Western Australian Junior Mathematics Olympiad 2009

Transcription

1 Western Australian Junior Mathematics Olympiad 2009 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions 1 to 9. Calculators are not permitted. Write your answers on the answer sheet provided. 1. Evaluate [1 mark] 2. From each vertex of a cube, we remove a small cube whose side length is one-quarter of the side length of the original cube. How many edges does the resulting solid have? [1 mark] 3. A certain 2-digit number x has the property that if we put a 2 before it and a 9 afterwards we get a 4-digit number equal to 59 times x. What is x? [2 marks] 4. What is the units digit of ? [2 marks] 5. At a pharmacy, you can get disinfectant at different concentrations of alcohol. For instance, a concentration of 60% alcohol means it has 60% pure alcohol and 40% pure water. The pharmacist makes a mix with 3 litres of alcohol at 90% and 1 litres of alcohol at 50%. 5 5 How many percent is the concentration of that mix? [2 marks] 6. If we arrange the 5 letters A, B, C, D and E in different ways we can make 120 different words. Suppose we list these words in alphabetical order and number them from 1 to 120. So ABCDE gets number 1 and EDCBA gets number 120. What is the number for DECAB? [3 marks]

2 7. Every station on the Metropolis railway sells tickets to every other station. Each station has one set of tickets for each other station. When it added some (more than one) new stations, 46 additional sets of tickets had to be printed. How many stations were there initially? [3 marks] 8. At a shop, Alice bought a hat for $32 and a certain number of hair clips at $4 each. The average price of Alice s purchases (in dollars) is an integer. What is the maximum number of hair clips that Alice could have bought? [3 marks] 9. The interior angles of a convex polygon form an arithmetic sequence: 143, 145, 147,.... How many sides does the polygon have? [4 marks] 10. For full marks, explain how you found your solution. A square ABCD has area 64 cm 2. Let M be the midpoint of BC, let d be the perpendicular bisector of AM, and let d meet CD at F. How many cm 2 is the area of the triangle AMF? D F C d M A B [4 marks]

3 Western Australian Junior Mathematics Olympiad 2009 Team Questions 45 minutes General instructions: Calculators are (still) not permitted. Crazy Computers Rebecca has an old Lemon brand computer which has a defective keyboard. When she types L the letters LOM appear on the screen, and when she types M she gets OL and when she types O she gets M. We will abbreviate this to: Lemon Computer: L LOM, M OL, O M. So if she types OLO, she gets MLOMM. A. What will she get on the screen if she presses the Enter key to get to a new line and then types MLOMM? B. Tom has a more modern Raincoat computer which also has a broken keyboard. Typing R puts RS on the screen and S puts R. Raincoat: R RS, S R. If Tom types R to give one line of the screen, types in this line to get a second line and so on until he has 5 lines, what will the last line be? (Be careful the first line on the screen is RS not R.) C. If Tom kept going till he had 12 lines, how many letters would there be in the final line? Try to calculate this without writing down the final line of letters (which is quite long). D. Sarah s computer uses software produced by the giant Megafloppy Corporation, and is as defective as the others. If she types in H to get a first line on the screen, then types that line in to get a second line, then types that to get a third line, she finds the third line is HGGHGHHG. Assuming that only the G and H keys are faulty, what would she get if she deleted everything and then typed G? In the next question, Rebecca again uses her Lemon computer.

4 E. If Rebecca starts by typing O on her Lemon and continues until there are 6 lines on the screen, how many Os will there be in the last line? For full marks you must show how to calculate this without actually writing down the final line. F. Ben has a Super-Useless Lap Bottom computer, which has mysterious problems with the letters U, X and Z of its keyboard. You will need to discover its rule by describing what should replace each box in answering this question (the size of the boxes does not indicate the number of replacement letters): Lap Bottom: U, X, Z. Ben started by entering the letter U, and then like the others entered what he saw on the screen onto the next line; doing this 4 times he finally obtained: UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU What letter(s) replace each box above? Explain how you got your answer. Below, we repeat the above sequence of letters several times, in the hope it might be helpful in your scratchwork for this problem. UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU

5 Individual Questions Solutions 1. Answer: = 52 (5 3 1) = = 62 [1 mark] 2. Answer: 84. Initially there are = 12 edges. By removing a small cube from a vertex (of which there are 8), we increase the number of edges by 12 3 = 9. Hence, the resulting solid has = 84 edges. [1 mark] 3. Answer: 41. Represent the 2-digit number x by #. Putting 2 before it and a 9 after it, we get 2 #9 = #0 = x. We are told that this number is 59x. Thus we have x = 59x 2009 = 49x 41 = x. [2 marks] 4. Answer: 6. Since = is just a power of 6 and 6 6 = 36 also ends in 6, any power of 6 ends in 6. So the answer is 6. Alternative. Observe that 6 2 = 36 6 (mod 10) = n 6 (mod 10) for any integer n 1 6 (mod 10) Hence, the last digit of is 6. [2 marks]

6 5. Answer: 80. The new concentration is the the total volume of alcohol over the total volume of liquid expressed as a percentage: total volume of alcohol total volume 3 = = ( ) = (27 + 5) = = = 80% So the number of percent of the new concentration is 80. [2 marks] 6. Answer: 95. There are 120/5 = 24 words beginning with A, 24 beginning with B and 24 beginning with C. These all come before DECAB. Of those beginning with D there are 24/4 = 6 beginning with DA, 6 beginning with DB and 6 beginning with DC. These also come before DECAB. Those beginning with DE go DEABC, DEACB, DEBAC, DEBCA and DECAB. There are 5 of these. So DECAB s number is = 95. Alternative. There s less counting if one starts from the other end. There are 24 words beginning with E. Then DECBA is the last word beginning with D, and the one before it is DECAB. So DECAB s number is 120 minus the number that follow it, i.e. 120 (24 + 1) = 95. [3 marks] 7. Answer: 11. If y stations are added to x already existing stations, each new station will require (x + y 1) sets of tickets; for y new stations this is y(x + y 1) sets. Each old station needs y sets. So: y(x + y 1) + xy = 46 y(2x + y 1) = 46. Thus y must be a positive integer which is a factor of 46, i.e. it is 1, 2, 23, or 46. But y > 1, and y = 23 or y = 46 imply x < 0. y = 2, x = 11. Therefore there were 11 old stations. [3 marks]

7 8. Answer: 27. Let x be the number of hair clips Alice bought. Then the total of her purchases is: 4x + 32 so that the average price of her purchases is 4x (x + 1) + 28 = = x + 1 x + 1 x + 1, which is an integer, if 28 is divisible by x + 1. Hence, x + 1 must be one of 1, 2, 4, 7, 14 or 28 (the divisors of 28), i.e. x must be one of 0, 1, 3, 6, 13 or 27. The largest of these is 27. [3 marks] 9. Answer: 18. Let n be the number of sides of the polygon. Then, (n 2) 180 = n ( ) (n 1) (n 2) = n(143 + n 1) n 2 38n = 0 (n 18)(n 20) = 0. = n(142 + n) So n = 18 or n = 20. Since the n-gon is convex, all its angles, in particular, the largest, must be less than 180. Now, for n = 20, the largest angle is = 181 > 180. So, n 20. On the other hand, for n = 18, the largest angle which is ok. So n = = 177 < 180, [4 marks] 10. Answer: 30. Let y = F D. Since d is the perpendicular bisector of AM, it is the locus of points equidistant from A and M. So AF = MF. Since the area of the square is 64 cm 2, its side length is 8 cm. Hence applying Pythagoras Theorem to F DA and CF M, we have y 2 = (8 y) y = 16 = y 2 16y + 16 y = 1 Take the parenthesising of the vertices of a figure, as a convenient shorthand for the figure s area, so that (XY Z) means the area of figure XY Z. Then (AMF ) = (ABCD) (ABM) (F DA) (CF M) = 64 1 ( ) 2 = = = 30. So AMF has area 30 cm 2.

8 Alternative 1. Instead, let x = F C. As before, deduce AF = MF, and that square has side length 8 cm. Applying Pythagoras Theorem to F DA and CF M, we have (8 x) 2 = x x + x 2 = x = 2 8x x = = 8 1 = 7. The rest of the solution proceeds like the first solution. Alternative 2. Since the area of the square is 64 cm 2, its side length is 8 cm. Since M is the midpoint of BC, MB = 4 cm. Applying Pythagoras Theorem to ABM, we have AM = = = 4 5 Let the midpoint of AM be X, i.e. XM = XA. Then F M 2 = F X 2 + XM 2 = F X 2 + XA 2 = F A 2 So, F M = F A. Now deduce x or y as above and hence deduce that F M 2 = F A 2 = 65 F X 2 = F A 2 XA 2 = 65 (2 5) 2 = = 45 (AMF ) = 1 2 AM F X = = = 6 5 = 30 [4 marks]

9 Crazy Computers Team Questions Solutions A. Answer: OLLOM M OLOL. Putting a little space between the replacement letters, we have MLOMM OL LOM M OL OL. [4 marks] B. Answer: RSRRSRSRRSRRS. R RS RSR RSRRS RSRRSRSR RSRRSRSRRSRRS (1 st line) (2 nd line) (3 nd line) (4 th line) (5 th line) [5 marks] C The numbers of letters increases by the number of Rs in the line, which is the same as the number of letters in the previous line. Let l n be the length of the n th line and let l 0 = 1 be the length of the first entered line (namely R). Then l n+1 = l n + l n 1, n 0, where l 0 = 1, l 1 = 2. So we have: l 2 = l 1 + l 0 = = 3 l 3 = l 2 + l 1 = = 5 l 4 = l 3 + l 2 = = 8 l 5 = l 4 + l 3 = = 13 l 6 = l 5 + l 4 = = 21 l 7 = l 6 + l 5 = = 34 l 8 = l 7 + l 6 = = 55 l 9 = l 8 + l 7 = = 89 l 10 = l 9 + l 8 = = 144 l 11 = l 10 + l 9 = = 233 l 12 = l 11 + l 10 = = 377. So the 12 th line has 377 letters.

10 It helps to recognise that l n = F n+1, where F n is the n th term of the Fibonacci sequence. [8 marks] D. Answer: GH. One can show that HGGHGHHG arises from the replacements H HG, G GH. [6 marks] E. Answer: 11. Let l n, m n, o n be the number of Ls, Ms and Os, respectively on line n, or initially input in the case when n = 0. Then l 0 = m 0 = 0, o 0 = 1 and for n 1, l n = l n 1 + m n 1 m n = l n 1 + o n 1 o n = l n 1 + m n 1 = l n. Representing this in a table we have n l n = l n 1 + m n 1 m n = l n 1 + o n 1 o n = l n 1 + m n So there are 11 Os in the sixth line. [12 marks] F. Answer: U UZU, X ZX, Z UX. Looking at the final line we see the following repetitions. UZUUXUZUUZU ZX UZUUXUZUUZUUXUZUUX ZX UZUUXUZUUZU ZX UZUUXUZU If U UZU then passing from the third to fourth line, we require Z UX. So the remaining letters must be the result of Z, i.e. we must have X ZX. Confirming this U UZU UZUUXUZU UZUUXUZUUZUZXUZUUXUZU UZUUXUZUUZUZXUZUUXUZUUZUUXUZUUXZXUZUUXUZUUZUZXUZUUXUZU [10 marks]

11 Western Australian Junior Mathematics Olympiad 2008 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions 1 to 9. Calculators are not permitted. Write your answers on the answer sheet provided. 1. Below are three different views of a child s building block and a single view of a different block Which is the different block? [1 mark] 2. Some horses and some jockeys are in a stable. In all, there are 71 heads and 228 legs. How many jockeys are in the stable? [1 mark] 3. Four friends go fishing and catch a total of 11 fish. Each person caught at least one fish. The following five statements each have a label from 1 to 16. What is the sum of the labels of all the statements which must be true? 1: One person caught exactly 2 fish. 2: One person caught exactly 3 fish. 4: At least one person caught fewer than 3 fish. 8: At least one person caught more than 3 fish. 16: Two people each caught more than 1 fish. [2 marks] 4. There are four throwers in the shot put final at the Olympic games. The distance thrown by the second thrower is 2% less than the first thrower while the third thrower achieves a distance 20% greater than the first thrower. The fourth thrower throws it 10% further than the third person. If the total distance of the four throwers is 90 m how many metres did the first thrower throw the shot put? [2 marks]

12 5. A cube of side 7 cm is painted green all over, then cut into cubes of side 1 cm. How many of these small cubes have exactly 2 faces painted green? [2 marks] 6. Ann is four times as old as Mary was when Ann was as old as Mary is now. Furthermore, Ann is twice as old as Mary was when Ann was six years older than Mary is now. How old is Ann? [3 marks] 7. A barrel contains a number of blue balls and a number of green balls which you take out one by one. Each time you take out a blue ball somebody puts 100 frogs into the barrel and each time you take out a green ball the person puts in 72 frogs. Finally, when you have removed all the balls, you find there are 2008 frogs in the barrel. How many green balls were there in the barrel initially? [3 marks] 8. In the figure, B is the mid-point of AD, C is the mid-point of DE, A is the mid-point of EF, and M is the midpoint of AF. If the area of AM B is 6 cm 2, how many cm 2 is the remaining area of DEF? F D M B A C E [3 marks] 9. Five grandmothers go to a cafe to eat cake. The cafe sells 4 different types of cake. Each grandmother chooses two different cakes. They find their bills are for $6, $9, $11, $12 and $15. The next day I go to the cafe and buy one of each type of cake. How much do I pay? [4 marks] 10. For full marks, explain how you found your solution. A square is divided into three x pieces of equal area as shown. The distance between the parallel lines is 1 cm. What is the area of the square in cm 2. z 1 y [4 marks]

13 Western Australian Junior Mathematics Olympiad 2008 Team Questions 45 minutes General instructions: Calculators are (still) not permitted. How to multiply on Titania The Titans are an intelligent race who live on the planet Titania. They use the same numerals as us for the integers and their addition and subtraction is the same as ours. However, they don t use multiplication. Instead of, they have an operation called star which has the following properties in common with : For all integers a, b and c, a c = c a and (a b) c = a (b c). However the other properties of star are completely different: (i) For all integers a, a 0 = a, and (ii) For all integers a and b, a (b + 1) = (a b) + (1 a), e.g. 7 2 = (7 1) + ( 6). A. Copy and complete the table shown. The entry in the row labelled by a and the column labelled by b should be a b, where each of a, b range over 0, 1,..., 4. Some entries have been filled in to get you started B. Show that for every integer a, a 1 = 1. C. Find ( 4) ( 3). D. Show that for any integer a, 4 a = 4 3a. E. Show that for all integers a and positive integers b, a b = a+b ab. Hint. Try a = 5 and try to generalise your argument.

14 F. Show how to express a b, for any (positive, negative or zero) integers a, b, in terms of our addition and multiplication. G. The Titans also use powers by defining a n to be a a a, where a appears n times, for n a positive integer, e.g. a 1 = a, a 2 = a a, etc. Find (2 3 ) (3 2 ) and (3 3 ) 3. H. Write the Titans power operation a n in terms of our operations, where n is a positive integer and a is any integer. Hint. To start, try a 1, a 2, a 3 and look for a pattern.

15 Individual Questions Solutions 1. Answer: 3. The views 1, 2 and 4 can be explained thus: the star face is opposite the square face, and the pentagon face is opposite the triangle (the face opposite the octagon is unknown). Then 2 is obtained from 1 by rotating in an axis perpendicular to the centre of the octagon face so that the triangle face becomes the front face. And 4 is obtained by flipping 1 upside down and then rotating in the vertical axis so that the octagon is the front face. Flipping 2 so that the octagon is on the top and the star on the right, the triangle face should be behind and the pentagon at the front. So 3 is the different block. [1 mark] 2. Answer: 28. Let j be the number of jockeys and let h be the number of horses. Then j + h = 71 = 2j + 2h = 142 2j + 4h = 228 = j + 2h = 114 Thus the number of jockeys is 28. = j = = 28 [1 mark] 3. Answer: 4. The example: 3 of the friends caught 1 fish and 1 caught 8 fish meets the criteria, so that the statements labelled 1, 2 and 16 need not be true. Also, the example: 3 of the friends caught 3 fish and 1 caught 2 fish, shows that the statement labelled 8 need not be true. Now, suppose for a contradiction that the statement labelled 4 is false, then all 4 friends caught 3 or more fish, which implies there are 12 or more fish, but there are only 11 fish (contradiction). Thus, at least one friend caught fewer than 3 fish. Hence the statement labelled 4 must be true, and since this is the only statement that must be true, the sum of such labels is 4. [2 marks] 4. Answer: 20. Let x be the distance thrown by the first person in metres. Then the second throws it 0.98x, the third, 1.2x and the fourth 1.32x. Thus 4.5x = 90 and so x = 20. [2 marks] 5. Answer: 60. The cube has 12 edges. Along each of those edges of the 7 cm cube, 5 will have exactly two faces painted green. So there are 12 5 = 60 such cubes. [2 marks]

16 6. Answer: 24. Let Ann s age be a and Mary s age be m, and write a 1, a 2 and m 1, m 2 be their ages at the two other times. Write a 1 = a+k (so that m 1 = m+k), and a 2 = a+l (so that m 2 = m+l. Then rewriting the statements with their ages, we have: Ann [is a and] is four times as old as Mary was [when she was m 1 and] when Ann [was a 1 and] was as old as Mary is now [namely m]. Furthermore, Ann [is a and] is twice as old as Mary [was when she was m 2 which] was when Ann [was a 2 and] was six years older than Mary is now [namely m]. Thus from the given information we have: a = 4m 1 = a = 4(m + k) = a 4m 4k = 0 (1) a 1 = m = a + k = m = 4a 4m + 4k = 0 (2) a = 2m 2 = a = 2(m + l) = a 2m 2l = 0 (3) a 2 = 6 + m = a + l = 6 + m = 2a 2m + 2l = 12 (4) Firstly, we eliminate k and l: 5a 8m = 0, (1) + (2) (5) 3a 4m = 12, (3) + (4) (6) a = 24, 2 (6) (5) Hence, Ann is 24 (and Mary is 5 24 = 15). 8 [3 marks] 7. Answer: 14. Say we started with b blue balls and g green balls. So we must find integer solutions to the equation 100b + 72g = Dividing through by 4 and rearranging gives 18g = b. We now try values of b until we find one that makes the right hand side divisible by 18. This happens when b = 10 since = 252 = So he had 14 green balls. [3 marks] 8. Answer: 42. Write ( XY Z) for the area of XY Z and let S = ( ABC). Then ABC and DBC have a common altitude to C.

17 Hence ( DBC) = ( ABC) = S Similarly, ( ECA) = ( DCA) = 2S ( F AD) = ( EAD) = 4S AM AF = 1 2 = AB AD MAB = F AD MAB F AD, ( MAB) = ( F AD) 2 2 = S by the Similar s SAS Rule Now S = 6 and ( DEF ) ( AMB) = 7S = 42. [3 marks] 9. Answer: 21. Say the cakes have prices a, b, c and d. The possible totals for the price of two cakes are the 6 = ( 4 2) sums a + b, c + d, a + c, b + d, a + d, b + c. Since the grandmothers all paid different amounts the amounts they paid are 5 of these 6 sums. Notice that the sum of the first and second pairs is a + b + c + d, so is the sum of the third and fourth, so is the sum of the fifth and sixth. This means that two pairs of grandmothers bills have the same total. The totals are: = = = = = = = = = = 27 We see that there is only one pair of equal sums: = 21 and = 21 so a + b + c + d = 21, which is the amount I pay for my 4 cakes. [4 marks]

18 10. Answer: 13. With x, y and z as shown, the middle strip is a parallelogram made up of two congruent triangles of height x and base y, and this is a third of the total area, i.e. By Pythagoras Theorem, xy = 1 3 x2 y = 1 3 x z 2 = x 2 + (x y) 2 = x 2 + ( 2 3 x)2 = 13 9 x2 z = x. Looking at the two congruent triangles making up the parallelogram another way: they have height 1 and base z, i.e. the area of the parallelogram is also given by: 1 z = x = 1 3 x2 13 = x Thus, the area of the square is x 2 = 13 cm 2. [4 marks]

19 How to multiply on Titania Team Questions Solutions A [8 marks] B. a 1 = a a, by (ii) = a + 1 a, by (i) = 1 [3 marks] C. Answer: 19. Rearranging (ii) we have Hence a b = a (b + 1) (1 a) ( 4) ( 3) = ( 4) ( 2) (1 4) = ( 4) ( 1) (1 4) (1 4) = ( 4) ( 1) 2(1 4) = 4 0 3(1 4) = 4 15, by (i) = 19. [4 marks] D. 4 a = a 4 = a a = a 2 + 2(1 a) = a 1 + 3(1 a) = a 0 + 4(1 a) = a + 4 4a = 4 3a. [5 marks]

20 E. a b = a (b 1) + (1 a) = a (b 2) + (1 a) + (1 a). = a (b k) + k(1 a) = a 0 + b(1 a) = a + b ab. [6 marks] F. Observe that a 0 = a = a + 0 a 0. i.e. for b = 0, the formula a b = a + b ab still works. Rearranging, a (b + 1) = (a b) + (1 a), we have a b = a (b + 1) (1 a). Suppose b = β, where β > 0. Then a β = a ( β + 1) (1 a) = a ( β + 2) (1 a) (1 a). = a ( β + k) k(1 a) = a 0 β(1 a) = a + β a( β) = a + b ab, again, since in this case b = β. Thus, since we already showed a b = a + b ab, for positive b, a b = a + b ab, a, b Z. [6 marks]

21 G. Answer: (2 3 ) (3 2 ) = 5, (3 3 ) 3 = = (2 2) 2 = 0 2 = 2, from table in A 3 2 = 3 3 = 3, from table in A (2 3 ) (3 2 ) = 2 ( 3) = , from rule in E = = (3 3) 3 = ( 3) 3 = , from rule in E = 9 (3 3 ) 3 = (9 9) 9 = ( ) 9 = ( 63) 9 = = 9( ) = 9 57 = 513 [6 marks] H. Method 1.Using the a b = a + b ab rule, again and again. a 1 = a = 1 (1 a) a 2 = a + a a a = 2a a 2 = 1 (1 a) 2 a 3 = (a + a a a) + a (a + a a a) a = 4a 6a 2 + 4a 3 a 4. ( n a n = na 2 = 1 (1 a) n ) a ( 1) k+1 ( n k ) a k + + ( 1) n+1 a n

22 Method 2. a 1 = a = 1 (1 a) a b = a + b ab = 1 (1 + ab a b) = 1 (1 a)(1 b) a 2 = a a = 1 (1 a)(1 a) = 1 (1 a) 2 a 3 = (a a) a ( = 1 1 ( 1 (1 a) 2)) (1 a) = 1 (1 a) 2 (1 a) = 1 (1 a) 3. a (k+1) = a k a ( = 1 1 ( 1 (1 a) k)) (1 a) = 1 (1 a) k (1 a) = 1 (1 a) k+1 a n = 1 (1 a) n [7 marks]

23 Western Australian Junior Mathematics Olympiad 2007 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions 1 to 9. Calculators are not permitted. Write your answers on the answer sheet provided. 1. I m a two digit number. I m one less than a multiple of 8 and three less than a multiple of seven. What is the least number I could be? [1 mark] 2. The diagram, which is not drawn to scale, shows a rectangle divided by a horizontal and a vertical line into four rectangles. The areas of three of them are shown. What is the area of the whole rectangle? [2 marks] 3. From a group of girls and boys, fifteen girls depart, leaving twice as many boys as girls. Then 45 boys depart, leaving five times as many girls as boys. How many girls were there originally? [2 marks] 4. A regular pentagon has five diagonals and they are all of the one length. A regular hexagon has nine diagonals and they are of two different lengths. If we consider all of the diagonals of a regular polygon which has twenty sides, how many different lengths will there be? [2 marks]

24 5. There are three cards on a table, each marked with a positive whole number. Alice says The sum of these two is 54. Bill points to another pair of cards and says The sum of these is 41. Finally, Cyril points to another pair and says The sum of these two is 33. What is the sum of all three cards? [2 marks] 6. Given that a, b, c and d are positive integers with a < 2b, b < 3c, c < 4d and d < 5, what is the largest possible value of a? [3 marks] 7. Gwen has four children, one is a teenager (13 to 19 years old) and the product of their ages is How old is the teenager? [3 marks] 8. An arithmetic progression is a sequence of numbers such that the difference of any two successive numbers is a constant. For example, 3, 5, 7 is an arithmetic progression of three numbers, with common difference 2. Also, 3, 5, 7 are prime. The sum of the numbers is = 15, which is the lowest possible sum for an arithmetic progression of primes of length 3. Find an arithmetic progression of four numbers, all of which are prime, which has the lowest sum of any arithmetic progression of primes of length 4. What is the sum of your four numbers? [3 marks] 9. The picture shows a board of nails on a 1 cm grid. Jane wants to put rubber bands around some of the nails to make squares. Two such squares are shown in the diagram (the larger square has side length 2 cm). How many square centimetres is the total area of all the squares she can make? [3 marks] 10. For this question you must show working to all parts. Two snails, Alfa and Romeo, both set out at the same time to go along the same road from X to Y. Alfa crawled at a constant speed of 12 m/h (metres per hour) till he reached Y. Romeo started out at 8 m/h but after two hours he realised he was falling behind so hitched a ride on a passing turtle called Toyota, who was on her way to Y at a constant speed of 20 m/h. Toyota and Romeo soon caught up with Alfa and two hours after doing so reached Y. (a) How many hours after leaving X did it take Romeo to catch Alfa? (b) How many hours after starting out from X did Romeo reach Y? (c) How many metres is it from X to Y? [4 marks]

25 Individual Questions Solutions 1. Answer: 39. The number has the form 8n 1 and so belongs to the set {7, 15, 23, 31, 39, 47,... }. It also has the form 7n 3 and so belongs to {4, 11, 18, 25, 32, 39, 46,... }. So it must be Answer: 35. The rectangles on the left have the same width, so their areas are proportional to their heights, namely 8 to 6. The rectangles on the right also have areas proportional to their heights, so the unknown area is 9 8/6 = 12, and the area of the whole rectangle is = Answer: 40. Say there were g girls and b boys originally. Then b = 2(g 15), after the first departure (1) g 15 = 5(b 45), after the second departure (2) g 15 = 5 ( 2(g 15) 45 ), substituting for b from (1) in (2) 5 45 = 9(g 15) 5 5 = g 15 g = Answer: 9. If we think about the diagonals starting at vertex number 1 we see that they increase from the shortest diagonal, from vertex 1 to 3, to the longest, from vertex 1 to vertex 11, then decrease again, in a symmetric fashion, down to the diagonal from vertex 1 to vertex 18. Thus there are 9 different lengths. 5. Answer: 64. If the numbers on the cards are a, b and c then we get the system of equations a + b = 54 (3) a + c = 41 (4) b + c = 33 (5) 2(a + b + c) = = 128, adding (3), (4) and (5) a + b + c = Answer: 87. The largest possible value of d is 4 = 5 1. So the largest possible value of c is 15 = So the largest possible value of b is 44 = Hence the largest possible value of a is 87 = Answer: 14. We have the factorisation 1848 = The only combination of factors yielding a teen is 2 7, so the teenager is 14. (There is more than one possibility for the other three ages!) 8. Answer: 56. The arithmetic progression 5, 11, 17, 23 consists of primes, has sum = 56 and has common difference 6. There is no arithmetic progression of length 4 starting at 3 and with common difference 2, 4 or 6. There is none starting at 5 with common difference 2 or 4. If there is one starting at 7 with smaller

26 sum it must have common difference 2 or 4, but no such progressions exist. There are none starting at 11 with difference 2 or 4, nor 13. Since 56/4 = 14, there can t be an arithmetic progression with smaller sum starting with a prime greater than 13. Thus 5, 11, 17, 23 has the smallest sum (56) of an arithmetic progression of primes of length Answer: 52. There are nine 1 1 squares with area 1, four 2 2 squares with area 4, one 3 3 square with area 9, four 2 2 squares with area 2 and two 5 5 squares with area 5. So the total area is = (a) Romeo covers 16 m in the first two hours. Suppose Romeo and Toyota overtake Alfa t hours later. Then they have travelled t m. meanwhile, Alfa has travelled 12(2 + t) m. Hence t = t. So t = 1. Hence it took a total of 3 hours to catch Alfa. (b) Romeo travelled 2 hours on his own and 3 hours on Toyota s back, a total of 5 hours. (c) Romeo travelled 16 m on his own and 60 m on Toyota s back, a total of 76 m.

27 Western Australian Junior Mathematics Olympiad 2007 Team Questions General instructions: Calculators are (still) not permitted. 45 minutes A teacher has a class of 100 students whom he numbers from 1 to 100. He has given each a t-shirt showing his or her number, and taken them to a very long corridor with 100 doors. The doors are also numbered from 1 to 100. When a student goes down the corridor for each door whose number is divisible by the number on his/her t-shirt, he/she either closes the door if it is open or opens the door if it is closed. Students don t touch doors with numbers that are not divisible by their t-shirt numbers. For example, if the doors are all closed to start with then when student 40 goes down the corridor he will open doors 40 and 80. If student 80 follows him she will close door 80. A. If all the doors are closed, and students 7, 28 and 84 go down the corridor, which doors will be open? B. All the doors are closed except number 42. Which students should he send down to get all the doors closed? C. Suppose all the doors are closed and all the students go down the corridor. Which of the doors 49, 51 and 53 will be open? D. If all the doors are closed, and students 1, 2, 4, 8, 16, 32 and 64 go down the corridor, explain how you can predict which doors will be closed? E. If all the doors are closed, and all the students go down the corridor, which doors will be open? Explain your answer not just by saying which doors are open and which are closed. F. The doors from 1 to 49 are closed but somebody has left all the others open. Which students should he send down to ensure all the doors are closed? G. Now the doors from 1 to 49 are open and the rest are closed.which students should he send down to get all the doors closed? H. All the doors were closed but the Number 1 student has just walked down the corridor and gone home. The teacher now has no way of closing door number 1. But is it possible to use the remaining students to close all the other doors? Explain your reasoning.

28 Team Questions Solutions A. 7, 14, 21, 35, 42, 49, 63, 70, 77, 84, 91, 98. [4 marks] B. 42 and 84. [4 marks] C. Only 49 will be open, because 49 has 3 factors, i.e. it is moved by students 1, 7, 49; and 51 and 53 will be closed because they have 4 and 2 factors, respectively. [6 marks] D. Doors 2, 6, 8, 10, 14, 18, 22, 24, 26, 30, 32, 34, 38, 40, 42, 46, 50, 54, 56, 58, 62, 66, 70, 72, 74, 78, 82, 86, 88, 90, 94, 96, 98 will be closed. To decide whether a given door is closed or not find the highest power of 2 which divides the door number. If this is an odd power of 2 (2, 8 or 32) then the door will be closed. Otherwise it will be open. [7 marks] E. The doors 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 be open because these are perfect squares and such numbers have an odd number of factors, and so will be opened or closed by an odd number of students. [6 marks] F. 50, 51, 52,..., 99 (not 100). [6 marks] G. 1 and 50, 51, 52,..., 99. [4 marks] H. Yes, send down the student whose number is that of the first open door, and continue doing this till all doors other than 1 are closed. This is the same as sending all students except those with square numbers on their shirts. [8 marks]

29 Western Australian Junior Mathematics Olympiad 2006 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions 1 to 9. Calculators are not permitted. Write your answers on the answer sheet provided. 1. The diagram below shows the train line from the outer suburb of A to the inner city station E, with the distance between stations shown in kilometres. After leaving a station the train travels at an average speed of 30 km/h for the first kilometre, then at 60 km/h until it reaches the next station. It spends 2 minutes at each station. How many minutes will elapse between the train leaving A and arriving at E? 3 km 4 km 6 km 1 km A B C D E [1 mark] Solution. 60 km/h = 1 km/minute. So when the train travels at 60 km/h, it takes 1 minute to cover a km, and when it travels at 30 km/h, it takes 2 minutes to cover a km. So the minutes that will elapse between the train leaving A and arriving at E is given by: (2 + 2) (2 + 3) (2 + 5) (2 + 0) = 24 where the first number (2) in each bracket is the time spent at 30 km/h, the second number in each bracket is the number of kilometres travelled at 60 km/h (which equals the time in minutes to cover that distance) and each unbracketed 2 is the waiting time at a station. Answer: Find the two-digit prime number that is 2 less than a perfect square that is 2 less than a prime. [1 mark] Solution. Let the prime be p. Then p has 2 digits and hence is odd (2 is the only even prime). It is 2 less than a square. So the square is odd; possibilities are: 25, 49, 81. Of these only 81 has a prime both 2 less than it and 2 more than it. So p must be 79. Answer: 79.

30 3. For how many positive integers (whole numbers) n are n, 3n and n/3 all three-digit integers? [2 marks] Solution. The least that n/3 can be is 100 (the smallest 3-digit positive integer) and the most that 3n can be is 999 (the largest 3-digit positive integer). If 3n = 999 then n/3 = 111. So n/3 can be any integer from 100 to 111 inclusive, a total of 12 possibilities, leading to 12 possibilities for n (300, 303,..., 333). Answer: A point E lies outside a square ABCD so that ABE is an equilateral triangle. What is the measure in degrees of the angle CED? E A B D C [2 marks] Solution. Since ABE is equilateral, its angles are all 60. Hence EAD = Now ADE is isoceles. Thus AED = 1 2 (180 EAD) = 15 Similarly, BEC = 15. So finally we have Answer: 30. CED = = A video store has a choice of 920 films to rent. You can rent some on DVD, some on video cassette and some on both. If the store owns a total of 1000 DVDs and video cassettes, how many films are available on both video cassette and DVD? [2 marks] Solution. By the pigeon hole principle, 80 films are available on both video cassette and DVD (the films are the pigeon holes; put 920 of the 1000 DVDs and video cassettes in the 920 available pigeon holes; the remaining 80 DVDs or video cassettes must go in a pigeon hole that already contains a DVD or video cassette). Answer: 80.

31 6. Brenda was sick on the day of the maths test so she had to sit for it the next day. Her score of 96 raised the class average from 71 to 72. How many students (including Brenda) took the test? [3 marks] Solution. Let n be the number of students in the class. Without Brenda the average is 71 and hence the total marks without Brenda is 71(n 1). The average with Brenda s mark of 96 is 72. So 71(n 1) = n 72n = 71(n 1) + 96 = 71n n = 25 So there are 25 students in the class. Answer: A plane flies in still air at an average speed of 810 km/h for the duration of its flights. When flying from Perth to Sydney it takes four hours while from Sydney to Perth it takes five hours. Assuming that the wind is at a constant speed and from the west (i.e. in the direction from Perth to Sydney) for both flights, what is the speed of the wind? [3 marks] Solution. Let w be the speed of the wind (from west to east). Then travelling east the plane s speed is (810 + w) km/h and travelling west its speed is (810 w) km/h. In general velocity v, distance d and time t are related by v = d/t. Let d now denote the distance from Perth to Sydney. Then w = d w = d 5 (1) (2) Eliminating d by dividing (1) by (2) we have w 810 w = 5 4 4(810 + w) = 5(810 w) 4w = 810 5w 9w = 810 w = 90 Therefore the wind speed w is 90 km/h. Answer: 90.

32 8. The five faces of a right square pyramid all have the same area. The height of the pyramid is 3 m. What is its total surface area in square metres? 3 m Solution. Let a be the sidelength of the square base, so that the area of the base is a 2 which is also the area of each isosceles triangular face. So each triangular face has height 2a, and the total surface area is 5a 2. Using Pythagoras Theorem and the pyramid height of 3 m, we have (2a) 2 (a/2) 2 = 3 2 (4 1 4 )a2 = 15 4 a2 = 3 4 5a2 = 9 5a 2 = = 12 So the total surface area of the pyramid is 12 m 2. Answer: 12. [3 marks] 3 m 9. Jasper glues together 504 cubes with 1 cm edges to make a solid rectangular-faced brick. If the perimeter of the base of the brick is 64 cm, what is the height of the brick? [4 marks] Solution. 504 = 9 56 = = The volume of the brick L B H = 504 cm 3, whereas the perimeter of the base 2(L + B) = 64 cm. So we need to write 504 as the product of three positive integers L, B, H such that L + B = 32. We see that = 504, so that (in cm) L = 18, B = 14, H = 2 is a solution. To see the solution H = 2 is unique (not required for the competition), observe first that L must satisfy 16 L < 32 (taking L B). Now L and B must both be even or both odd since there sum is even. If L and B are both odd then their product is 63 leading to L = 21 (to satisfy the inequality) and B = 3 which do not sum to 32. Hence L and B must both be even, whence H = 1 or 2. If H = 1 then the least L + B could be is > 2 22 = 44 > 32. Thus H = 2 cm. Answer: 2.

33 10. In the AFL Grand Final between the Knockers and the Beagles the Knockers won by 3 points. This was despite the Beagles kicking 16 scoring shots compared to the Knockers 14. It was also noticed that the number of behinds kicked by the Beagles was greater than the number of goals kicked by the Knockers, and the number of goals kicked by the Beagles was greater than the number of behinds kicked by the Knockers. How many points did the Beagles score? Note: in the AFL game there are two ways to score: behinds score 1 point each, and goals score 6 points each. For full marks, explain how you found your solution. Solution. Let... B = the number of goals kicked by the Beagles, b = the number of behinds kicked by the Beagles, K = the number of goals kicked by the Knockers, and k = the number of behinds kicked by the Knockers. From the given information, we have: [4 marks] B + b = 16 (3) K + k = 14 (4) 6K + k = 6B + b + 3 (5) B > k (6) b > K (7) Since B and b are integers, from (6) and (7) we have B k + 1 (8) b K + 1 (9) Adding the inequalities (8) and (9) and rearranging we get (B + b) (K + k) 2. But subtracting (3) and (4) gives (B + b) (K + k) = 2. So the inequalities (8) and (9) must actually be equalities: B = k + 1 (10) b = K + 1 (11) Substituting (10) in (3) and rearranging we have Rearranging (4) we have B = 15 K (12) k = 14 K (13)

34 Substituting (11), (12) and (13) in (5) we have 6K + 14 K = 6(15 K) + K K = 80 K = 8 (14) Substituting (14) in (12) and (11), we have: so that the Beagles scored B = 7 (15) b = 9 (16) 6B + b = 51 points Note the problem can also be solved by systematically listing the possible scores for the Beagles (17 of them) and the possible scores for the Knockers (15 of them) and then doing a careful elimination... and there are many other ways. The more satisfying solutions are ones that set up some inequality at the beginning that reduce the possibilities to a small list prior to doing an elimination. Answer: 51.

35 Western Australian Junior Mathematics Olympiad 2006 Team Questions 45 minutes General instructions: Calculators are (still) not permitted. Consider a garden table made of 15 square tiles in a 5 3 arrangement. The table has a straight crack along a diagonal. Seven of the individual tiles are broken. Now consider a 6 4 rectangle. This time eight tiles are broken. Try some other sizes. Use the squared paper provided. A. How many tiles get broken when an 8 6 table is cracked along a diagonal? Solution. 12 tiles get broken. B. Give the dimensions of two different rectangular tables that get nine tiles broken when they are cracked along a diagonal. Note. Remember that a square is also a rectangle. Also, note that for this and subsequent questions, an 8 6 table, for example, is considered the same as a 6 8 table. Solution. The possible dimensions are: 9 1, 7 3, 9 3, 9 9. C. How many different rectangular tables can you find that get ten tiles broken when they are cracked along a diagonal? Write down their dimensions. Solution. The possible dimensions are: 10 1, 10 2, 10 5, 10 10, 9 2, 8 3, 7 4, 6 5. D. Try some square tables. Describe what happens. Solution. The number of broken tiles is the same as the length of the side of the square.

36 E. What happens when the shorter dimension of the table is 1? Solution. Every tile is broken, i.e. the number of tiles is broken is the length of the rectangle. F. For what sort of dimensions does the crack go through corners of tiles inside the rectangle? Solution. For dimensions that have a common factor (larger than 1). G. How many tiles are cracked when the diagonal does not go through any corner of a tile inside the rectangle? Explain your reasoning. Solution. If the diagonal does not go through any corner of a tile inside the rectangle, a tile is cracked when and only when the diagonal enters a new column or enters a new row. Say the table has m rows and n columns. The first tile broken is in the first row and the first column. Then there are m 1 further rows and n 1 further columns. Therefore the number of tiles cracked is 1 + (m 1) + (n 1) = n + m 1. H. Predict the number of broken tiles in a rectangle. Solution. Remove the highest common factor 8. Then consider a 7 4 rectangle which has = 10 broken tiles. Reintroduce the removed common factor and the number of broken tiles is 10 8 = 80. Alternatively: In general, the number of cracked tiles in a table with dimensions m n is m+n hcf(m, n) where hcf(m, n) counts the number of times the diagonal enters a new column and new row at the same time, which, if we think of the first tile broken as being the bottom lefthand tile, is the number of times the crack goes through the bottom lefthand corner of a tile. So for a rectangle, hcf(56, 32) = = 80 tiles are broken. I. Explain how you can predict the number of broken tiles in any size of table. Solution. If the two dimensions have no common factor, add the two numbers and subtract 1. If the two dimensions do have a common factor, remove the highest common factor), consider the reduced table as above, then multiply the result by the removed HCF. Algebraically, the number of tiles broken is ( m hcf(m, n) + n ) hcf(m, n) 1 hcf(m, n) = m + n hcf(m, n). Alternatively: use the general argument in H to obtain the general formula m + n hcf(m, n).

37 Western Australian Junior Mathematics Olympiad 2005 Individual Questions 100 minutes General instructions: Each solution in this part is a positive integer less than 100. No working is needed for Questions 1 to 9. Calculators are not permitted. Write your answers on the answer sheet provided. (1) ABCD is a trapezium with AB parallel to DC, AB = 4 cm, DC = 11 cm and the area of triangle ABP is 12 cm 2. What is the area of the trapezium ABCD in square centimetres? D A 4 cm 12 cm 2 P B 11 cm C [1 mark] (2) A library has 6 floors. There are more books on the second floor than the first. The number of books on the third floor is the same as the number on the second. There are fewer books on the fourth floor than the third and twice as many books on the fifth floor as there are on the fourth. On the sixth floor there are fewer books than on the fifth. Coincidentally the number of books on the sixth floor is the same as the number on the first. Altogether, how many thousands of books are there in the library? [1 mark] (3) XY Z is a three-digit number. Given that XXXX + Y Y Y Y + ZZZZ = Y XXXZ, what is X + Y + Z? [2 marks]

38 (4) Mathilda and her friends have 20 1-metre sticks out on the basketball court to make triangles. Below is a diagram of a triangle using all 20 sticks. It has sides of 5 m, 7 m and 8 m. All triangles with sides of 5 m, 7 m and 8 m (even mirror images as shown) are considered to be the same. 5 m 7 m 7 m 5 m 8 m 8 m Altogether, how many different triangles could they make, one at a time, each time using all 20 sticks? [2 marks] (5) What is the size of the angle, in degrees, between the hands of a clock when the time is ten past eleven? [2 marks] (6) A plane is due to leave Perth at midnight and arrive at Tokyo at 12:29 pm, but it departs 49 minutes late. What percentage increase in speed is required in order that it will arrive exactly on time? [3 marks] (7) Esther has 20 coins in her purse. They are 10c, 20c and 50c coins and the total value is $5. If she has more 50c coins than 20c coins, how many 10c coins has she? [3 marks] (8) Alice spent all her money in five shops. In each shop, she spent $1 more than half of what she had when she entered that shop. How many dollars did Alice have when she entered the first shop? [3 marks]

39 (9) Given a square ABCD, circular arcs centred at B and D are drawn from A to C. Now draw diagonal BD to cut these arcs at X and Y, respectively. If XY = , what is the area of the square ABCD? A X Y B D C [4 marks] (10) Farmer Brown runs a dairy farm with cows, sheep and goats. Dabbling in mathematics in his spare time, he noticed that the numbers of each animal were different prime numbers. He also observed that if he multiplied the number of cows by the total number of cows and sheep, he obtained a number just 120 greater than the number of goats. How many goats are there? For full marks, explain how you found your solution. [4 marks]

40 Western Australian Junior Mathematics Olympiad 2005 Team Questions General instructions: Calculators are (still) not permitted. 45 minutes A number appears on your computer screen. If you push the A key the number is decreased by 3, if you push the B key it s increased by 3 and if you press the C key the number is halved. You want to use a sequence of key strokes to change the number to 1. For example, there are three ways to change 16 to 1: Use CCCC giving the number sequence 16, 8, 4, 2, 1. Use AAAAA giving 16, 13, 10, 7, 4, 1. Use CCA giving 16, 8, 4, 1. So the shortest path from 16 to 1 requires 3 key strokes. A. Find a starting number between 30 and 40 whose shortest path is 4, and show the path. B. Find a starting number between 50 and 60 whose shortest path is 5, and show the path. C. Find a starting number between 30 and 40 whose shortest path is 6, and show the path. D. Which starting number between 40 and 50 has the longest shortest path? E. Find a number which has two shortest paths, one beginning with the A key and one with the B key. F. Find a starting number where the add 3 operation must be used to get the shortest path. G. What starting numbers can never get to 1? H. What starting number between 500 and 1000 would have the shortest path? I. In this question, the numbers on the screen don t have to be integers. If we start with any number then these two sequences of key strokes always produce the same final number: ACCB and BCBC. Explain why.

41 Solutions Solutions to Individual Questions (1) Let h be the height of the trapezium. This is also the height of triangle ABP. Thus 12 = 1 4h so h = 6. The area of the 2 trapezium is then AB + DC Area trap = h = = (2) Let there be x thousand books on the first floor. On the second floor there are x + 10 thousand books. On the third floor there are x + 10 thousand books. On the fourth floor there are x thousand books. On the fifth floor there are 2x thousand books. On the sixth floor there are 2x 4 thousand books. Thus we have 2x 4 = x and so x = 4. The total number of books, in thousands, is therefore x + x x x + 2x + x = 7x + 20 (3) Writing the sum in the traditional way, XXXX Y Y Y Y Z Z Z Z Y XXX Z = = 48. we observe from the last column that X + Y + Z = Z + 10k, where k is the carry to the second column, and so X +Y = 10k. Now X + Y can t be as big as 20, and, since XY Z is a 3-digit number, X 0. Thus X + Y = 10 (1) and the carry k = 1. Looking at the second column now, we have X + Y + Z + 1 = X + 10m, i.e. Y + Z + 1 = 10m, where m is the carry to the third column. Again, Y + Z + 1 can t be as big as 20, so Y + Z + 1 = 10 (2) and m = 1. We see that this pattern is repeated for the third and fourth columns, and so the carry to the fifth column which is just Y, implies Y = m = 1. Adding equations (1) and (2) we have X + Y + Y + Z + 1 = 20 X + Y + Z = 20 1 Y = 18 (Of course, we had enough information to evaluate that X = 9 and Z = 8.)