Solutions to the December 2005 Putnam Exam
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چکیده
Problem A1 We call a finite set of positive integers feasible if every member of S has the form 23, and no member of S divides any other member of S. We call a positive integer n feasible if n = ∑ S for some feasible set S. We are required to show that every positive integer is feasible. If not, then there exists a smallest infeasible number n. If n is even, say n = 2k for some positive integer k < n then we have k = ∑ S for some feasible set S and n = ∑ 2S where the set 2S = {2a : a ∈ S} is clearly feasible, contrary to assumption. Otherwise n is odd. Let s ≥ 0 be maximal such that 3 ≤ n, so that 3 > 13n and n − 3 = 2k for some positive integer k. We have k = ∑ S for some feasible set S, and since k < 13n < 3 , no member of S (or of 2S) is divisible by 3. In particular 3 / ∈ 2S and n = ∑S ′ where S ′ = {3s} ∪ 2S. To check that S ′ is feasible, it remains only to observe that since 3 is odd, it is not divisible by any member of 2S. Thus n is feasible, a contradiction.
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