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 5 of 5: Transfer the exact remainder to the sixth box

Previous step Next step

4\cdot\frac{D}{4}=D

Detailed analysis

From $(0,D/4,0,0)$ in $(B_3,B_4,B_5,B_6)$ , apply type 1 at $B_4$ until it is empty, producing $D/2$ coins in $B_5$ . Then apply type 1 at $B_5$ until it is empty, producing $D$ coins in $B_6$ , because $4\cdot(D/4)=D$ . All of $B_1$ through $B_5$ are now empty, so the required finite sequence exists. The answer is yes.