MMM25 Sol.pdf
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… 6}, {2, 3}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {4, 5}, {4, 6}, {5, 6}, {1, 3, 5}, {1, 4, 6}, {2, 5, 6}, {3, 4, 5}, {4, 5, 6} (b) {3, 4, 5, 6} (c) n/2 if n is even, and (n+ 1)/2 is n is odd (d) {bn/2c+ 1, . . . , n} 1 Problem 2. Game 1 Solution. Player 1 can guarantee a win: Round Player 1 Player 2 1 1 2 2 3 3 4 4,5 5 6 6 7,8,9,10,11 7 12 8 13 ≤ x ≤ 23 9 25 10 26 ≤ x ≤ 49 11 50 12 51 ≤ x ≤ 99 13 100 Game 2 Solution. Player 1 always wins by selecting the best of even and odd coins Game 3 Solution. There … 2,5,8 or 2,6,7 or 3,4,8 or 3,5,7 or 3,4,8 or 4,5,6 8 1 6 3 5 7 4 9 2 Game is equivalent to tic-tac-toe: no winning strategy 2 Problem 3. Solution. We start by ’unwrapping’ the cone. This gives us part of a circle. In this case that would be a circle with radius 60, where the arc length is the length of the circumference of the circle that formed the base of the cone. That length is 2 · π · R = 2 · π · 20 = 40π. With the radius be- ing 60 the length of the circumference of the entire circle would … sum. Together, in the long run the whole pie is shared, making pA+pB = 1. This gives the same answer, but faster. (b) In the first round, Alice gets 1/3, Bob gets (1 − 1/3)b = (2/3)b, and Carol gets (1–1/3)(1− b)(1/5) = 2(1− b)/15. In each later round they get the same individual proportions of the amount that was there at the beginning of the round. This means that the total amounts they collect have the same relative sizes: 1/3, 2b/3, and 2(1 − b)/15, but they add up to 1. Therefore: pA = M · 1/3, …