radius of the pool will be 10 2 = 5 m. The area of the pool can be calculated using the formula A = πr 2 so A = π (5)2 = 25π 78.5 m 2. Since 78.5 > 72, the area of the pool is greater than one-sixth of the total area of the backyard. Therefore, the plan, as is, will not be approved. Problem of the Week Problem C and Solution Pool Anyone? Problem Solution The dimensions of the backyard are 24 m by 18 m so the area of the entire yard is 24 18 = 432 m 2. To comply with the by-law, the area of the pool cannot be more than 1 6 of 432 = 1 6 432 = 72 m2. Since we know the length of the backyard is 24 m and there are 3 rectangles across, the length of one rectangle is 24 3 = 8 m. Similarly, since we know the width of the backyard is 18 m and the backyard is 3 rectangles wide, the width of one rectangle is 18 3 = 6 m. Two adjacent sides of the middle rectangle and a diagonal form a right-angled triangle. Let d represent the diameter of the pool. We know that d is also the length of the hypotenuse of the right-angled triangle with sides 8 m and 6 m. Using the Pythagorean Theorem, d 2 = 82 + 6 2 = 64 + 36 = 100 d = 100, since d > 0 d = 10 m

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Puzzling Products In the product shown below, the letters F and L represent different digits from 0 to 9. Determine the value of F and L. F 8 3 L 2 7 3 0 Strand Number Sense and Numeration

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Puzzling Products Problem In the product shown below, the letters F and L represent different digits from 1 to 9. Determine the value of F and L. F 8 3 L 2 7 3 0 Solution In a multiplication question there are three parts: the multiplier, multiplicand and product. In our problem, F 8 is the multiplier, 3L is the multiplicand, and 2730 is the product. The units digit of the product 2730 is 0. The units digit of a product is equal to the units digit of the result obtained by multiplying the units digit of the multiplier and multiplicand. So 8 L must equal a number with units digit 0. The only choices for L are 0 and 5 since no other single digit times 8 produces a number ending in zero. However, if L = 0, the units digit of the product is 0 and the remaining three digits of the product are produced by multiplying 3 F 8. But 3 F 8 produces a number ending in 4, not 3 as required. Therefore L 0 and L must equal 5. We are then multiplying F 8 by 35 to create the product 2730. To determine F we can work backwards. When we divide 2730 by 35, the quotient is 78. But 78 and F 8 are the same number so F = 7. F = 7 and L = 5.

is on the right. The numbers can be read in increasing order from top row to bottom row and from left-to-right within a row. Notice that we can get from 0 to Specifically, below every number there are two numbers: one on the left and one on the right. For example, below 3, the number 7 is on the left, and the number 8 WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Tree Travel Your friend writes down all of the integers starting from 0 in the following way: Strands Number Sense and Numeration, Patterning and Algebra If row 1 has the number 0, what is the rightmost number in row 12? 12 by going right (R), left (L) then right (R).

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem Problem of the Week Problem C and Solution Tree Travel Your friend writes down all of the integers starting from 0 as shown in the diagram to the right. Specifically, below every number there are two numbers: one on the left and one on the right. For example, below 3, the number 7 is on the left, and the number 8 is on the right. The numbers can be read in increasing order from top row to bottom row and from left-to-right within a row. Notice that we can get from 0 to 12 by going right (R), left (L) then right (R). If row 1 has the number 0, what is the rightmost number in row 12? Solution Solution 1 One approach to solving the problem would be to write out the first 12 rows of the chart and read off the rightmost number in row 12. You would discover that the number is 4094. This solution may work in this example but it is certainly not ideal. It would not be practical if you were asked for the last number in row 50. Observations There are many patterns in the chart. The solutions provided will look at some of the different patterns which can be used to solve the problem. Solution 2 Row 1 contains 1 number. Row 2 contains 2 numbers, twice the number of numbers in row 1. Row 3 contains 4 numbers, twice the number of numbers in row 2. Row 4 contains 8 numbers, twice the number of numbers in row 3. Each new row in the tree has twice as many numbers as the previous row. Using this, we could find the number of numbers in the first 12 rows of the chart. There are 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4095 numbers in the first 12 rows of the chart. So there are 4095 numbers in the twelve rows but the first number in the chart is 0. Therefore, it follows that the last number in the 12th row is 1 less than the number of numbers in 12 rows. The last number is 4094. In some ways, this solution may not be much simpler than the first one except we did not need to write out all of the numbers.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Solution 3 This solution is similar to solution 2 but only looks at the rightmost number in each row. To get from the top number to the rightmost number in row 2 add 2. To get from the rightmost number in row 2 to the rightmost number in row 3 add 4. To get from the rightmost number in row 3 to the rightmost number in row 4 add 8. These numbers which are added correspond to the number of numbers in the next row. We must add 11 of these numbers to 0. 0 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 = 4094 The rightmost number in row 12 is 4094. Solution 4 Since we can get from one row to a row immediately following by multiplying the number of numbers in the previous row by 2, there is some connection to powers of 2 in the problem. The following table shows the row number, the rightmost number in that row, the power of 2 with the row number as the exponent and how this power is related to the last number in the row. Row Number Rightmost Number in Row Power of 2 Connection 1 0 2 1 = 2 2 1 2 = 2 2 = 0 2 2 2 2 = 4 2 2 2 = 4 2 = 2 3 6 2 3 = 8 2 3 2 = 8 2 = 6 4 14 2 4 = 16 2 4 2 = 16 2 = 14 n?? 2 n 2 n 2 =?? It would appear that the last number in row 5 should be 2 5 2 = 32 2 = 30. We could write out the fifth row to confirm that this is correct. It would also appear that the last number in row n should be 2 n 2. By recognizing the pattern, we predict that the last number in row 12 should be 2 12 2 = 4096 2 = 4094. We know from earlier solutions that this is correct. It should be noted that this relationship works for all of the rows we have sampled but we have not proven it true in general. You will have to wait for some higher mathematics to be able to prove that this is true in general. The pattern used in this solution is not an obvious one but by discovering it the solution actually is fairly straight forward. In fact, if we accept the result as true we can quickly state the value of the rightmost number in any row of the table with a simple calculation.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C Pushy Pushy Manny Kerr runs a company that specializes in grooming and caring for lawns. Manny has a push mower and powerful riding mower. At one location it takes him five hours to cut the entire lawn with the push mower but only 70 minutes with the powerful riding mower. After 90% of the lawn was cut using the powerful riding mower, the remainder was cut using the push mower. How many minutes did it take Manny to cut the entire lawn? Strands Number Sense and Numeration, Patterning and Algebra

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution Pushy Pushy Problem Manny Kerr runs a company that specializes in grooming and caring for lawns. Manny has a push mower and powerful riding mower. At one location it takes him five hours to cut the entire lawn with the push mower but only 70 minutes with the powerful riding mower. After 90% of the lawn was cut using the powerful riding mower, the remainder was cut using the push mower. How many minutes did it take Manny to cut the entire lawn? Solution It takes Manny 70 minutes to cut 100% of the lawn with the powerful riding mower. It would take him 90% of 70 minutes or 0.90 70 = 63 minutes to cut 90% of the lawn with the powerful riding mower. Since he cuts 90% with the powerful riding mower, he cuts 100% 90% = 10% using the push mower. It takes him 5 hours or 5 60 = 300 minutes to cut 100% of the lawn with the push mower. It would then take him 0.10 300 = 30 minutes to cut 10% of the lawn with the push mower. It would then take him a total of 63 + 30 = 93 minutes to cut the entire lawn using the powerful riding mower for 90% of the job and the push mower for 10% of the job.

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C More for Less Harry Wirks is currently paid $567 for working a 45 hour week. His weekly salary is to be increased by 10% but his hours are to be reduced by 10%. Calculate the change from his old hourly rate of pay to his new hourly rate of pay. Strands Number Sense and Numeration, Patterning and Algebra

WWW.CEMC.UWATERLOO.CA The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the Week Problem C and Solution More for Less Problem Harry Wirks is currently paid $567 for working a 45 hour week. His weekly salary is to be increased by 10% but his hours are to be reduced by 10%. Calculate the change from his old hourly rate of pay to his new hourly rate of pay. Solution Solution 1 To calculate the hourly rate of pay divide the weekly salary by the number of hours worked. Harry s old hourly rate of pay is $567 45 h = $12.60/h. New Weekly Salary = Old Weekly Salary + 10% of Old Weekly Salary = $567 + 0.1 $567 = $567 + $56.70 = $623.70 New Number of Hours Worked = Old Hours Worked 10% of Old Hours Worked = 45 h 0.1 45 h = 45 h 4.5 h = 40.5 h Harry s new hourly rate of pay is $623.70 40.5 h = $15.40/h. The change in his hourly rate of pay is $15.40/h $12.60/h = $2.80/h. Harry s hourly rate increased $2.80/h. Solution 2 In the second solution we will use a more concise calculation. Harry s weekly salary is 10% more than his old weekly salary. So Harry earns 110% of his old weekly salary. Harry s hours are reduced by 10% of his old hours so he now works 90% of his old hours. To calculate his change in hourly rate we can take his new hourly rate and subtract his old hourly rate. Change in Hourly Rate = New Hourly Rate Old Hourly Rate = New Salary Hours Worked Old Salary Old Hours Worked = ($567 1.10) (45 0.9) $567 45 = $623.70 40.5 $567 45 = $15.40/h $12.60/h = $2.80/h Harry s hourly rate increased $2.80/h.

Problem of the Week Problem C Prime Picking A natural number greater than 1 is said to be prime if its only factors are 1 and itself. For example, the number 7 is prime since its only factors are 1 and 7. A perfect square is an integer created by multiplying an integer by itself. The number 25 is a perfect square since it is 5 5 or 5 2. Determine the smallest perfect square that has three different prime numbers as factors. Strands Number Sense and Numeration, Patterning and Algebra

Problem of the Week Problem C and Solution Prime Picking Problem A natural number greater than 1 is said to be prime if its only factors are 1 and itself. For example, the number 7 is prime since its only factors are 1 and 7. A perfect square is an integer created by multiplying an integer by itself. The number 25 is a perfect square since it is 5 5 or 5 2. Determine the smallest perfect square that has three different prime numbers as factors. Solution The problem itself is not very difficult once you determine what it is asking. So before looking at the solution, let us examine some perfect squares. The numbers 4 and 9 are both perfect squares that have only one prime number as a factor, 4 = 2 2 and 9 = 3 2. The number 36 is a perfect square since 36 = 6 2. However, the number 6 = 2 3 so 36 = (2 3) 2 = 2 3 2 3 = 2 2 3 2. So 36 is the product of the squares of two different prime numbers. In fact, since 2 and 3 are the smallest prime numbers, 36 is the smallest perfect square that has two different prime factors. Using this idea, we can create the smallest perfect square with three different prime factors by squaring each of the three smallest primes, 2, 3, and 5, and then multiplying these squares together. So the smallest perfect square that uses three different prime factors would be 2 2 3 2 5 2 = 4 9 25 = 900. It should be noted that 900 = 30 2 = (2 3 5) 2. the smallest perfect square with three different prime factors is 900.

Problem of the Week Problem C A Pressing Issue Steve produces apple juice. One day, after pressing many apples, he filled a 45 litre container with pure apple juice. He found that the taste of the pure mixture was too strong for his liking. He removed 9 litres of pure apple juice and replaced it with 9 litres of water. The new mixture still seemed too strong so he removed 9 litres of the mixture and replaced it with 9 litres of water. Determine the ratio of pure apple juice to water in Steve s final 45 litre mixture. Strand Number Sense and Numeration

Problem of the Week Problem C and Solution A Pressing Issue Problem Steve produces apple juice. One day, after pressing many apples, he filled a 45 litre container with pure apple juice. He found that the taste of the pure mixture was too strong for his liking. He removed 9 litres of pure apple juice and replaced it with 9 litres of water. The new mixture still seemed too strong so he removed 9 litres of the mixture and replaced it with 9 litres of water. Determine the ratio of pure apple juice to water in Steve s final 45 litre mixture. Solution We need to determine the amount of pure apple juice and the amount of water in the final mixture. Steve starts with 45 litres of pure apple juice and no water. After removing 9 litres of pure apple juice and adding 9 litres of water, he has 45 9 = 36 litres of pure apple juice and 9 litres of water. So 36 45 = 4 5 of the new mixture is pure apple juice and 9 45 = 1 of the new 5 mixture is water. He then removes 9 litres of the new mixture, 4 5 of which is pure apple juice and 1 of which is 5 water. So Steve removes 4 5 9 = 36 5 or 71 5 litres of pure apple juice and 1 5 9 = 9 5 or 14 5 litres of water. Before adding another 9 litres of water he has 36 7 1 5 = 180 5 36 5 = 144 5 or 284 litres of pure 5 apple juice and 9 1 4 5 = 45 5 9 5 = 36 5 or 71 litres of water. 5 After adding the additional water he has 9 + 7 1 5 = 9 + 36 5 = 45 5 + 36 5 = 81 5 or 161 litres of 5 water. The final ratio of pure apple juice to water is 28 4 5 : 161 5 = 144 5 : 81 = 144 : 81 = 16 : 9. 5 the final ratio of pure apple juice to water is 16:9.

Problem of the Week Problem C Check It Out Debit and credit cards contain account numbers which consist of many digits. Often, when purchasing items online you are asked to type in your account number. Because there are so many digits it is easy to type the number incorrectly. The last digit of the number is a specially generated check digit which can be used to quickly verify the validity of the number. A common algorithm used for verifying numbers is called the Luhn Algorithm. A series of operations are performed on the number and a final result is produced. If the final result ends in zero, the number is valid. Otherwise, the number is invalid. The steps performed in the Luhn Algorithm are outlined in the flowchart below. Two examples are provided. Number: 135792 Reversal: 297531 A = 2 + 7 + 3 A = 12 2 9 = 18 2 5 = 10 2 1 = 2 B = (1+8)+(1+0)+2 B = 9 + 1 + 2 = 12 C = 12 + 12 = 24 C does not end in zero. The number is not valid. Number: 1357987 Reversal: 7897531 A = 7 + 9 + 5 + 1 A = 22 2 8 = 16 2 7 = 14 2 3 = 6 B = (1+6)+(1+4)+6 B = 7 + 5 + 6 = 18 C = 22 + 18 = 40 C ends in zero. The number is valid. The number 2853 P 9Q9 367 is a valid card number when verified by the Luhn Algorithm. P and Q are each single digits of the number such that P is less than Q. Determine all possible values of P and Q. Strands Number Sense and Numeration, Patterning and Algebra

Problem Problem of the Week Problem C and Solution Check It Out Debit and credit cards contain account numbers which consist of many digits. Often, when purchasing items online you are asked to type in your account number. Because there are so many digits it is easy to type the number incorrectly. Most companies use a check digit to quickly verify the validity of the number. One commonly used algorithm for verifying numbers is called the Luhn Algorithm. A series of operations are performed on the number and a final result is produced. If the final result ends in zero, the number is valid. Otherwise, the number is invalid. The steps performed in the Luhn Algorithm are illustrated in the flowchart to the right. The number 2853 P 9Q9 367 is a valid card number when verified by the Luhn Algorithm. P and Q are each single digits of the number such that P is less than Q. Determine all possible values of P and Q. Solution When the digits of the card number are reversed the resulting number is 763 9Q9P 3582. The sum of the digits in the odd positions is A = 7 + 3 + Q + P + 5 + 2 = 17 + P + Q. When the digits in the remaining positions are doubled, the following products are obtained: 6 2 = 12; 9 2 = 18; 9 2 = 18; 3 2 = 6; and 8 2 = 16 When the digit sums from each of the products are added, the sum is: B = (1 + 2) + (1 + 8) + (1 + 8) + 6 + (1 + 6) = 3 + 9 + 9 + 6 + 7 = 34 C is the sum of A and B, so C = 17 + P + Q + 34 = 51 + P + Q. To satisfy the Luhn Algorithm, the units digit of C must be zero. The closest number greater than 51 ending in a zero is 60. It follows that P + Q = 60 51 = 9. We want values of P and Q that sum to 9 with P less than Q. Possible combinations are P = 0, Q = 9; P = 1, Q = 8; P = 2, Q = 7; P = 3, Q = 6; and P = 4, Q = 5. The next closest number greater than 51 ending in a zero is 70. It follows that P + Q = 70 51 = 19. But the largest value possible for Q is 9 and the largest value for P would then be 8. It follows that the largest possible value for P + Q = 8 + 9 = 17. This means that there are no values of P and Q so that C could equal 70. Therefore, there are only 5 possible pairs of digits for P and Q so that 2853 P 9Q9 367 is a valid card number. The possibilities are P = 0, Q = 9; P = 1, Q = 8; P = 2, Q = 7; P = 3, Q = 6; and P = 4, Q = 5. The valid card numbers are 2853 0999 367, 2853 1989 367, 2853 2979 367, 2853 3969 367 and 2853 4959 367.

Problem of the Week Problem C Mmmm, I Love Tortoises! Jonathan the tortoise is 183 years old according to several sources (BBC.com, Wikipedia, live science.com, to name a few). That makes Jonathan the oldest living terrestrial animal. Two other tortoises, Terrance and his grandfather Tyrone share the same birthday. Tyrone is older than 100 but not as old as Jonathan. This year, Tyrone s age is exactly fifteen times Terrance s age. How old will Terrance and Tyrone be when Tyrone s age is exactly eleven times Terrance s age? Strand Number Sense and Numeration

Problem Problem of the Week Problem C and Solution Mmmm, I Love Tortoises! Jonathan the tortoise is 183 years old according to several sources (BBC.com, Wikipedia, live science.com, to name a few). That makes Jonathan the oldest living terrestrial animal. Two tortoises, Terrance and his grandfather Tyrone share the same birthday. Tyrone is older than 100 but not as old as Jonathan. This year, Tyrone s age is exactly fifteen times Terrance s age. How old will Terrance and Tyrone be when Tyrone s age is exactly eleven times Terrance s age? Solution Solution 1 Since Tyrone s age is fifteen times Terrance s age, this means Tyrone s age is a multiple of 15 that is greater than 100 and less than 183. Tyrone s possible ages are 105, 120, 135, 150, 165 and 180. The corresponding possible ages for Terrance are 7, 8, 9, 10, 11 and 12. At some time in the future, Tyrone s age will be eleven times Terrance s age. This means that Tyrone s future age is a multiple of 11. The possibilities are 110, 121, 132, 143, 154, 165, 176, 187 and (possibly) 198. We will look at the possible cases. 1. Could Terrance s age now be 7 and Tyrone s age now be 105? The closest multiple of 11 greater than 105 is 110. This is 5 years from now. Terrance would then be 12 and 110 12 < 11. At this point, Tyrone s age is less than eleven times Terrance s age. It is not possible for Terrance to be 7 and Tyrone to be 105. 2. Could Terrance s age now be 8 and Tyrone s age now be 120? The closest multiple of 11 greater than 120 is 121. This is 1 year from now. Terrance would then be 9 and 121 9 > 11. At this point, Tyrone s age is still more than eleven times Terrance s age. Let s check the next multiple of 11 past 120. That is, consider 132. This is 12 years from now. Terrance would then be 20 and 132 20 < 11. At this point, Tyrone s age is less than eleven times Terrance s age. It is not possible for Terrance to be 8 and Tyrone to be 120. 3. Could Terrance s age now be 9 and Tyrone s age now be 135? The closest multiple of 11 greater than 135 is 143. This is 8 years from now. Terrance would then be 17 and 143 17 < 11. At this point, Tyrone s age is less than eleven times Terrance s age. It is not possible for Terrance to be 9 and Tyrone to be 135. 4. Could Terrance s age now be 10 and Tyrone s age now be 150? The closest multiple of 11 greater than 150 is 154. This is 4 years from now. Terrance would then be 14 and 154 14 = 11. At this point, Tyrone s age is eleven times Terrance s age. (We need to check the two remaining possibilities for completeness.)

5. Could Terrance s age now be 11 and Tyrone s age now be 165? The closest multiple of 11 greater than 165 is 176. This is 11 years from now. Terrance would then be 22 and 176 22 < 11. At this point, Tyrone s age is less than eleven times Terrance s age. It is not possible for Terrance to be 11 and Tyrone to be 165. 6. Could Terrance s age now be 12 and Tyrone s age now be 180? The closest multiple of 11 greater than 180 is 187. This is 7 years from now. Terrance would then be 19 and 187 19 < 11. At this point, Tyrone s age is less than eleven times Terrance s age. It is not possible for Terrance to be 12 and Tyrone to be 180. We have considered all of the possible cases. Terrance s age will be one-eleventh of Tyrone s age in 4 years when Terrance is 14 and Tyrone is 154. Although it was not asked, today Terrance is 10 and Tyrone is 150. Solution 2 This solution is much more algebraic and may be beyond your present experience. Let c represent Terrance s age today and g represent Tyrone s age today. Since Tyrone s age is fifteen times Terrance s age, it follows that g = 15c. Let n be the number of years until Tyrone s age is eleven times Terrance s age. In n years, Terrance s age will be c + n and Tyrone s age will be g + n. Then g + n = 11(c + n) But g = 15c so 15c + n = 11(c + n) 15c + n = 11c + 11n 4c = 10n 2c = 5n 2c 5 = n Since both c and n are whole years, it follows that c is divisible by 5. But in the first solution, the possible ages for Terrance were 7, 8, 9, 10, 11 and 12. The only multiple of 5 in this list is 10. Therefore c = 10 and then n = 4. But g = 15c so g = 15 10 = 150. Terrance is 10 today and Tyrone is 150 today. In 4 years, Terrance will be 14, Tyrone will be 154 and Tyrone s age will be eleven times Terrance s age.

Problem of the Week Problem C Powerful Factorials at Work The product of the positive integers 1 to 3 is 3 2 1 = 6 and can be written in an abbreviated form as 3!. We say 3 factorial. So 3! = 6. The product of the positive integers 1 to 17 is 17 16 15 3 2 1 and can be written in an abbreviated form as 17!. We say 17 factorial. The represents the product of all the missing integers between 15 and 3. In general, the product of the positive integers 1 to n is n!. Note that 1! = 1. Determine the tens digit in the sum 1! + 2! + 3! + + 18! + 19! + 20!. Strands Number Sense and Numeration, Patterning and Algebra

Problem of the Week Problem C and Solution Powerful Factorials at Work Problem The product of the positive integers 1 to 3 is 3 2 1 = 6 and can be written in an abbreviated form as 3!. We say 3 factorial. So 3! = 6. The product of the positive integers 1 to 17 is 17 16 15 3 2 1 and can be written in an abbreviated form as 17!. We say 17 factorial. The represents the product of all the missing integers between 15 and 3. In general, the product of the positive integers 1 to n is n!. Note that 1! = 1. Determine the tens digit and units digits in the sum 1! + 2! + 3! + + 18! + 19! + 20!. Solution At first glance seems like there is a great deal of work to do. However, by examining several factorials, we will discover otherwise. 1! = 1 2! = 2 1 = 2 3! = 3 2 1 = 6 4! = 4 3 2 1 = 24 5! = 5 4 3 2 1 = 120 Now 6! = 6 (5 4 3 2 1) = 6 5! = 6(120) = 720, 7! = 7 (6 5 4 3 2 1) = 7 6! = 7(720) = 5040, 8! = 8 (7 6 5 4 3 2 1) = 8 7! = 8(5040) = 40320, 9! = 9 (8 7 6 5 4 3 2 1) = 9 8! = 9(40320) = 362 880, and 10! = 10 (9 8 7 6 5 4 3 2 1) = 10 9! = 10(362880) = 3 628 800. An interesting observation surfaces, 9! = 9 8!, 10! = 10 9!, 11! = 11 10!, and so on. Furthermore, the last two digits of 10! are 00. Every factorial above 10! will also end with 00 since multiplying an integer that ends with 00 by another integer produces an integer product that ends in 00. So all factorials above 10! will end with 00 and will not change the tens digit or the units digit in the required sum. We can determine the last two digits of the required sum by adding the last two digits of each of the factorials from 1! to 9!. The sum of the last two digits of 1! to 9! will equal 1 + 2 + 6 + 24 + 20 + 20 + 40 + 20 + 80 = 213. Therefore, for the sum 1! + 2! + 3! + 18! + 19! + 20!, the tens digit will be 1 and the units digit will be 3. From what we have done, we do not know the hundreds digit. For Further Thought: If we had only been interested in the units digit in the required sum, how many factorials would we need to calculate? If we wanted to know the last three digits, how many more factorials would be required?

Problem of the Week Problem C Palindrome Hunt A palindrome is a word, phrase, number, or other sequence of characters which reads the same backward or forward. The number 9876789 is a seven-digit palindrome. Find all seven-digit palindromes that satisfy each of the following conditions: four of the digits are different; and the number is divisible by 45; and the sum of the digits of the number is equal to 45. The number 9876789 only satisfies the first condition. Divisibility fact: An integer is divisible b is divisible by 9.

Problem of the Week Problem C and Solution Palindrome Hunt AProblem palindrome is a word, phrase, number, or other sequence of characters which reads the same backward or forward. Find all seven-digit palindromes that satisfy each of the following conditions: four of the digits are different; the number is divisible by 45; and the sum of the digits of the number is 45. Solution We will start with the second condition first. For a number to be divisible by 45, it must be divisible by 5 and 9. For a number to be divisible by 5, the units digit must be either a 0 or 5. If the units digit is 0, then the first digit of the palindrome would be a 0. The number would no longer be a seven-digit number. Therefore, the first and last digit of the seven-digit palindrome must be a 5. The number looks like 5abcba5 where a b c 5 and 5 + a + b + c + b + a + 5 = 45. This simplifies to 2a + 2b + c = 35. The number is divisible by 5. To also be divisible by 9, the sum of the digits must be divisible by 9. If we let a = 9 and select values for b so that b < a, then we can generate the following possibilities: a b Number Digit Sum Value of c to add to 45 Valid or Invalid 9 8 598c895 44 + c 1 5981895, valid 9 7 597c795 42 + c 3 5973795, valid 9 6 596c695 40 + c 5 invalid, c cannot equal 5 9 4 594c495 36 + c 9 invalid, c cannot equal a 9 3 593c395 34 + c none By switching the positions of a and b in the valid palindrome, we generate two more valid palindromes, 5891985 and 5793975. Using a = 9 with b = 0, 1, or 2 would produce digit sums which could not reach 45. If we let a = 8 and select values for b so that b < a, then we can generate the following possibilities: a b Number Digit Sum Value of c to add to 45 Valid or Invalid 8 7 587c785 40 + c 5 invalid, c cannot equal 5 8 6 586c685 38 + c 7 5867685, valid 8 4 584c485 34 + c none By switching the positions of a and b in the valid palindrome, we generate one more valid palindrome, 5687865. Using a = 8 with b = 0, 1, 2, or 3 would produce digit sums which could not reach 45.

If we let a = 7 and select values for b so that b < a, then we can generate the following possibilities: a b Number Digit Sum Value of c to add to 45 Valid or Invalid 7 6 576c675 36 + c 9 5769675, valid 7 4 574c475 32 + c none Since 5769675 is valid, 5679765 is also valid. Using a = 7 with b = 0, 1, 2, or 3 would produce digit sums which could not reach 45. If we let a = 6 and select values for b so that b < a, then we can generate the following possibility: a b Number Digit Sum Value of c to add to 45 Valid or Invalid 6 4 564c465 30 + c none Using a = 6 with b = 0, 1, 2, or 3 would produce digit sums which could not reach 45. We cannot use a = 5 since the digit 5 has already been used. If we let a = 4 and select values for b so that b < a, no possibilities can be generated. The maximum value of b is 3. The maximum digit sum in this case is 5 + 4 + 3 + c + 3 + 4 + 5 = 24 + c. There is no possible value of c available so that the digit sum could equal 45. There are no other remaining possibilities to check. Therefore, there are 8 valid palindromes which satisfy all of the conditions of the problem. They are 5981895, 5973795, 5891985, 5793975, 5867685, 576967, 5687865, and 5679765.