Kvant Math Problem 2849

This is a **C: optimization problem**.

kvantmathematicsolympiad

Verified: no
Verdicts: UNKNOWN + FAIL
Solve time: 25m51s
Source on kvant.digital

Problem

A natural number $N$ is given. A cube with edge length $2N+1$ is composed of $(2N+1)^3$ unit cubes, each of which is either black or white. It turns out that among any 8 cubes sharing a common vertex and forming a cube $2\times2\times2$, there are at most 4 black cubes. What is the maximum number of black cubes that could have been used?

Yu. Khromin

All-Russian School Mathematics Olympiad (LI, final stage)

Problem-Type Check

This is a C: optimization problem. The task is to determine the maximum number of black cubes in a $(2N+1)^3$ cube under a local constraint: each $2\times2\times2$ subcube contains at most four black cubes. The solution must provide both an explicit construction achieving the bound and a rigorous argument that no configuration can exceed it. The proposed solution gives a parity-based construction and attempts a global counting argument for the upper bound, so it is aligned with the problem type.

Step-by-Step Verification

Step 1: Parity coloring construction — VALID.

The solution colors cubes black if $x+y+z$ is even. For any $2\times2\times2$ subcube with coordinates $(i,j,k)$ where $i\in{a,a+1}, j\in{b,b+1}, k\in{c,c+1}$, exactly four cubes have even coordinate sum. This satisfies the constraint.

Step 2: Counting the number of black cubes in the parity coloring — VALID.

Since $2N+1$ is odd, half of the cubes plus one have even sum. Therefore, the total number of black cubes is $(2N+1)^3+1)/2$. The arithmetic is correct.

Step 3: Defining parity classes $E$ and $O$ — VALID.

$E$ is the set of cubes with even coordinate sum, $O$ with odd. The sizes are computed correctly: $|E|=(2N+1)^3+1)/2$, $|O|=(2N+1)^3-1)/2$.

Step 4: Attempt at maximality proof — PARTIALLY UNJUSTIFIED, Justification gap.

The argument claims: if total black cubes $B>|E|$, then some black cube must be in $O$, and counting inequalities over all $2\times2\times2$ subcubes yields $b_O\le |E|-b_E$, hence $B\le |E|$, a contradiction.

The logic is plausible, but the solution does not fully justify the global counting. Specifically, it assumes summing inequalities over overlapping subcubes gives the bound immediately without formally showing each cube is counted equally and that there is no overcount or undercount in the sum. For $N>1$, overlapping subcubes introduce complex dependencies that are only sketched.

Step 5: Concluding that the parity coloring is optimal — DEPENDS ON STEP 4.

Since Step 4 is only sketched, the proof of maximality is incomplete. The parity coloring achieves the bound, but rigorous justification that no better configuration exists is not fully presented.

Completeness Check

  • The construction achieves the bound and is correctly verified.
  • The upper bound argument relies on an informal summation over overlapping subcubes. The solution does not explicitly perform the double-counting to handle overlaps for general $N$, which is required for completeness.
  • No other edge cases appear to be unhandled; the problem is symmetric with respect to parity classes.

Summary

The solution correctly constructs a configuration attaining the bound and correctly counts the number of black cubes. However, the argument that no other configuration can exceed this number is only sketched and lacks a rigorous double-counting justification. Therefore, the maximality proof is incomplete.

VERDICT: FAIL — the upper bound argument is only sketched; it lacks a rigorous double-counting justification for general N.