IMO 2010 · Problem 5

In each of six boxes \(B_1,B_2,B_3,B_4,B_5,B_6\) there is initially one coin. There are two types of operation allowed:

Type 1: Choose a nonempty box \(B_j\) with \(1\le j\le5\). Remove one coin from \(B_j\) and add two coins to \(B_{j+1}\).

Type 2: Choose a nonempty box \(B_k\) with \(1\le k\le4\). Remove one coin from \(B_k\) and exchange the contents of (possibly empty) boxes \(B_{k+1}\) and \(B_{k+2}\).

Determine whether there is a finite sequence of such operations that results in boxes \(B_1,B_2,B_3,B_4,B_5\) being empty and box \(B_6\) containing exactly \(2010^{2010^{2010}}\) coins. (Note that \(a^{b^c}=a^{(b^c)}\).)

Official text of the problem. Official problem paper (PDF)

Answer

Yes, such a finite sequence exists.

Solution

Write \[ N=2010^{2010^{2010}}. \] We will first produce a sufficiently large pile in \(B_4\), with all other boxes empty. We can then remove surplus coins from \(B_4\) and obtain exactly \(N\) coins in \(B_6\).

Apply Type 1 once at \(B_1\), and then three times at \(B_2\). This empties \(B_1,B_2\) and leaves seven coins in \(B_3\). From now on, write \((a,b,c,d)\) for the contents of \(B_3,B_4,B_5,B_6\); the first two boxes remain empty.

Applying Type 1 at \(B_5\), then Type 2 at \(B_4\), and then Type 2 at \(B_3\), gives \[ (7,1,1,1)\longrightarrow(7,1,0,3) \longrightarrow(7,0,3,0)\longrightarrow(6,3,0,0). \]

We next prove that, for any integers \(a\ge1\) and \(b\ge2\), we can achieve \[ (a,b,0,0)\longrightarrow(a-1,2^b,0,0). \tag{1} \]

First apply Type 1 once at \(B_4\), then twice at \(B_5\): \[ (a,b,0,0)\longrightarrow(a,b-1,2,0) \longrightarrow(a,b-1,0,4). \] Whenever the configuration is \((a,r,0,c)\) with \(r\ge1\), applying Type 2 at \(B_4\), followed by \(c\) applications of Type 1 at \(B_5\), gives \[ (a,r,0,c)\longrightarrow(a,r-1,c,0) \longrightarrow(a,r-1,0,2c). \] Repeat this procedure \(b-2\) times. The resulting configuration is \[ (a,1,0,4\cdot2^{b-2})=(a,1,0,2^b). \] Finally apply Type 2 at \(B_4\), and then at \(B_3\): \[ (a,1,0,2^b)\longrightarrow(a,0,2^b,0) \longrightarrow(a-1,2^b,0,0). \] This proves (1), including \(b=2\), when the repeated procedure is performed zero times. Every operation is legal, and each sequence is finite.

Define \[ t_0=3,\qquad t_{i+1}=2^{t_i}\quad(0\le i<6). \] Starting from \((6,3,0,0)\), apply (1) six times. We obtain \[ (0,t_6,0,0). \] Thus only \(B_4\) is nonempty, containing \(M=t_6\) coins.

To check that this is sufficient, observe that \[ t_1=8,\qquad t_2=256,\qquad t_5=2^{2^{2^{256}}}. \] Since \(2010<2^{11}\), \(11<2^4\), and \(22114<2^{15}<2^{256}\), we have \[ \begin{aligned} N &<(2^{11})^{(2^{11})^{2010}}\\ &=2^{11\cdot2^{22110}}\\ &<2^{2^{22114}}\\ &<2^{2^{2^{256}}} =t_5<t_6=M. \end{aligned} \]

Because \(2010\) is even and \(2010^{2010}\ge2\), the number \[ q=\frac N4 \] is a positive integer. Apply Type 2 at \(B_4\) exactly \(M-q\) times. Both \(B_5\) and \(B_6\) are empty, so each operation simply removes one coin from \(B_4\). We are left with exactly \(q\) coins there.

Finally, apply Type 1 at \(B_4\) exactly \(q\) times, producing \(2q\) coins in \(B_5\), and then apply Type 1 at \(B_5\) exactly \(2q\) times. All of \(B_1,\ldots,B_5\) are now empty, and \(B_6\) contains \[ 4q=N=2010^{2010^{2010}} \] coins. The entire sequence is finite, as required.

The solution was checked against the official IMO shortlist solutions.