MathLabs

Problem 5

Each of the six boxes B1,B2,B3,B4,B5,B6B_1,B_2,B_3,B_4,B_5,B_6 initially contains one coin. A type 1 operation chooses a nonempty box BjB_j with 1≤j≤51\le j\le5, removes one coin from it, and adds two coins to Bj+1B_{j+1}. A type 2 operation chooses a nonempty box BkB_k with 1≤k≤41\le k\le4, removes one coin from it, and exchanges the contents of (possibly empty) boxes Bk+1B_{k+1} and Bk+2B_{k+2}. Determine whether a finite sequence of operations can leave B1,B2,B3,B4,B5B_1,B_2,B_3,B_4,B_5 empty and B6B_6 containing exactly 2010201020102010^{2010^{2010}} coins. Here abca^{b^c} means a(bc)a^{(b^c)}.
Step 1 of 5: Reach a useful seed configuration
(1,1,1,1,1,1)→(0,0,5,11,0,0)(1,1,1,1,1,1)\to(0,0,5,11,0,0)
Detailed analysis

Write a configuration as (B1,B2,B3,B4,B5,B6)(B_1,B_2,B_3,B_4,B_5,B_6). Applying type 1 at B5B_5 and then at B1B_1 gives (0,3,1,1,0,3)(0,3,1,1,0,3). The following legal moves, with unchanged coordinates suppressed, are (0,3,1,1,0,3)→(0,0,7,0,0,7)→(0,0,6,2,0,7)→(0,0,6,1,2,7)→(0,0,6,1,0,11)→(0,0,6,0,11,0)→(0,0,5,11,0,0)(0,3,1,1,0,3)\to(0,0,7,0,0,7)\to(0,0,6,2,0,7)\to(0,0,6,1,2,7)\to(0,0,6,1,0,11)\to(0,0,6,0,11,0)\to(0,0,5,11,0,0). The first arrow uses type 1 at B2B_2 three times, then at B4B_4 once and B5B_5 twice; the other arrows use, respectively, type 1 at B3B_3, type 1 at B4B_4, type 1 at B5B_5 twice, type 2 at B4B_4, and type 2 at B3B_3. Hence the displayed seed configuration is reachable.