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Weekly Dose 21 - Maths Olympiad Practice

Weekly Dose 21 - Maths Olympiad Practice

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Weekly Dose 21 - Maths Olympiad Practice

  1. 1. Solution: Let the number be 𝑎𝑏𝑐𝑑𝑒 and 𝑎, 𝑏, 𝑐, 𝑑, 𝑒 are all different and > 0 𝑎𝑏𝑐𝑑𝑒 = multiple of 4  𝑑𝑒 is multiple of 4 𝑎𝑏𝑐𝑑 = multiple of 5  𝑑 is 5, and 𝑑𝑒 is either 52 or 56 𝑎𝑏𝑐 = multiple of 6 𝑎𝑏 = multiple of 7  𝑎𝑏 can be 14, 21, 28, 42, 49, 63, 84, 91, 98 (35,56 cannot because d is 5, 70 cannot because all digits > 0, 77 cannot because all digits are different) 𝑎𝑏𝑐 = multiple of 6  𝑎𝑏𝑐 can be 216, 426, 492, 498, 846, 912, 918, 984 (144, 282, 288, 636 cannot because all digits are different, 210, 630 cannot because all digits > 0, ) There is a five-digit number with all the digits different and all are greater than zero. It is a multiple of 4. If the ones digit is cancelled, then the four- digit number left will become a multiple of 5. If the ones digit of the newly-formed four-digit number is cancelled again, then the three-digit number left will become a multiple of 6. If the ones digit of this three-digit number is cancelled again, the number formed by this two-digit number will become multiple of 7. How many such kind of five-digit numbers are there? So we have: 𝑎𝑏𝑐 can be 216, 426, 492, 498, 846, 912, 918, 984 𝑑𝑒 is either 52 or 56 Possible number: 49256, 49852, 49856, 84652, 91256, 91852, 91856, 98452, 98456 Answer: 9
  2. 2. Four friends are lined-up in a row from the oldest to the youngest such that the difference of every two of the neighbouring ages in the row is the same and the product of their ages is 1680, what is the sum of their ages? Solution: Let their age be a, b, c, d, and a > b > c > d. Given a – b = b – c = c – d = k And a x b x c x d = 1680 First, find the prime factorization for 1680: 1680 = 2 x 2 x 2 x 2 x 3 x 5 x 7 = 5 x 6 x 7 x 8 or = 2 x 6 x 10 x 14 From above, we can tell a = 8, b = 7, c = 6, d = 5 or a = 14, b = 10, c = 6, d = 2 a + b + c + d = 8 + 7 + 6 + 5 = ___ or a + b + c + d = 14 + 10 + 6 + 2 = ___ Answer:26 𝑜𝑟 32
  3. 3. The sum of all the remainders of 3 consecutive two-digit numbers when divided by 5 is 7. The sum of all the remainders of those numbers when divided by 7 is 9, and the sum of all the remainders of those numbers when divided by 9 is 15. What is the sum of the remainders of those numbers when divided by 11? Solution: Let the 3 consecutive two-digit numbers as x, y, z The possible remainders for 5: 1, 2, 3, 4, 0 The possible remainders for 7: 1, 2, 3, 4, 5, 6, 0 The possible remainders for 9: 1, 2, 3, 4, 5, 6, 7, 8, 0 Since x, y, and z are consecutive numbers, their remainder are also ‘consecutive’ Given: x y z sum of remainder divided by 5 is 7, the three remainders are: 3, 4, 0 sum of remainder divided by 7 is 9, the three remainders are: 2, 3, 4 sum of remainder divided by 9 is 15, the three remainders are: 4, 5, 6 Look at the ‘z’ column: since the remainder is 0 when divided by 5, z is multiple of 5. And when divided by 7, it has a remainder of 4. we can figure that possible values for z are 25 or 60 or 95. Also it must fulfill the condition that when divided by 9, the remainder is 6. Only 60 fulfill this condition. Therefore, 𝑥 = 58, 𝑦 = 59, 𝑧 = 60 Answer: the sum of remainders divided by 11 is 12
  4. 4. Solution: Five kids A, B, C, D and E are sitting around a circular table (refer to the given figure) with some candies. Each of them gets an even number of candies. The quantities are 10, 30, 20, 20 and 40 respectively. In the first round, each of them gives one half of their candies to the kid to their right. At this time, the amounts of their candies become 25, 20, 25, 20 and 30 respectively. If the kid’s number of candies is odd, then he/she should pick one from the table. Is it possible that the kids have the same number of candies after several rounds? How many pieces does everyone have?

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