Source: IMO-2010-P5
Category: Operations and Strategies

Problem:
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin.  
The following two types of operations are allowed:
\begin{enumerate}
  \item Choose a non‑empty box $B_j$, $1\le j\le 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;
  \item Choose a non‑empty box $B_k$, $1\le k\le 4$, remove one coin from $B_k$ and swap the contents (possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.
\end{enumerate}
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.

Solution:
Yes, such a finite sequence exists.

First, we apply the allowed operations as follows:
\begin{align*}
  (1,1,1,1,1,1) &\to (0,3,1,0,3,1) \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).
\end{align*}
Henceforth, we ignore boxes $B_1$ and $B_2$, looking at just the last four boxes; so we write the current position as $(5,11,0,0)$.

We prove a lemma:
\begin{lemma*}
  Let $k \ge 0$ and $n > 0$. From $(k,n,0,0)$, we may reach $(k-1,2^n,0,0)$.
\end{lemma*}
\begin{proof}
  Working with only the last three boxes for now,
  \begin{align*}
    (n,0,0) &\to (n-1, 2, 0) \to (n-1, 0, 4) \\
    &\to (n-2, 4, 0) \to (n-2, 0, 8) \\
    &\to (n-3, 8, 0) \to (n-3, 0, 16) \\
    &\to \dots \to (1, 2^{n-1}, 0) \to (1, 0, 2^n) \to (0, 2^n, 0).
  \end{align*}
  Finally, we have $(k,n,0,0) \to (k,0,2^n,0) \to (k-1,2^n, 0,0)$.
\end{proof}

Now from $(5,11,0,0)$ we go as follows:
\begin{align*}
  (5,11,0,0) &\to (4, 2^{11}, 0, 0)
  \to \left(3, 2^{2^{11}}, 0, 0\right)
  \to \left(2, 2^{2^{2^{11}}}, 0, 0\right) \\
  &\to \left( 1, 2^{2^{2^{2^{11}}}}, 0, 0\right)
  \to \left(0, 2^{2^{2^{2^{2^{11}}}}}, 0, 0\right).
\end{align*}
Let $A = 2^{2^{2^{2^{2^{11}}}}} > 2010^{2010^{2010}} = B$. Then by using the second type of operation repeatedly on the fourth box (i.e., throwing away several coins by swapping the empty $B_5$ and $B_6$), we go from $(0,A,0,0)$ to $(0,B/4,0,0)$. From there we reach $(0,0,0,B)$.

Answer: Yes, such a finite sequence exists

Grading guidelines:
(Partial) 1. Found and correctly justified a nontrivial reachable configuration that simplifies the state, e.g. exhibited a valid sequence of moves from (1,1,1,1,1,1) to a state with B1=B2=0 (such as (0,0,5,11,0,0)), and explained why one may then focus on the last four boxes. 2. Proved a key “exponentiation step” of the form: from a state (k,n,0,0) (in the last four boxes) one can reach (k−1,2^n,0,0), by explicitly giving/justifying the repeated use of move type 1 to create powers of 2 and move type 2 to swap empties. 3. Established a correct reduction that once a state (0,m,0,0) is reached, one can transfer all coins into B6 and empty the other boxes (e.g. (0,m,0,0) -> (0,0,0,4m) or an equivalent final-transfer mechanism), even if the exact target value is not yet matched. (Almost) 1. Gave the full construction idea: reach a state (0,A,0,0) with A a power tower 2^(2^(…)) larger than 2010^(2010^2010), then ‘discard’ a controlled amount via swaps with an empty box, and finally transfer to (0,0,0,2010^(2010^2010)); but left a localized gap in justifying the discard step (e.g. why repeated type-2 moves can reduce A to the desired value or to B/4). 2. Successfully implemented the iterative lemma (k,n,0,0)->(k−1,2^n,0,0) enough times to produce a tower exceeding the target, and described the final adjustment and transfer to B6, but made a minor arithmetic/divisibility slip (e.g. a missing factor 4 in the last transfer) while the rest of the argument is correct. 3. Solution is almost complete, but made minor mistakes which are not negligible.