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Weekly Dose 14 - Maths Olympiad Practice

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- 1. How many natural numbers less than 1000 are there, so that the sum of its first digit and last digit is 13? Solution: Since it’s less than 1000, our number may have two or three digits. For two digits, the numbers that meet the condition are: 49, 58, 67, 76, 85, 94. A total of 6 numbers. For three digits, 4𝑥9, 5𝑥8, 6𝑥7, 7𝑥6, 8𝑥5, 9𝑥4 where 𝑥 can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Total numbers: 10 + 10 + 10 + 10 + 10 + 10 = 60 Altogether, there are ____ + ____ numbers. Answer: 66
- 2. In the equation below, N is a positive whole number. 𝑁 = □ + □ − □ A numbered card is placed in each box. If three cards numbered 1, 2, 3 are used, we get 2 different answers for N, that is 2 and 4. How many different answers for N can we get if four cards numbered 1, 2, 3, 5 are used? Solution: The maximum number we can get is 5 + 3 − 1 = 7. So we can discard the numbers greater than 7 and check out numbers smaller than 7. 6 = 5+3-2 or 5+3-1 possible answer 5 not possible to get 4 = 5+2-3 or 5+1-2 or 3+2-1 possible answer 3 = … 2 = … Answer: 5 different answers
- 3. While B is riding a bicycle from Point X to Point Y, C is driving a car from Point Y to Point X, each at a steady speed along the same road. They start at the same time and, after passing each other, B takes 25 times longer to complete the journey as C. Find the ratio of the speed of the bicycle to the speed of the car. Solution: Let’s name: - the time taken for B and C to meet as 𝑥, - the time for C to reach Point X from the place they meet as 𝑎 - the time for B to reach Point Y from the place they meet as 25𝑎. The ratio of the time for B and C to cover the distance from the place they meet to Point X is 𝑥 ∶ 𝑎, which mean the ratio of the speed is 𝑎 ∶ 𝑥. The ratio of the time for B and C to cover the distance from the place they meet to Point Y is 25𝑎 ∶ 𝑥, which mean the ratio of the speed is 25𝑎 ∶ 𝑥. As they are traveling in steady speed, 𝑎 ∶ 𝑥 = 𝑥 ∶ 25𝑎 ⇒ 𝑥2 = 25𝑎2 ⇒ 𝑥 = 5𝑎 Ratio of speed is 𝑎 ∶ 5𝑎 = __ ∶ __ Answer: 1 ∶ 5
- 4. The following figures show a sequence of equilateral triangles of 1 square unit. The unshaded triangle in Pattern 2 has its vertices at the midpoint of each side of the larger triangle. If the pattern is continued as indicated by Pattern 3, what is the total area of the shaded triangles in Pattern 5, in square units? Solution: Starting from the second triangle, the total area of the shaded triangles in each triangle is equal to ¾ the area of the preceeding triangle. It follows that the total area of the fifth triangle is 1 × 3 4 × 3 4 × 3 4 × 3 4 = ____ Answer: 81 256

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