Two Zero Four Eight

Number Theory Level 4

2048 is a single-player online and mobile game in which the objective is to slide tiles on a grid to combine them and create a tile with the number 2048. It is played on a simple gray 4×4 grid with tiles of varying colors overlaid that slide smoothly when a player moves them. If two tiles of the same number collide while moving, they will merge into a tile with the total value of the two tiles that collided. Higher-scoring tiles emit a soft glow. Every turn, a new tile (with a value of 2 or 4) will randomly appear in an empty spot on the board. When the player has no legal moves (there are no empty spaces and no adjacent tiles with the same value), the game ends.

If the largest possible tile value that can be achieved in the game is $a^b$ , where $a$ is a prime number and $b$ is a positive integer, find the value of $a+b$ .

Details and assumptions:
Assume the player is skillful enough and keep playing the game after they reach the target tile value, 2048.
The tiles that appear automatically on the board are all with a value of 2 or 4 only.

17 16 18 19

1 solution

Tristan Shin
May 4, 2014

To get $2^{n}$ , you need two cells of $2^{n-1}$ . Using this recursion, start with two 4's. This adds to 8, so there should be another 8. Continuing in this fashion, you can get $16, 32, 64, ..., 2^{13}, 2^{14}, 2^{15}, and 2^{17}$ , but not higher. In an attempt to get a higher power( $2^{18}$ ), the board will be filled with one of each tile( $2^{2}, 2^{3}, ..., 2^{17}$ ). Thus, the highest it can go is $2^{17}$ , so $a = 2$ and $b = 17$ so $a + b = \boxed {19}$ .

Moderator note:

When the problem was first created, the correct answer was 18.

There were several changes made to the problem, and the correct answer is now 19. Sorry for the confusion caused.

However, in the real game, the answer would be different, because a $4$ can also be dropped. The highest obtainable tile is $2^{17}$ , because there could be a board which has the tiles $2^{2}, 2^{3}, ... , 2^{16}$ , and then a $4$ could be dropped, making a $2^{17}$ tile possible.

Siva M. - 7 years, 1 month ago

I agree with Siva. ,the correct answer is $2^{17}$ .Take a look at the following configuration

$\huge{\begin{array} {|c|c|c|c|} \hline 16& 32 & 4096& 8192 \\ \hline 8 & 64 & 2048 & 16384 \\ \hline 4 & 128& 1024 & 32768 \\ \hline & 256 & 512 & 65536 \\ \hline \end{array}}$

A $2$ or $4$ maybe spawned in the empty spot ,even though there is a chance that it might be a $2$ a $4$ is also equally plausible,therefore the highest obtainable tile is $2^{17}$

Thaddeus Abiy - 7 years, 1 month ago

@Tristan Shin @Thaddeus Abiy @Kho Yen Hong

Several changes had been made to the question since it was first introduced.
1) The tile of 4 could be dropped. (previously only the tile of 2) 2) He asked for the value of the 2nd highest tile. (previously the highest)

I have since removed "2nd highest" tile and replaced it with "highest", as it is easy to overlook in the problem. I believe that the answer should be 19, as per Tristan's solution above. Let me know if this is not the case.

I have removed the comments from yesterday, to keep the solution discussion clean.

Calvin Lin Staff - 7 years, 1 month ago

Ok, thanks for help me to improve the problem.

Kho Yen Hong - 7 years, 1 month ago

The question is second largest possible tile value, right? it should be 2^{16} ie. 18???

Chandan Sankar P.S. - 7 years, 1 month ago

Sorry, there were several changes made to this problem, and as such the answers were changing. Those who answered 18 have been marked correct.

Calvin Lin Staff - 7 years, 1 month ago

That's okay. The answer becomes more reasonable. By the way, this is a interesting problem. :D

Jaydee Lucero - 7 years ago

Two Zero Four Eight

1 solution

Moderator note:

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