Division S4 Singapore and Asian Schools Math Olympiad 2015 Full Name: Index Number: Class: School: SASMO 2015 Secondary 4 Contest INSTRUCTIONS 1. Please DO NOT OPEN the contest booklet until the Proctor has given permission to start. 2. TIME: 1 hour 30 minutes. 3. Attempt all 25 questions. Questions 1 to 15 score 2 points each, no points are deducted for unanswered question and 1 point is deducted for wrong answer. Questions 16 to 25 score 4 points each. No points are deducted for unanswered or wrong answers. 4. Shade your answers neatly using a pencil in the answer sheet. 5. PROCTORING: No one may help any student in any way during the contest. 6. No electronic devices capable of storing and displaying visual information is allowed during the course of the exam. 7. Strictly No Calculators are allowed into the exam. 8. All students must fill and shade in their Name, Index number, Class and School in the answer sheet and contest booklet. 9. MINIMUM TIME: Students must stay in the exam hall at least 1h 15 min. 10. Students must show detailed working and transfer answers to the answer sheet. 11. No exam papers and written notes can be taken out by any contestant. 10 SASMO 2015, Secondary 4 Contest SASMO 2015 Secondary 4 Starting Score = 15 marks (to avoid negative marks); Max Possible Score = 85 marks Section A (Correct answer = 2 marks; no answer = 0; incorrect answer = minus 1 mark) 1. The diagram shows two overlapping squares. The length of the bigger square is 14 cm and the length of the smaller square is 7 cm. Find the difference between the area of the two unshaded regions X and Y. X Y (a) (b) (c) (d) (e) 137 cm2 147 cm2 157 cm2 167 cm2 None of the above ________________________________________________________________ 2. In a class of 40 students, 7 study both Physics and Chemistry, 16 study Physics and 14 study Chemistry. How many students do not study either Physics or Chemistry? (a) (b) (c) (d) (e) 3 7 10 17 None of the above 12 SASMO 2015, Secondary 4 Contest 3. Find the range of values of 𝑘 if the curve 𝑦 = 𝑘𝑥 2 2 𝑥 + (2 𝑘 – 1) lies completely above the 𝑥-axis. 1 <𝑘<1 (a) (b) 𝑘< (c) (d) (e) 0<𝑘<1 𝑘>1 None of the above 2 1 2 or 𝑘 > 1 ________________________________________________________________ 4. A number gives a remainder of 9 when divided by 10. Another number gives a remainder of 8 when divided by 10. The sum of these two numbers is multiplied by 12 to give the third number. What is the remainder when this third number is divided by 10? (a) (b) (c) (d) (e) 4 7 8 9 None of the above 13 SASMO 2015, Secondary 4 Contest 5. In the figure below, the ratio of the trapezium ABCD to the area of the 1 triangle DEF to the area of parallelogram GHJK is 4 : 2 : 3. Given that 3 of the area of ∆DEF is shaded, find the ratio of the area of the shaded region to the total area of the unshaded regions of the figure. B C F G A H D E J K (a) (b) (c) (d) (e) 2 : 11 2 : 17 2 : 21 2 : 27 None of the above ________________________________________________________________ 6. Which of the following statement(s) is or are correct? Statement A: A cubic equation can have 3 real and distinct roots. Statement B: A cubic equation can have 2 real roots. Statement C: A cubic equation can have 1 real root and 2 non-real roots (a) (b) (c) (d) (e) All the three statements are correct. Only Statements A and B are correct. Only Statements A and C are correct. Only Statement A is correct. None of the above 14 SASMO 2015, Secondary 4 Contest 7. A big cube is made up of 125 small cubes. All the faces of the big cube are then painted. How many of the small cubes have no painted face? (a) (b) (c) (d) (e) 1 8 27 64 None of the above ________________________________________________________________ 8. In ABC, AB = 14 cm, BC = 10 cm and AC = 7 cm. Find the value of (a) (b) (c) (d) (e) 0.5 1 2 Cannot be found None of the above 15 sin 𝐵 sin 𝐶 . SASMO 2015, Secondary 4 Contest 9. All the match sticks in the diagram are identical. Find the total number of squares in the diagram? (a) (b) (c) (d) (e) 6 7 8 9 None of the above ________________________________________________________________ 10. Given that 𝑥𝑦𝑧 = 2015, and 𝑥 , 𝑦 and 𝑧 are positive integers, how many possible triples (𝑥, 𝑦, 𝑧) are there? (a) (b) (c) (d) (e) 5 15 27 2015 None of the above 16 SASMO 2015, Secondary 4 Contest 11. 𝑥 Given that 8 𝑥 12 and 4 𝑦 2, find the least possible value of 𝑦. (a) (b) (c) (d) (e) 4 2 3 6 12 ________________________________________________________________ 12. How many four-digit numbers of the form X56Y are divisible by 24? (a) (b) (c) (d) (e) 1 3 4 6 8 17 SASMO 2015, Secondary 4 Contest 13. A rectangular floor of 1540 cm by 1440 cm is to be covered completely by identical square tiles. What is the least possible number of square tiles? (a) (b) (c) (d) (e) 616 5544 22 176 88 704 None of the above ________________________________________________________________ 14. Johnny has 35 toys. He divides them into 5 piles so that each pile has a different number of toys. Find the smallest possible number of toys in the biggest pile. (a) (b) (c) (d) (e) 7 8 9 10 None of the above 18 SASMO 2015, Secondary 4 Contest 15. The diagram shows a circle with two chords AB and CD intersecting at E. Given that AE = 12 cm, BE = 3 cm and CE = 9 cm, find the length of DE. C 9 3 12 E B D A (a) (b) (c) (d) (e) 2 cm 3 cm 4 cm 6 cm None of the above ________________________________________________________________ Section B (Correct answer = 4 marks; incorrect or no answer = 0) 16. A man buys 30 metres of fence to build a rectangular garden at the back of his house. He uses the wall XY at the back of his house as one side of the garden ABCD as shown in the diagram below. Find the largest possible area of the garden. X wall A D B C 19 Y SASMO 2015, Secondary 4 Contest 17. In a school hall, 7 31 of the chairs are arranged in rows of 5, and 11 31 of the chairs are arranged in rows of 13. The rest of the chairs are stacked up. If there are less than 4000 chairs in the hall, find the total number of chairs in the hall. ________________________________________________________________ 18. Polite numbers are numbers that can be expressed as the sum of two or more consecutive positive integers, e.g. 5 = 2 + 3; 9 = 2 + 3 + 4 = 4 + 5. The degree of politeness of a number is the number of ways a number can be expressed as the sum of two or more consecutive positive integers, e.g. the degree of politeness of 2, 5 and 9 is 0, 1 and 2 respectively. Find the smallest number with a degree of politeness of 3. 20 SASMO 2015, Secondary 4 Contest 19. Find the value of √12 + √12 + √12+. . . ________________________________________________________________ 20. The figure shows a circle with 4 points on its circumference. Each point is joined to every other point by a line (called a chord). The chords divide the circle into 8 regions. Find the maximum number of regions formed by the chords for a circle with 7 points. 21 SASMO 2015, Secondary 4 Contest 21. Find the values of 𝑘 for which the equation 𝑘𝑥 2 2015𝑥 + (𝑘 – 2015) = 0 has one positive and one negative root. ________________________________________________________________ 22. A circle and a triangle are drawn on a rectangular sheet of paper. What is the biggest number of regions that can be formed on the paper? 22 SASMO 2015, Secondary 4 Contest 23. 3 4 2 2 Find the sum of the coefficients in the expansion of 6 x 5 x 2 3 2 x x . ________________________________________________________________ 24. Albert and Bernard just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates. May 15 June 17 July 14 August 14 May 16 June 18 July 16 August 15 May 19 August 17 Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. Albert: Bernard: Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too. At first I don’t know when Cheryl’s birthday is, but I know now. Then I also know when Cheryl’s birthday is. So when is Cheryl’s birthday? 23 SASMO 2015, Secondary 4 Contest 25. Find the last six digits of 20152015. END OF PAPER 24 SASMO 2015, Secondary 4 Contest SASMO 2015 Secondary 4 Solution Section A 1. The diagram shows two overlapping squares. The length of the bigger square is 14 cm and the length of the smaller square is 7 cm. Find the difference between the area of the two unshaded regions X and Y. X Y 137 cm2 147 cm2 [Ans] 157 cm2 167 cm2 None of the above (a) (b) (c) (d) (e) Solution Let the area of the overlapping region be A. Area of X Area of Y = (Area of big square A) (Area of small square A) = Area of big square A Area of small square + A = Area of big square Area of small square = 142 72 = 196 49 = 147 cm2 2. In a class of 40 students, 7 study both Physics and Chemistry, 16 study Physics and 14 study Chemistry. How many students do not study either Physics or Chemistry? (a) (b) (c) (d) (e) 3 7 10 17 [Ans: 40 9 7 7] None of the above Solution Fill in the following Venn diagram in this order: No. of students who study both Physics (P) and Chemistry (C) = 7 No. of students who study Physics only = 16 7 = 9 No. of students who study Chemistry only = 14 7 = 7 No. of students who do not study either Physics or Chemistry = 40 9 7 7 = 17 25 SASMO 2015, Secondary 4 Contest P C 9 7 7 10 no. of students who do not study either Physics or Chemistry = 17 3. Find the range of values of k if the curve y = kx2 2x + (2k – 1) lies completely above the x-axis. 1 (a) 2<k<1 (b) (c) (d) (e) k < 2 or k > 1 0<k<1 k > 1 [Ans] None of the above 1 Solution If the curve y = kx2 2x + (2k – 1) lies completely above the x-axis, then b2 – 4ac < 0. 4 – 4k(2k – 1) < 0 1 – k(2k – 1) < 0 1 – 2k2 + k < 0 2k2 – k – 1 > 0 (2k + 1)(k – 1) > 0 1 k < 2 or k > 1 But the coefficient of x2 must be positive for y = kx2 2x + (2k – 1) to lie completely above the x-axis. k > 1. 4. A number gives a remainder of 9 when divided by 10. Another number gives a remainder of 8 when divided by 10. The sum of these two numbers is multiplied by 12 to give the third number. What is the remainder when this third number is divided by 10? (a) (b) (c) (d) (e) 4 [Ans: 9 + 8 = 17 7 2 = 14] 7 8 9 None of the above 26 SASMO 2015, Secondary 4 Contest Solution Method 1 1st number = 10 + 10 + 10 + … + 10 + 9 (it does not matter how many 10s are there) 2nd number = 10 + 10 + 10 + … + 10 + 8 (it does not matter how many 10s are there) Sum of first two numbers = 10 + 10 + 10 + … + 10 + 17 = 10 + 10 + 10 + … + 10 + 10 + 7 rd 3 number = 12 sum of first two numbers = 10 + 10 + 10 + … + 10 + 84 (it does not matter how many 10s are there) = 10 + 10 + 10 + … + 10 + 10 + 10 + 10 + … + 10 + 4 when the 3rd number is divided by 10, the remainder is 4. Method 2 Let the first number be 10x + 9 and the second number be 10y + 8, where x and y are whole numbers. Then the sum of the first two numbers is (10x + 9) + (10y + 8) = 10(x + y) + 17 = 10(x + y + 1) + 7 = 10z + 7 for some whole number z. Thus the third number is 12(10z +7) = 120z + 84 = 120z + 80 + 4 = 10(12z + 8) + 4 = 10w + 4 for some whole number w. when the third number is divided by 10, the remainder is 4. 5. In the figure below, the ratio of the trapezium ABCD to the area of the triangle DEF to the area of parallelogram GHJK is 4 : 2 : 3. Given that 1 3 of the area of DEF is shaded, find the ratio of the area of the shaded region to the total area of the unshaded regions of the figure. B C F G A D E J K (a) (b) (c) (d) (e) H 2 : 11 2 : 17 [Ans] 2 : 21 2 : 27 None of the above 27 SASMO 2015, Secondary 4 Contest Solution Since the number 2, in the ratio 4 : 2 : 3, is not divisible by 3, we use the equivalent ratio 12 : 6 : 9 because 6 is divisible by 3. Let the area of DEF be 6 units. 1 Area of shaded region = 6 = 2 units 3 Area of trapezium ABCD = 12 units Area of parallelogram GHJK = 9 units Area of trapezium ABCD that is unshaded = 12 2 = 10 units Area of parallelogram GHJK that is unshaded = 9 2 = 7 units Total area of unshaded regions = 10 + 7 = 17 units ratio of area of shaded region to total area of unshaded regions = 2 : 17 6. Which of the following statement(s) is or are correct? Statement A: A cubic equation can have 3 real and distinct roots. Statement B: A cubic equation can have 2 real roots. Statement C: A cubic equation can have 1 real root and 2 non-real roots (a) (b) (c) (d) (e) All the three statements are correct. Only Statements A and B are correct. Only Statements A and C are correct. [Ans] Only Statement A is correct. None of the above Solution A cubic equation has exactly 3 roots: 3 real and distinct roots (when the curve cuts the x-axis at 3 distinct points) 2 real and equal roots, and another distinct real root (when the curve just touches the x-axis at 1 point, which gives rise to 2 real and equal roots; and the curve cuts the x-axis at another point) 1 real root and 2 non-real roots (when the curve cuts the x-axis at only 1 point) only Statements A and C are correct. 7. A big cube is made up of 125 small cubes. All the faces of the big cube are then painted. How many of the small cubes have no painted face? (a) (b) (c) (d) (e) 1 8 27 [Ans: 33 = 27] 64 None of the above 28 SASMO 2015, Secondary 4 Contest Solution Since 125 = 5 5 5, then the dimensions of the big cube are 5 by 5 by 5. The only small cubes that have no painted face are those that are not visible from outside. there are (5 2)3 = 33 = 27 small cubes that have no painted face. 8. In ABC, AB = 14 cm, BC = 10 cm and AC = 7 cm. Find the value of (a) (b) (c) (d) (e) sin 𝐵 sin 𝐶 . 0.5 [Ans] 1 2 Cannot be found None of the above Solution Using Sine Rule, 9. sin 𝐵 7 = sin 𝐶 14 sin 𝐵 , i.e. = 𝑏 sin 𝐵 sin 𝐶 sin 𝐶 𝑐 = 7 14 , where b = AC = 7 and c = AB = 14. = 𝟏 𝟐 or 0.5 All the match sticks in the diagram are identical. Find the total number of squares in the diagram? (a) (b) (c) (d) (e) 6 7 8 [Ans] 9 None of the above Solution No. of 1 1 squares = 5 No. of 2 2 squares = 2 No. of 3 3 squares = 1 total no. of squares = 8 29 SASMO 2015, Secondary 4 Contest 10. Given that xyz = 2015, and x, y and z are positive integers, how many possible triples (x, y, z) are there? (a) (b) (c) (d) (e) 5 15 27 [Ans] 2015 None of the above Solution 2015 = 5 13 31, where 5, 13 and 31 are prime numbers. 2015 = 1 1 2015 (3 arrangements) = 1 5 403 (3! = 6 arrangements) = 1 13 155 (3! = 6 arrangements) = 1 31 65 (3! = 6 arrangements) = 5 13 31 (3! = 6 arrangements) there are 27 possible triples (x, y, z). 11. 𝑥 Given that 8 x 12 and 4 y 2, find the least possible value of . 𝑦 (a) (b) (c) (d) (e) 4 2 3 6 [Ans] 12 Solution Because the values of x and y can be negative, we have to test different combinations of the greatest and least possible values of x and y as follow: −8 −4 −8 −2 12 −4 12 −2 =2 =4 = 3 = 6 least possible value of 𝑥 𝑦 = 6 30 SASMO 2015, Secondary 4 Contest 12. How many four-digit numbers of the form X56Y are divisible by 24? (a) (b) (c) (d) (e) 1 3 4 6 [Ans: 1560, 4560, 7560, 2568, 5568, 8568] 8 Solution If a number is divisible by 24 (= 3 8), then it is also divisible by 3 and 8, since 3 and 8 are relatively prime. Using the divisibility test for 8, the last three digits 56Y is also divisible by 8. But there are still 10 possibilities for Y. If a number is divisible by 8, then the number is also divisible by any factor of 8, i.e. X56Y is also divisible by 2 and by 4. Since X56Y is divisible by 2, then Y must be even, i.e. Y = 0, 2, 4, 6 or 8. Since X56Y is divisible by 4, using the divisibility test for 4, the last two digits 6Y is also divisible by 4. Since only 62 and 66 are not divisible by 4, then Y = 0, 4 or 8. Now we use the divisibility test for 8. Since only 564 is not divisible by 8, then Y = 0 or 8. Using the divisibility test for 3, X + 5 + 6 + Y = X + Y + 11 is also divisible by 3. If Y = 0, then X + Y + 11 = X + 11 is also divisible by 3, i.e. X = 1, 4 or 7. If Y = 8, then X + Y + 11 = X + 19 is also divisible by 3, i.e. X = 2, 5 or 8. Thus four-digit numbers of the form X56Y that are divisible by 24 are: 1560, 4560, 7560, 2568, 5568, 8568. no. of four-digit numbers of the form X56Y that are divisible by 24 = 6 13. A rectangular floor of 1540 cm by 1440 cm is to be covered completely by identical square tiles. What is the least possible number of square tiles? (a) (b) (c) (d) (e) 616 5544 [Ans] 22 176 88 704 None of the above Solution 1540 = 22 5 7 11 1440 = 25 32 5 Possible lengths of the square tiles must be factors of both 1540 cm and 1440 cm. Greatest possible length of the square tiles = HCF of 1540 cm and 1440 cm = 22 5 = 20 cm The least possible number of square tiles occurs when the square tiles are the largest. least possible number of square tiles = 31 1540×1440 20×20 SASMO 2015, Secondary 4 Contest = 22 ×5×7×11×25 ×32 ×5 22 ×5×22 ×5 = 7 11 23 32 = 5544 14. Johnny has 35 toys. He divides them into 5 piles so that each pile has a different number of toys. Find the smallest possible number of toys in the biggest pile. (a) (b) (c) (d) (e) 7 8 9 [Ans] 10 None of the above Solution For each pile to have a different number of toys, and the biggest pile to have the smallest possible number of toys, put 1 toy in the 1st pile, 2 toys in the 2nd pile, 3 toys in the 3rd pile, 4 toys in the 4th pile and 5 toys in the 5th pile. So the biggest pile is the 5th pile, but there are only 1 + 2 + 3 + 4 + 5 = 15 toys. The 16th toy will have to go to the 5th pile so that each pile will have a different number of toys. The 17th toy cannot go to the 5th pile because we want to find the smallest possible number of toys in the biggest pile, so the 17th toy will have to go to the 4th pile. Similarly, the 18th, 19th and 20th toys will go to the 3rd, 2nd and 1st piles respectively. The 21st toy will then go to the 5th pile again, and the 22nd, 23rd, 24th and 25th toys will go to the 4th, 3rd, 2nd and 1st piles respectively. Similarly, the 26th to 30th toys will go to the 5th to 1st piles respectively. Similarly, the 31st to 35th toys will also go to the 5th to 1st piles respectively. the largest pile (which is the 5th pile) will contain 5 + 1 + 1 + 1 + 1 = 9 toys. 32 SASMO 2015, Secondary 4 Contest 15. The diagram shows a circle with two chords AB and CD intersecting at E. Given that AE = 12 cm, BE = 3 cm and CE = 9 cm, find the length of DE. C 9 3 12 E B D A (a) (b) (c) (d) (e) 2 cm 3 cm 4 cm [Ans] 6 cm None of the above Solution Method 1 DBE and ACE are similar (angles in same segment and vertically opposite angles). 𝐷𝐸 𝐵𝐸 = 𝐴𝐸 𝐶𝐸 , i.e. 𝐷𝐸 3 = 12 9 , so DE = 4 cm. Method 2 If you know about the intersecting chords theorem, i.e. DE CE = AE BE, then DE 9 = 12 3, so DE = 4 cm. Note: To prove the intersecting chords theorem, use similar triangles as in Method 1. 33 SASMO 2015, Secondary 4 Contest Section B 16. A man buys 30 metres of fence to build a rectangular garden at the back of his house. He uses the wall XY at the back of his house as one side of the garden ABCD as shown in the diagram below. Find the largest possible area of the garden. X wall A D B C Y Solution Let BC = x m. 30 x 30 x x 2 2 30 x m . Then AB = m and area of garden ABCD = x = 2 2 2 30 x Let y x. 2 This is a quadratic function and its graph is an inverted-U parabola which cuts the xaxis at x = 0 and x = 30 as shown in the diagram below. y 0 15 30 x 0 30 = 15, i.e. BC = 15 m. 2 AB = 7.5 m and area of garden = 15 7.5 = 112.5 m2 By symmetry, the maximum point occurs when x = 34 SASMO 2015, Secondary 4 Contest 17. 7 11 In a school hall, of the chairs are arranged in rows of 5, and of the chairs are 31 31 arranged in rows of 13. The rest of the chairs are stacked up. If there are less than 4000 chairs in the hall, find the total number of chairs in the hall. Solution Let the total number of chairs in the hall be n. Since 7 31 n chairs are arranged in rows of 5, and the numerator 7 is not divisible by 5, then n must be divisible by 5. 11 Since n chairs are arranged in rows of 13, and the numerator 11 is not divisible 31 by 13, then n must also be divisible by 13. 7 11 Since n and n are whole numbers, then n must also be divisible by 31. 31 31 Since 5, 13 and 31 have no common factors other than 1, then n must be a multiple of 5 13 31 = 2015. Since 2 2015 = 4030 is already more than 4000, then n = 2015. total no. of chairs in the hall = 2015 18. Polite numbers are numbers that can be expressed as the sum of two or more consecutive positive integers, e.g. 5 = 2 + 3; 9 = 2 + 3 + 4 = 4 + 5. The degree of politeness of a number is the number of ways a number can be expressed as the sum of two or more consecutive positive integers, e.g. the degree of politeness of 2, 5 and 9 is 0, 1 and 2 respectively. Find the smallest number with a degree of politeness of 3. Solution Method 1 1 2 3=1+2 4 5=2+3 6=1+2+3 7=3+4 8 9=4+5=2+3+4 10 = 1 + 2 + 3 + 4 11 = 5 + 6 12 = 3 + 4 + 5 13 = 6 + 7 14 = 2 + 3 + 4 + 5 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 the smallest number with a degree of politeness of 3 is 15. 35 SASMO 2015, Secondary 4 Contest Method 2 1+2=3 2+3=5 3+4=7 ⋮ 1+2+3=6 2+3+4=9 3 + 4 + 5 = 12 ⋮ 1 + 2 + 3 + 4 = 10 2 + 3 + 4 + 5 = 14 3 + 4 + 5 + 6 = 18 ⋮ 1 + 2 + 3 + 4 + 5 = 15 2 + 3 + 4 + 5 + 6 = 20 3 + 4 + 5 + 6 + 7 = 25 Each polite number is obtained from the previous one by adding a constant. all the polite numbers are: 3, 5, 7, 9, 11, 13, 15, 17, … 6, 9, 12, 15, 18, … 10, 14, 18, … 15, 20, 25, … 21, 27, 33, … Hence, the smallest number with a degree of politeness of 3 is 15. 19. Find the value of √12 + √12 + √12+. . . Solution Let x = √12 + √12 + √12+. . . Then x2 = 12 + √12 + √12+. . . x2 = 12 + x x2 x 12 = 0 (x + 3)(x 4) = 0 x = 4 or 3 (rejected because x is the positive square root) √12 + √12 + √12+. . . = 4 20. The figure shows a circle with 4 points on its circumference. Each point is joined to every other point by a line (called a chord). The chords divide the circle into 8 regions. Find the maximum number of regions formed by the chords for a circle with 7 points. 36 SASMO 2015, Secondary 4 Contest Solution The maximum number of regions occurs when no three chords intersect one another inside the circle. Draw and count. maximum no. of regions formed by the chords for a circle with 7 points = 57. 𝑛 𝑛 𝑛−1 Note: The formula is not 2n1 = 271 = 64, but ( ) + ( ) + ( ). 4 1 2 21. Find the range of values of k for which the equation kx2 2015x + (k – 2015) = 0 has one positive and one negative root. Solution Method 1 Let the two roots of the equation kx2 2015x + (k – 2015) = 0 be and . Since one of the roots is positive and the other negative, then < 0. Now = 𝑐 𝑎 = 𝑘−2015 𝑘 < 0. If k = 0, then the coefficient of x2 = 0, i.e. the equation will become a linear equation with only one root, contradicting that there is a positive root and one negative root. If k < 0, then k – 2015 > 0, i.e. k > 2015, which contradicts that k < 0 in the first place. If k > 0, then k – 2015 < 0, i.e. k < 2015. 0 < k < 2015. Method 2 The curve y = kx2 2015x + (k – 2015) cuts the x-axis at two values, one of which is positive and the other negative. If k = 0, then the coefficient of x2 = 0, i.e. the equation will become a linear equation with only one root, contradicting that there is a positive root and one negative root. If k < 0, then the coefficient of x2 will be negative, i.e. the curve will cut the y-axis at a value that is positive, i.e. the y-intercept of the curve, k – 2015, will be positive. This means that k > 2015, which contradicts that k < 0 in the first place. If k > 0, then the coefficient of x2 will be positive, i.e. the curve will cut the y-axis at a value that is negative, i.e. the y-intercept of the curve, k – 2015, will be negative. This means that k < 2015. 0 < k < 2015. 37 SASMO 2015, Secondary 4 Contest 22. A circle and a triangle are drawn on a rectangular sheet of paper. What is the biggest number of regions that can be formed on the paper? Solution To form the biggest number of regions, the triangle and the circle should intersect as often as possible. Since a line can cut a circle at most two times, then the following diagram shows the biggest number of regions that can be formed between a triangle and a circle. 3 4 2 8 7 5 1 6 The next step is to divide the 8th region into as many regions as possible. This is achieved by using the 3 vertices of the triangle, and the 3 curved parts of the circle that are outside the triangle, to touch the edges of the rectangular sheet of paper, as shown in the diagram below. biggest number of regions that can be formed on the paper = 13 3 10 9 4 2 7 11 1 12 23. 8 6 5 13 Find the sum of the coefficients in the expansion of 6 x 2 5 x 2 3 2 x x 2 . 3 Solution 6 x 3 4 5 x 2 3 2 x x 2 = 𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 +. . . + 𝑎14 𝑥14 Let x = 1. Then 𝑎0 + 𝑎1 + 𝑎2 +. . . + 𝑎14 = (6 5 + 2)3(3 2 + 1)4 = 33 24 = 432 2 38 4 SASMO 2015, Secondary 4 Contest 24. Albert and Bernard just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates. May 15 June 17 July 14 August 14 May 16 June 18 July 16 August 15 May 19 August 17 Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively. Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too. Bernard: At first I don’t know when Cheryl’s birthday is, but I know now. Albert: Then I also know when Cheryl’s birthday is. So when is Cheryl’s birthday? Solution Out of the 10 dates, the day ranges from 14 to 19, with only 18 and 19 each occurring once. If the day of Cheryl’s birthday is 18 or 19, then Bernard would have known when Cheryl’s birthday is since Cheryl has told him the day of her birthday. But why does Albert know that Bernard does not know? If Cheryl has told Albert that her birth month is May or June, then it is possible that her birthday may be May 19 or June 18. This means that Bernard may know when Cheryl’s birthday is. The fact that Albert knows that Bernard does not know means that Cheryl has told Albert that her birth month is either July or August. Initially, Bernard does not know when Cheryl’s birthday is, but how did he know after Albert has first spoken? Out of the 5 remaining dates in July and August, the day ranges from 15 to 17, with only 14 occurring twice. If Cheryl has told Bernard the day of her birthday is 14, then Bernard would not have known. The fact that Bernard knows means the day of her birthday is not 14. So now we are left with 3 possible dates: July 16, August 15 and August 17. After Bernard has spoken, Albert now knows when Cheryl’s birthday is. If Cheryl has told Albert her birth month is August, then Albert would not have known because there are two possible dates in August. Cheryl’s birthday is on July 16. 39 SASMO 2015, Secondary 4 Contest 25. Find the last six digits of 20152015. Solution Since the last 6 digits of a product ab depends only on the last 6 digits of a and of b, then 2015 = 002 015 Last 6 digits of 002 015 2015 = 060 225 Last 6 digits of 060 225 2015 = 353 375 Last 6 digits of 353 375 2015 = 050 625 Last 6 digits of Last 6 digits of 009 375 2015 = 890 625 Last 6 digits of 890 625 2015 = 609 375 Last 6 digits of 609 375 2015 = 890 625 the last 6 digits repeat with a period of 2, with the exception of the first 5 powers. Since the index 2015 is odd, then the last 6 digits of 152015 are 609 375. 40