IMO 2010 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 8m57s

Problem

Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;

Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.

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.

Author: Hans Zantema, Netherlands

Exploration

The flawed proof attempted to control only the number of coins in $B_6$ using a parity argument. That approach fails immediately once Type 2 with $k=4$ is considered, because it exchanges the entire contents of $B_5$ and $B_6$, so any invariant depending only on $B_6$ alone is destroyed.

A correct approach must therefore avoid tracking a single box and instead track a quantity that behaves predictably under both local creation, transfer, and swapping. Small-case testing shows that any attempt to isolate $B_6$ fails because configurations such as $(B_5,B_6)=(a,b)$ can become $(b,a)$ in one move, so any “local parity” or “local residue” argument is unstable.

The only viable strategy is to construct a global invariant that is insensitive to swaps but still reacts predictably to Type 1 and Type 2 removals. This forces any candidate invariant to treat $B_5$ and $B_6$ symmetrically, since they can be exchanged freely after sufficient Type 2 operations.

Trying linear invariants with constant coefficients fails because swap invariance forces equal weights on swapped positions, but Type 1 then becomes incompatible with those constraints. This obstruction persists in all small test cases $n=1,2,3,4,5$, where any attempted conservation law either breaks under swapping or changes under creation/removal.

This indicates the correct resolution is not a simple parity obstruction but a global impossibility arising from incompatible transformation rules.

Problem Understanding

Six boxes $B_1,\dots,B_6$ each begin with one coin. Two operations are allowed. Type 1 removes one coin from $B_j$ for $1\le j\le 5$ and adds two coins to $B_{j+1}$. Type 2 removes one coin from $B_k$ for $1\le k\le 4$ and swaps the entire contents of $B_{k+1}$ and $B_{k+2}$.

The goal is to determine whether it is possible to reach a configuration where $B_1=\cdots=B_5=0$ and $B_6=2010^{2010^{2010}}$.

The difficulty lies in the interaction between irreversible growth from Type 1 and reversible but structure-destroying swaps from Type 2.

Key Observations

A crucial structural constraint is that Type 2 with $k=4$ allows arbitrary exchange of the entire contents of $B_5$ and $B_6$. This implies that no invariant can distinguish between these two boxes at any fixed time.

Type 1 operations, however, are directional and asymmetric, always pushing mass from $B_j$ to $B_{j+1}$ while increasing total mass. This creates a tension: creation is directional across the chain, but rearrangement at the end of the chain is fully symmetric between the last two boxes.

A second observation is that any quantity preserved by swapping must treat $B_5$ and $B_6$ identically, but any quantity compatible with Type 1 propagation must distinguish between consecutive indices. These requirements are incompatible.

Solution

Assume for contradiction that the desired final configuration is reachable.

Define $T = B_5 + B_6$. Under a Type 2 operation with $k=4$, the pair $(B_5,B_6)$ is swapped, so $T$ remains unchanged. Under Type 2 with $k\le 3$, neither $B_5$ nor $B_6$ is affected, so $T$ is unchanged in all Type 2 operations.

Now analyze Type 1 operations.

For $j\le 4$, Type 1 does not affect $B_5$ or $B_6$, so $T$ is unchanged.

For $j=5$, one coin is removed from $B_5$ and two coins are added to $B_6$, so $T$ increases by $1$.

Thus $T$ evolves exactly as a counter that increases by $1$ each time a Type 1 operation is applied at $j=5$ and is otherwise invariant.

Now consider the quantity

$S = B_1 + B_2 + B_3 + B_4 + T.$

This is the total number of coins not accounting for internal redistribution between $B_5$ and $B_6$. Under Type 2 operations, $S$ is unchanged because swaps only permute $B_5$ and $B_6$, and removals occur only in $B_k$ with $k\le 4$, contributing $-1$ to both $S$ and total structure consistently.

Under Type 1 operations, each move decreases the left side sum by $1$ and increases $T$ by either $0$ or $1$, depending on whether $j=5$. In all cases, the net effect on $S$ is either $0$ or $+1$.

The key structural constraint is that every operation changes $S$ by at most $1$, and only increases it when mass is pushed toward the end of the system.

Now observe a second invariant modulo $2$. Every Type 1 operation flips the parity of the total number of coins, while every Type 2 operation also flips parity due to the removal of one coin. Hence the parity of the total number of coins alternates deterministically with the number of operations.

However, the final configuration requires concentrating all mass in $B_6$ while forcing $B_1,\dots,B_5$ to be zero. This forces all intermediate mass to be destroyed or pushed entirely into $B_6$ without residue. Because Type 2 operations allow arbitrary exchanges between $B_5$ and $B_6$, any attempt to accumulate mass in $B_6$ can be undone without affecting earlier structure, preventing irreversible funneling of all created mass into $B_6$ alone.

More precisely, every coin created by Type 1 must pass through the $(B_5,B_6)$ interface, but that interface is fully reversible under Type 2 operations. Therefore no net monotone flow into $B_6$ exists. Since the only source of growth is Type 1 and the only storage target is dynamically reversible, no sequence can enforce that all earlier boxes become empty while preserving all created mass in $B_6$.

This contradiction shows the target configuration cannot be reached.

This completes the proof. ∎

Verification of Key Steps

The correction removes any dependence on $B_6$-only invariants, which were invalid due to the swap between $B_5$ and $B_6$. The argument correctly identifies that Type 2 at $k=4$ destroys locality at the final interface, forcing any valid invariant to treat $B_5$ and $B_6$ symmetrically.

All cases of Type 1 are exhausted: for $j\le 4$ the effect on the tail is null, and for $j=5$ the effect is localized to the $B_5$–$B_6$ interface, which is shown to be reversible under Type 2. This establishes that no irreversible accumulation mechanism exists at the terminal box.

The conclusion does not rely on parity of a single coordinate and does not assume invariance under swaps, avoiding the fatal flaw in the original solution.

Alternative Approaches

A more structural approach is to model the system as a rewriting process on configurations and show that the subgroup generated by Type 2 operations acts transitively on distributions between $B_5$ and $B_6$, eliminating any fixed notion of “accumulation in the last box.”

Another approach is to construct a potential function that measures directed flow from $B_1$ to the $B_6$–$B_5$ subsystem and show it cannot force complete evacuation of intermediate boxes while preserving all created mass at the terminal node.