The University of Melbourne BHPBilliton School Mathematics Competition, 2007 JUNIOR DIVISION, QUESTIONS & SOLUTIONS

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1 The University of Melbourne BHPBilliton School Mathematics Competition, 2007 JUNIOR DIVISION, QUESTIONS & SOLUTIONS Flower power. Rose is a teacher at Kinder and has 12 children in her class. She would like to make flowers with the children by gluing together used toilet rolls. One roll in the middle (the heart ) is surrounded by a ring of other rolls (the petals) such that each petal touches the heart and two neighbouring petals. heart petal How many rolls does Rose need for her class? (Carefully explain your answer.) Two neighbouring petals and the heart form the configuration The centers of the three circles form an equilateral triangle. Hence the two lines below make an angle of 60 degrees. Each child therefore needs 360/60 = 6 petals to complete one flower. 1

2 2 With 7 rolls per flower and 12 children, Rose needs to bring 7 12 = 84 rolls. John Howard s walkabout. Our Prime Minister John Howard always starts his day with a good brisk walk. During a visit to Canada the PM is staying in the Fantasyland Hotel, located in the world s largest shopping mall, West Edmonton Mall. There is no way for the PM to get out of the huge mall before his first scheduled meeting with Candian counterpart Stephen Harper. The PM thus decides to take his walk inside by walking up and down an escalator that is going up, hoping he will not be arrested for dangerous behaviour. Assume that the PM walks at a constant pace of 5 km per hour. How many times in 1 hour does he go up and down a 100 metres long escalater that is moving up at a speed of 4 km per hour? The PM starts at the bottom of the escalator. The PM walks 5 km per hour. Since the escalator is going up at a speed of 4 km per hour he is effectively going up at a speed of 9 km per hour. Similarly, he is effectively going down at a speed of 5 4 = 1 km per hour. To cover 100 metres at 9 km per hour takes 1 hour divided by 90 is 40 seconds. To cover 100 metres at 1 km per hour takes 1 hour divided by 10 is 6 minutes. Hence going up and down takes 6 minutes and 40 seconds which is 400 seconds. Since there are 3600 seconds in the hour the PM can go exactly 9 times up and down. A fancier way to get to the same result is by working in the units of 100 metres per hour. Then the PM is going up at a speed of 90 and down at a speed of 10. Hence the PM needs 1/90 + 1/10 = 1/9th of an hour to do one lap. Again it follows he can do exactly 9 laps. Pythagoras. Pythagoras who lived around 500 BC was one of the wisest men ever. He rightly believed that everything was related to mathematics and that numbers were the ultimate reality. Pythagoras discovered that in a right-angled triangle the square of the hypotenuse equals the sum of the squares of the other two sides. In other words a 2 + b 2 = c 2. c b a One day Pythagoras buys a triangular block of land on the Greek island of Samos. The sides of the (right-angled) triangle are 3, 4 and 5 orguia. 1 Despite his love for right-angled triangles Pythagoras realises that a triangular house is not practical and he decides to build a house on the square block indicated below An orguia is a unit of measurement used by the ancient Greek and is approximately 1.85 metres.

3 What is the area (in square orguia) occupied by Pythagoras house? The three small triangles are all similar to the original right-angled triangle. Hence the ratio of their sides is 5:4:3. If the square has side-length x then 3 5 x 4 x 3 5 Since we must have 5 3 x x = 4 it follows that x = The area of Pythagoras house is thus ( 60 ) 2 square orguia. 37 x Paris and Nicole s maths-mayhem. In their frequent shopping sprees Paris and Nicole often have trouble checking the bill. To address their less than impressive numeracy skills they decide to return to school. After a year of maths training the teacher wants to test the class of 25 students and lines them up in a queue such that Paris and Nicole are standing next to each other. The teacher then writes a whole number on the board and the first person in the queue says That number is divisible by 1. Then the second person in the queue says That number is divisible by 2, and so on till the final student in the queue says That number is divisible by 25. After all this the teacher exclaims Well done! Except for Paris and Nicole everyone made a correct statement. Find where Paris and Nicole were standing in the queue. We do not need to know what number the teacher actually wrote on the board. If we denote this number by n then we know that n is divisible by all the numbers 1 to 25 except for two consecutive numbers (corresponding to the positions of Paris and Nicole). We know that n must be divisible by 1, 2, 3,..., 12 because (1) all whole numbers are divisible by 1 (2) if n is not divisible by 2 it is also not divisible by 4, 6, 8,..., 24 which cannot be true. (3) if n is not divisible by 3 it is also not divisible by 6, 9,..., 24 which cannot be true. (4)... (5) if n is not divisible by 12 it is also not divisible by 24 which cannot be true. This leaves us with the numbers 13, 14,..., 25, but (6) because n is divisible by 2 and 7 it must also be divisible by 14. (7) because n is divisible by 3 and 5 it must also be divisible by 15. (8) because n is divisible by 2 and 9 = 3 2 it must also be divisible by 18. (9) because n is divisible by 4 = 2 2 and 5 it must also be divisible by 20. (10) because n is divisible by 3 and 7 it must also be divisible by 21.

4 4 (11) because n is divisible by 2 and 11 it must also be divisible by 22. (12) because n is divisible by 3 and 8 = 2 3 it must also be divisible by 24. This only leaves us with the numbers 13, 16, 17, 19, 23, 25. The only pair of consecutive numbers is 16, 17 so that Paris and Nicole were 16th and 17th in the queue. The smallest number the teacher could have written on the board is in fact Magic square. An n n magic square is a square grid filled with the numbers 1, 2,..., n 2 such that the sum of the numbers on the (two) diagonals, in each row and in each column are all the same. The following is an example of a 3 3 magic square since all three rows, three columns and two diagonals sum up to Complete the following 5 5 magic square: In a 5 5 magic square the numbers 1, 2,..., 25 are all used exactly once. Hence the sum of all the numbers is = (1 + 25) + (2 + 24) + + ( ) + 13 = (26) + (26) + + (26) + 13 = = = 325. Since each row adds up to the same number the row-sum must be 325/5 = 65. Using this we immediately find that the missing number in the 4th column is leading to = 17

5 Next we note that the two missing numbers in the last row must sum to = 44. Since the largest possible number is 25 the only 3 ways to achieve this is as , , But 19 and 23 are already used in the first row so that the two missing numbers in the last row must be 24 and 20. The question remains which of these should go in the south-west corner and which in the south-east corner. The two missing numbers in the first column must sum to = 39. Hence 20 cannot go in the south-west corner because = 39 but 19 is already used in the first row. Therefore Next we can complete the third row and the north-west to south-east diagonal: Next we can complete the second row and last column:

6 6 Next we can complete the first row and the third column: The only number we have not yet used is 16. Amazingly it is exactly what is needed to complete the second row, second column and north-east to south-west diagonal! Year of the pig. Let a 1, a 2,...,a n be a sequence of positive integers such that a 1 + a a n = 2007 and such that the a i are ordered from smallest to largest. Two examples of allowed sequences are 1, 1, 3, 3, 999, 1000 and 1, 2, 2, 6, 11, 31, Given the sequence a 1, a 2,...,a n we can also compute the product a 1 a 2 a n. Find which sequence gives the largest possible product. In our two examples the product of the numbers 1, 1, 3, 3, 999, 1000 is smaller than the product of the numbers 1, 2, 2, 6, 11, 31, 1954, but one can do much, much better. To maximise the product all the numbers a i must be less than 5; if a 5 would occur we could replace this by a 2 and a 3 creating a larger product: 2 3 > 5, if a 6 would occur we could replace this by a 2 and a 4 creating a larger product: 2 4 > 6, and so on. We also cannot have more than one 4 since can be replaced by creating a larger product: > 4 4. We also cannot have more than two 2 s since can be replaced by creating a larger product: 3 3 > We also cannot have both a 2 and a 4 since can be replaced by creating a larger product: 3 3 > 2 4. We also cannot have more than one 1 since can be replaced by a 2 creating a larger product: 2 > 1 1. From the above we conclude that we must have a lot of 3 s among our numbers. But if we have a 3 we cannot have a 1 since can be replaced by creating a larger product: 2 2 > 1 3.

7 The conclusion is that we have only 3 s with possibly a single or a double 2 or a single 4. Since 2007 is odd and in fact divisible by 3: 669 3, all our numbers must be a 3 so that the wanted sequence is a 1,...,a 669 = 3, 3, 3,..., 3. Don t ask your calculator to compute the product, is a number with 320 digits! 7