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Weekly Dose 20 - Maths Olympiad Practice

Weekly Dose 20 - Maths Olympiad Practice

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Weekly Dose 20 - Maths Olympiad Practice

  1. 1. Solution: From ๐ถ + ๐ผ = ๐ถ, and since ๐ผ cannot be 0, ๐ผ can either be 7, 8 or 9. Then from ๐ถ + ๐ถ + ๐ถ + ๐ถ = ๐ผ, ๐ผ can only be 8. No multiple of 4 ends with 7 or 9. Alphametric is a puzzle which replaces numerical digits in arithmetic problems with letters. Each letter represents a digit and no digits are represented by two different letters. What does the three-digit number ๐ผ๐‘€๐ถ equal to ? Letโ€™s try ๐ถ = 2: Because 2 + 8 = ๐‘€2, then 2 + 8 + ๐ธ + ๐ธ ๐‘š๐‘ข๐‘ ๐‘ก ๐‘๐‘’ 22 , thus ๐ธ = 6, M = 1, S = 4 Answer: 812 As such, ๐ถ can either be 2 or 7.
  2. 2. Neal uses the month of his birthday and multiply it by 31; then the day by 12. He then adds the two products and get a sum of 400. On what month and day is Nealโ€™s birthday? Solution: Let ๐‘š be the birthday month and ๐‘‘ be the birthday day, 31๐‘š + 12๐‘‘ = 400, 1 โ‰ค ๐‘š โ‰ค 12, 1 โ‰ค ๐‘‘ โ‰ค 31 12๐‘‘ always returns even number, 400 is also an even number, therefore 31๐‘š must be an even number as well. To get an even number out of 31๐‘š, ๐‘š must be even. Also, since 12๐‘‘ and 400 are multiple of 4, 31๐‘š must be multiple of 4 as well, Which means ๐‘š can either be 4, or 8 or 12 Since ๐‘‘ must be natural number .. Answer: April 23 ๐‘š 31๐‘š 12๐‘‘ = 400 โ€“ 31๐‘š ๐‘‘ 4 124 276 23 8 248 152 12.67 12 372 28 2.33
  3. 3. The sum of two different numbers is 43.67. If the decimal point of the smaller number moves one place towards the right side, then it will equal to the bigger one. What is the bigger number. Solution: Assume the two numbers are: ๐‘  (for smaller one), and ๐‘ Given ๐‘  + ๐‘ = 43.67 If ๐‘  = ๐‘Ž๐‘. ๐‘๐‘‘, then ๐‘ = ๐‘Ž๐‘๐‘. ๐‘‘ , but ๐‘Ž๐‘. ๐‘๐‘‘ + ๐‘Ž๐‘๐‘. ๐‘‘ โ‰  43.67, so ๐‘  โ‰  ๐‘Ž๐‘. ๐‘๐‘‘ If ๐‘  = ๐‘Ž. ๐‘๐‘, then ๐‘ = ๐‘Ž๐‘. ๐‘ , then it is possible that ๐‘Ž. ๐‘๐‘ + ๐‘Ž๐‘. ๐‘ = 43.67 From above, we can derive from right: ๐‘ = 7, ๐‘ = 9, ๐‘Ž = 3 Answer: 39.7
  4. 4. Solution: 2012 pupils are arranged in one row and are supposed to call numbers according to a certain pattern. If one student calls a one-digit number, then the student next to him must call a number twice of it. When a student call a two-digit number, then the next pupil after him must call the sum of 8 and the ones digit of that two-digit number. If the first pupil calls 1, then what number must the last pupil in the row call? Pattern starts at 4th call, 2012 โ€“ 3 = 2009 2009 รท 5 = 401 โ€ฆ 4 Remaining is 4, therefore the last pupil call is: ___ Answer: 12

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