IMO Shortlist 2010

[...]... and exchange stacks Sk1 and Sk2 Decide whether it is possible to achieve by a sequence of such moves that the rst ve stacks 2010 are empty, whereas the sixth stack S6 contains exactly 20102 010 coins 2010 ẵ C4 Same as Problem C4, but the constant 20102 010 is replaced by 20102 010 (Netherlands) Answer Yes (in both variants of the problem) There exists such a sequence of moves ễaẵ1 , aẵ2, , aẵnế... 0ế ễ1, 1, 0, 3, 0, 0ế ễ1, 0, 3, 0, 0, 0ế ễ0, 3, 0, 0, 0, 0ế ễ0, 0, P3, 0, 0, 0ế ễ0, 0, 16, 0, 0, 0ế ễ0, 0, 0, P16, 0, 0ế We already have more than A coins in stack S4 , since A 2010 201 02010 ễ211 2010 2010 211 2010 2010 2011 22010 11 2011 2ễ2 ế Ô 215 11 2011 22 22 P16 To decrease the number of coins in stack S4 , apply Move 2 to this stack repeatedly until its size decreases to Aò4 (In every step,... , an coins, then it is possible to perform several allowed moves such that the stacks contain aẵ1 , , aẵn coins respectively, whereas the contents of the other stacks remain unchanged 2010 Let A 20102 010 or A 20102 010 , respectively Our goal is to show that ễ1, 1, 1, 1, 1, 1ế ễ0, 0, 0, 0, 0, Aế First we prove two auxiliary observations Lemma 1 ễa, 0, 0ế ễ0, 2a, 0ế for every a 1 Proof We prove by... around 5 and 6 stacks the maximal number of coins 14 explodes With 5 stacks it is possible to achieve more than 22 coins With 6 stacks the maximum is greater than PP 14 2 2010 It is not hard to show that the numbers 20102 010 and 20102 010 by any nonnegative integer up to PP214 in the problem can be replaced Comment 2 The simpler variant C4ẵ of the problem can be solved without Lemma 2: ễ1, 1, 1, 1, 1,... a1 x b2 c1 b1 a2 b4 a y a3 b1 b3 b1 b3 b5 c2 b2 a4 a6 y x a8 a9 (30) (30) b) a10 Fig 4 a6 a7 x a8 c9 Fig 5 c10 c11 a11 (2010) a) b4 a5 a7 y b2 (2010) Fig 6 2 One can merge the examples for 30 and 67 shown in Figs 4b and 3 in a smarter way, obtaining a set of 13 singers representing 2010 This example is shown in Fig 5; an arrow from/to group ỉb1 , , b5 means that there exists such arrow from each... that each of these orders can be complemented by x and y in exactly 67 ways, hence obtaining 30 Ô 67 2010 good orders at all Analogously, one can merge the examples in Figs 13 to represent 2010 by 13 singers as is shown in Fig 6 25 a1 a5 a2 a3 b1 b4 a4 a4 b3 a3 b5 b2 a2 b6 b3 a6 b2 a1 (67) Fig 7 b4 b1 (2010) Fig 8 3 Finally, we will present two other improvements; the proofs are left to the reader The... are exactly 2010 orders of the singers such that all their wishes are satised? (Austria) Answer Yes, such an example exists Solution We say that an order of singers is good if it satised all their wishes Next, we say that a number N is realizable by k singers (or k-realizable) if for some set of wishes of these singers there are exactly N good orders Thus, we have to prove that a number 2010 is 20-realizable... Hence, the number of good orders is n1 n2 é In view of Lemma, we show how to construct sets of singers containing 4, 3 and 13 singers and realizing the numbers 5, 6 and 67, respectively Thus the number 2010 6 Ô 5 Ô 67 will be realizable by 4 3 13 20 singers These companies of singers are shown in Figs 13; the wishes are denoted by arrows, and the number of good orders for each Figure stands below in... Fig 8 3 Finally, we will present two other improvements; the proofs are left to the reader The graph in Fig 7 shows how 10 singers can represent 67 Moreover, even a company of 10 singers representing 2010 can be found; this company is shown in Fig 8 26 C2 On some planet, there are 2N countries (N 4) Each country has a ag N units wide and one unit high composed of N elds of size 1 Â 1, each eld being . Problems sh ould be kept strictly confidential until IMO 2011. Contributing Countries The Organizing Committee and the Problem Selection Committee of IMO 2010 thank the following 42 countries for contributing.