Problem 5

Each of the six boxes

B_1,B_2,B_3,B_4,B_5,B_6

initially contains one coin. A type 1 operation chooses a nonempty box

B_j

with

1\le j\le5

, removes one coin from it, and adds two coins to

B_{j+1}

. A type 2 operation chooses a nonempty box

B_k

with

1\le k\le4

, removes one coin from it, and exchanges the contents of (possibly empty) boxes

B_{k+1}

and

B_{k+2}

. Determine whether a finite sequence of operations can leave

B_1,B_2,B_3,B_4,B_5

empty and

B_6

containing exactly

2010^{2010^{2010}}

coins. Here

a^{b^c}

means

a^{(b^c)}

Step 4 of 5: Discard the surplus exactly

Previous step Next step

(0,A,0,0)\to\left(0,\frac{D}{4},0,0\right)\qquad D=2010^{2010^{2010}}

Detailed analysis

Use type 2 repeatedly at the box containing $A$ , with the next two boxes empty; each such operation simply removes one coin and leaves the other contents unchanged. Because $A>D/4$ , stop exactly when that box contains $D/4$ . This is possible since $D=2010^{2010^{2010}}$ is divisible by $4$ (indeed $2010$ is even). The configuration is then $(0,D/4,0,0)$ in the last four boxes.