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 3 of 5: Five compound moves exceed the target

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A=2^{2^{2^{2^{2^{11}}}}}\ >\frac{2010^{2010^{2010}}}{4}

Detailed analysis

Starting from $(0,0,5,11,0,0)$ , apply the compound move five times. The successive nonzero pair is $(5,11)\to(4,2^{11})\to(3,2^{2^{11}})\to(2,2^{2^{2^{11}}})\to(1,2^{2^{2^{2^{11}}}})\to(0,A)$ , where $A=2^{2^{2^{2^{2^{11}}}}}$ . To compare sizes, $2010<2^{11}$ gives $2010^{2010^{2010}}<2^{11\cdot2^{11\cdot2^{11}}}$ . On the other hand, $2^{2^{2^{11}}}>11\cdot2^{11\cdot2^{11}}$ , so $A>2010^{2010^{2010}}$ and in particular the displayed inequality holds. (The stronger inequality is harmless.)