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Section A 1. Arithmetic Series: A series of numbers that have a common difference between terms Sum of an Arithmetic Series = Number of terms = 1 (first term + last term)(number of terms) 2 1 (last term - first term) + 1 2 The series is an arithmetic series. ( First term = 12, Last term = 78, Common difference = 2 ) No. of terms 78 12 1 34 2 The required sum 2. 1 12 7834 1530 2 15 x 3 y 17 6 x 7 y 15 1515 x 3 y 176 x 7 y 225 x 45 y 102 x 119 y 123x 164 y x 4 y 3 3. Notice that 1 2 3 1 1 2 1 3 1 1 2 3 1 61 3 3 3 6 9 1 3 2 3 3 3 1 2 3 3 63 3 3 5 10 15 1 5 2 5 3 5 1 2 3 5 65 3 3 2009 4018 6027 1 2009 2 2009 3 2009 1 2 3 2009 62009 3 Therefore, 1 2 3 3 6 9 5 10 15 2009 4018 6027 6 13 33 53 2009 3 3 Similarly, 1 4 7 3 12 21 5 20 35 2009 8036 14063 28 13 33 53 2009 3 The required answer is 281 3 2009 14 6 13 33 53 2009 3 3 3 5 3 3 3 Alternative Method 1 2 3 3 6 9 5 10 15 7 14 21 ... 2009 4018 6027 1 4 7 3 12 21 5 20 35 7 28 49 ... 2009 8036 14063 = (1 2 3)(1 3 5 ... 2009) (1 4 7)(1 3 5 ... 2009) = 1 2 3 1 4 7 = 3 14 4. Consider the powers of 3: 3, 9, 27, 81, 243, 729, 2187, 6561, … The unit digits are: 3, 9, 7, 1, 3, 9, 7, 1, … The sequence repeats with a period of 4. ∵ 167 41 4 3 ∴ Unit digit of 583167 = Unit digit of 33 =7 5. Triangle Inequality: The sum of lengths of two sides of a triangle must be greater than the length of the third side. Let x be the length of the third side. By the triangle inequality, 78 x 15 x 8 x 7 => x 1 => 1 x 15 x7 8 x 1 Since x is an integer, x = 2, 3, 4, …, 13 or 14 The sum 1 2 1413 104 (Sum of Arithmetic Sequence) 2 6. If the product of a group of numbers is divisible by an integer n, then at least one of the numbers in the group is a multiple of n. Among n3–1, n3 and n3+1, at least one is divisible by 5. By observation, n = 4 7. If Mr. Fu does not buy the 2.5L bottle, he needs 7 bottles of 1L, cost = $6.5×7 = $45.5 If Mr. Fu buy 1 bottle of 2.5L, he needs 4 bottles of 1L, cost = $15 + $6.5×4 = $41 If Mr. Fu buy 2 bottles of 2.5L, he needs 2 bottles of 1L, cost = $15×2 + $6.5×2 = $43 If Mr. Fu buy 3 or more bottles of 2.5L, cost ≥ $15×3 = $45 ∴ Mr. Fu must pay at least $41. 8. possible (number of outcomes) total Probability = Total number of outcomes of rolling two dice 6 6 36 Outcomes with sum = 8: 2,6, 3,5, 4,4, 5,3, 6,2 (5 outcomes) The probability required 5 36 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 9. Let A, B, C, D be the vertices of the quadrilateral, O be the intersection of the diagonals. Area of ABCD = Area of △ABC + Area of △ADC = 1 AC OD 1 AC OB 2 2 1 = ( AC )(OD OB ) 2 1 1 = ( AC )( BD ) 68 24 2 2 10. Cow Eat Grass Problem: Common Question in Lower Level Competitions Assumed cows eat grass and grass grow at constant rates. Let the original amount of grass on the farm be 1 unit. Let x be the units of grass eaten by a cow per week, y be the units of grass grown per week. 1 ........ (1) 32 1 40 x y ........ (2) 8 28 x y (2) – (1), 12 x Sub x 1 1 3 1 , x 8 32 32 128 1 3 1 1 into (1), 28 , y y 128 16 32 128 3 1 ∴ The maximum no. of cows can be kept y x 24 16 128 Section B 1. Definition: Function y=f(x) y is a function of x if and only if each value of x corresponds to only one value of y. f 0 2 a0 b0 c 2 3 c2 f 3 1 a3 b3 2 1 27a 3b 1 3 f 3 a 3 b 3 2 2 7a 3b 2 1 2 3 3 2. x 1# x 3 x 3# x 1 x 1x 3 x 1 x 3 4 x 2 4 x 3 2 x 4 4 x 2x 3 2 x 2 2x 3 x 2 x 2 2 x 3 4 2x 3 2 x 2 2x 1 0 x 12 0 x 1 3. In Fill in the Blanks Questions, look for same digits in the equation, and note that ‘9’ can usually be found at the leftmost digits in addition. Suppose I ≤ 7, TIME + IS + UP ≤ T799 + 99 + 99 < T999 This is impossible since the thousands-digit of the sum is S. Suppose I = 8, The carry from the tens is at most 2, so the hundreds-digit of the sum T must be 0, but this does not satisfy the requirement of the question. (T is a digit from 1 to 9) Therefore, I = 9. It follows that T = 1 and S = 2. Consider the unit digit, ∵ ‘E + S + P = P’ ∴ E + S = 10, E = 10 – 2 = 8 Consider the tens-digit, M + I + U + 1 (carry from the ones) = O + 20 (carry to the hundreds) M + U = O + 10 Since 1, 2, 7, 8 are already used, M, U, O = 3, 4, 5, 6 or 7. The only solutions are (M = 6, U = 7, O = 3) or (M = 7, U = 6, O = 3) In both cases, O = 3 4. Note that in Secondary School, we express answers in terms of instead of taking it as a specific numerical value such as 3.14. Area not covered = Area of the square – Area of the circle 1 2 1 2 2 1/2 1 1 4 1 5. n n 1 n 2 2009 3n 3 2009 3n 2006 n 668 2 3 For n = 1, 2, 3, … , 668 , 3n 3 can be represented as the sum of three consecutive positive integers. Therefore, there are 668 such numbers in total. Section C 1. In figure (I), it is easy to see that the areas of regions A, B, C, D are the same. For convenience, we can move region A to region C, and move region B to region D, without changing the area of the shaded region. (as shown as Figure (II)) Without loss of generality, let r be the radius of the quarter-circle. The required ratio 1 r 2 1 r 2 2 Area of the shaded region 4 1 22 Area of the quarter - circle 4 r 2.) Let x, y, z be the no. of roosters, hens and chicks respectively. x y z 100......(1) 5 x 3 y 0.2 z 100......(2) (2) × 5 – (1), 24 x 14 y 400 , y 400 24 x 200 12 x 450 3x 14 7 7 ∵ x, y are integers. ∴ 50 - 3x is divisible by 7. (x, y) = (5, 20) or (12, 8) Sub (x, y) = (5, 20) into (1), z 100 5 20 75 Solution 1: Derek has 5 roosters, 20 hens and 75 chicks. Sub (x, y) = (12, 8) into (1), z 100 12 8 80 Solution 2: Derek has 12 roosters, 8 hens and 80 chicks.