+ 2^1 cannot be written as a sum of anything less than exactly 16 distinct powers of two (using unique representation in binary). Recursively adding together pairs of equal elements of S until all the elements are distinct (i.e., adding them as if they had combined in the game), we obtain a set R with at most 16 distinct powers of two adding to get 131070. Then at this time there was a set S of at most 16 powers of two adding to get 131070. Thus at some point in time, the sum of all the tiles was exactly 131070 (= 131072 - 2). That at this time the sum of all the tiles is at least 131072. Suppose towards contradiction that at some point in time the 131072 tile is attained. It is easy to adapt the proof to show that 131072 is the maximum when 4 may also appear.Īt every step of the game, the total value among all tiles goes up by 2, or stays the same. I will prove that 65536 is the maximum when 2 is the only tile which appears newly on the board. Next is 131072 which will require at least 17 tiles which can't fit on the board.ĮDIT: Since the game pops up some 4's at times, 131072 is still theoretically possible if the last tile that pops up is a 4 making the board look like this:ġ31072: 65536, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 4īlueFlame gives an essentially correct answer, but I wanted to add a rigorous proof (for anyone looking for it). Since that's less than 16 (max number of tiles that can fit), 4096 is possible. These can be combined from right to left to get to 4096. We've 16 squares in total and considering the new tile popped up at the position we chose, to make it to 4096, we need minimum these tiles:Ģ048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 2 Let's ignore that possibility for the sake of simplicity. In the original game, sometimes the new tile that pops up is a 4.
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