000RM.dvi
926 Some geometry problems from recent journals Crux 2847, proposed by G. Tsintsifas, Thessaloniki, Greece Theinscircleinscribed ...
927 Crux 2849, proposed by Toshio Seimiya, Kawasaki, Japan In a convex quadrilateralABCD,wehave∠ABC=∠BCD= 120◦. Suppose thatAB^2 ...
928 Some geometry problems from recent journals Mathematics Magazine, Problem 1669, proposed by A. N. Duman, Bilkent University, ...
929 Mathematics Magazine, Problem 1671, proposed by M. N. Desh- pande, Institute of Science, Nagpur, India LetTbe the set of tri ...
930 Some geometry problems from recent journals American Mathematical Monthly, Problem 11006, proposed by B. Suceava, ̆ Californ ...
931 Pi Mu Epsilon Journal, Problem 1058, proposed by P. A. Lindstrom, Batavia, NY Suppose that triangleABChas an interior pointP ...
932 Some geometry problems from recent journals Pi Mu Epsilon Journal, Problem 1060, proposed by A. B. Ayoub, Pennsylvania State ...
Chapter 37 The Josephus problem and its generalization 37.1 The Josephus problem nprisoners are arranged in a circle. In success ...
1002 The Josephus problem and its generalization Theorem 37.1.Letf(n)be the survivor in the Josephus problem ofn prisoners. f(2n ...
37.2 Generalized Josephus problemJ(n, k) 1003 37.2 Generalized Josephus problemJ(n, k) J(10,3): 1 2 4 3 5 6 7 8 9 10 6 4 * 1 8 2 ...
1004 The Josephus problem and its generalization Exercise 1.For what values ofnisf(n)=n? 2.For what values ofnisf(n)=n− 1? 3.Mak ...
Chapter 38 Permutations 38.1 The universal sequence of permutations For convenient programming we seek an enumeration of the per ...
1006 Permutations Example To find the 12345th permutation, we write 12344 = 0×1! + 1×2! + 1×3! + 4×4! + 0×5! + 3×6! + 2×7!. The ...
38.2 The position of a permutation in the universal sequence 1007 38.2 The position of a permutation in the universal sequence q ...
1008 Permutations Exercise 1.Find the one-billionth permutation in the universal sequence. 2.The inverse permutation of (5,1,3,7 ...
Chapter 39 Cycle decompositions 39.1 The disjoint cycle decomposition of a permutation Theorem 39.1.Every permutation can be dec ...
1010 Cycle decompositions 39.3 Dudeney’s Canterbury puzzle Take nine counters numbered 1 to 9, and place them in a row in the na ...
39.2 2 1011 However, there is one which can be made inthreemoves. Can you find it? 123456789 ...
1012 Cycle decompositions The pandigital case of Dudeney’s puzzle In the pandigital case, there are 4 ways to move 1234567890 (3 ...
39.3 Dudeney’s Canterbury puzzle 3 1013 39.3 Dudeney’s Canterbury puzzle 3 7 × 2 8 = 1 9 6 = 3 4 × 5. While 7 ×28 = 196, it is n ...
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