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November Mathematics Challenge 2016 Set theory is at the heart of mathematics but it is rarely discussed at schools or sixth forms. Once you know the rules and notation, set theory is actually fairly simple. Some research might be required to understand the symbols, and therefore translate the hieroglyphs into English. Your task is to define a set S that obeys the rules below. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. π β β βͺ {0} βπ β π π . π‘. 8|π π β β, 2π β π β π β€ 4 π, π β π, π β π β gcd(π, π) = 1 ππ βπ β β π . π‘. gcd(π, π) = 2π π β© {3} β β 22 β π β 23 β π βπ, π β π π . π‘. π β π = 2 πππ π + π = 24 βπ β π π < 17 β β, π β π π . π‘. βπ β π βπ = π + π = π πΏππ‘ π β {2,5,7} πβππ π β π β π β‘ 2 πππ 4 Worked Solution Hint 1 says that the set is a subset of the natural or counting numbers with 0 included. {0,1,2,3,4,...} Hint 8 says that all members of the set are less than 17 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 9 says that the multiplicative and additive identity are members of the set, so 1 and 0 are included. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Solve the simultaneous equation in Hint 7 to realise that 11 and 13 are in the set. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 5 says that the intersection of our set and 3 is non empty therefore 3 must be in our set. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 10 says that any of 2,5 & 7 can be in our set if and only if they are 2 more than a multiple of 4, only 2 is, so 2 is in our set and 5 and 7 are not. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 3 can be translated as any even number in our set can be at most 8, so we can remove any even number greater than 8. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 2 says that there exists (at least) one number which is a multiple of 8, as we have removed 16, 8 must be in the set. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 6 says that 4 and 8 are either both in or both not in the set. We have established that 8 is in the set so 4 is also included. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Hint 4 says that for any two different numbers in the set either they are co-prime (they share no common factor other than 1) or their highest common factor is a power of 2. e.g. 6 and 8 have a common factor of 2 which is allowed, but 6 and 3 have a common factor of 3 which is not so 6 cannot be in the set, in fact 6, 9 and 15 are all multiples of 3 and 3 is in the set so 6, 9 and 15 cannot be in the set. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Therefore π = {0, 1, 2, 3, 4, 8, 11, 13}