2048: much harder

Algebra Level 5

Assuming a perfect game, what's highest possible SCORE that can be achieved in a game of 2048 if only two's spawn.

Hint: it's not 2^16

How the game works for those who aren't familiar (although I suggest playing the game first or watching a video of it to fully understand it)

You may only combine tiles of the same number.

The grid is 4x4 and can have a maximum of 16 tiles at once.

You may not "delete" a piece from the board.

The game ends when no more tiles can be combined.

When combining two tiles of value x, the resulting tile will have a value of 2x.

Exg: if you combine two two's to make a four, then combine two more two's to create another four, then combine those two fours to get an eight, your total score will be 4+4+8=16.

When two tiles are combined, then they increase your score by the number on the new tile created.

The answer is 1835012.

1 solution

Trevor Arashiro
Nov 10, 2014

We must evaluate the "total value" of each tile (total value meaning if you have only that tile on the board, then your total score is equivalent to the total value of that tile)

A two has a value of 0, since you don't gain any points from spawning a two.

A four has a value of 4 because you have to combine two two's and it increases your score by 4

Now it gets a little more complicated. To evaluate the value of an 8. We must start from two's. To make two fours, we already established that our score will be 8, thus after combining the two fours, our score will be increased by 8 once more. Therefore, the total value of an 8 is 16.

The total value of a 16 is then the total value of two eights +16, which sums to be 48.

We can see a pattern evolving. The total value of the tile represented by $2^n$ is equivalent to $(n-1)(2^n)$ .

The ideal ending scenario will be when the board looks like this

$\begin{array}{l|c|r} 2 & 4 & 8 & 16 \\ \hline 256 & 128 & 64 & 32 \\ \hline 512 & 1024 & 2048 & 4096 \\ \hline 65536 & 32768 & 16384 & 8192\end{array}$

Thus we are looking for $\displaystyle \sum_1^{16} (n-1)(2^n)=\boxed{1835012}$

Hey Trevor this problem is far too easy to be a level 5. You should have made it a hard level 3 or maybe a easy level 4.

Boris Barron - 6 years, 6 months ago

I originally had the level at 4 when I posted it because I too thought it was a level four. But the low number of attempts and even lower number of solvers brought it's level all the way up to 400 (27% attempted it, 2% solved it)

Trevor Arashiro - 6 years, 6 months ago

@Trevor Arashiro did you created it

U Z - 6 years, 6 months ago

@U Z – yes, I did

Trevor Arashiro - 6 years, 6 months ago

Yeah! Exactly! Still solving @Trevor's question on first shot boosted me up! :)

Pranjal Jain - 6 years, 6 months ago

I got that pattern, but on the third attempt did I figure out that the summation was required. Thanks for explaining it so beautifully

Mrinmay Dhar - 6 years, 7 months ago

2048: much harder

The answer is 1835012.

1 solution

0 pending reports