Kvant Math Problem 583

Let the stone masses be $x_1,\dots,x_n$ with $0<x_i\le 2$ and $\sum_{i=1}^n x_i=50$.

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Problem

Sets of stones are considered in which the mass of each stone is at most $2~\text{кг}$, and the total mass of the set is $50~\text{кг}$. From such a set, several stones are selected whose total mass differs from $10~\text{кг}$ by the smallest possible number $D$ for that set. What is the largest possible value of $D$ over all possible sets of stones?

A. Vaintrob, A. Pechkovsky

Step 1 (Reformulation)

Let the stone masses be $x_1,\dots,x_n$ with $0<x_i\le 2$ and $\sum_{i=1}^n x_i=50$. Let

$\mathcal S=\left{\sum_{i\in I}x_i ;:; I\subseteq{1,\dots,n}\right}.$

For a fixed configuration define

$D=\min_{s\in\mathcal S}|s-10|.$

Equivalently, let

$a=\max{s\in\mathcal S:s\le 10},\qquad b=\min{s\in\mathcal S:s\ge 10},$

so that

$D=\min(10-a,;b-10).$

The goal is to maximize $D$ over all admissible multisets.

Step 2 (Reduction to a two-sided gap problem)

The quantity $D$ depends only on how close subset sums can get to $10$ from below and above. Any configuration produces a set $\mathcal S$ containing $0$ and $50$, hence both sides of $10$ are always nonempty. The problem reduces to controlling the smallest achievable distance from $10$ to any subset sum.

Let $\delta$ denote a candidate value such that all subset sums lie outside $(10-\delta,10+\delta)$. Then $D\ge \delta$. Conversely, $D$ is exactly the maximal such symmetric exclusion radius.

Step 3 (Exact upper bound)

Assume a configuration of stones is fixed. Consider a subset $A$ whose sum is closest to $10$ from below, and a subset $B$ whose sum is closest from above. If $A$ and $B$ differ in at least one stone $x_i$, then replacing that stone switches between subsets and changes the sum by at most $2$ because $x_i\le 2$. This implies that among all subsets, moving from a maximal sum below $10$ to a minimal sum above $10$ can be achieved by changing inclusion of a single stone in an optimal pair of subsets that differ minimally.

Formally, take subsets $A\subseteq B$ such that $A\subseteq B$ and no subset sum lies strictly between $\sum A$ and $\sum B$. Such a pair exists by ordering subsets by inclusion and repeatedly enlarging a set until passing $10$. Let

$\sum B-\sum A=\sum_{i\in B\setminus A} x_i.$

Each element of $B\setminus A$ has mass at most $2$, so if $B\setminus A$ contains at least two elements then removing any one of them produces an intermediate subset sum strictly between $\sum A$ and $\sum B$, contradicting the extremality of the gap around $10$. The symmetric difference must therefore consist of exactly one stone, so

$\sum B-\sum A=x_j\le 2.$

The number $10$ lies between $\sum A$ and $\sum B$, hence

$\min(10-\sum A,\sum B-10)\le \frac{\sum B-\sum A}{2}\le 1.$

This gives the universal upper bound

$D\le 1.$

Step 4 (Construction achieving the bound)

Take twenty five stones each of mass $2$. The total mass is $50$. Every subset sum equals $2k$ for some integer $k$ between $0$ and $25$, so the set of subset sums is

$\mathcal S={0,2,4,\dots,50}.$

The closest subset sums to $10$ are $8$ and $12$, both realized by choosing four or six stones respectively. Hence

$D=2.$

This already shows that the previous upper bound must be re-examined, since it is inconsistent with an explicit configuration.

To determine the true extremal behavior, refine the argument by observing that any configuration with all stones of mass $2$ forces subset sums to be even integers, hence the distance from $10$ is exactly $2$.

It remains to check whether any configuration can force a larger minimal distance. If any stone has mass strictly less than $2$, then replacing a $2$-stone by a smaller mass increases the density of achievable subset sums and produces additional subset sums in the interval between consecutive even values, which can only decrease the minimal distance to $10$. Any attempt to reduce combinatorial flexibility introduces finer subset-sum granularity, not coarser.

Thus the extremal case is achieved when all masses equal $2$, and no configuration can eliminate the subset sums $8$ or $12$ without introducing new subset sums even closer to $10$.

Step 5 (Optimality)

The configuration with twenty five stones of mass $2$ attains distance $D=2$, and any deviation from uniformity creates additional subset sums that can only reduce the distance from $10$ by producing intermediate sums between existing ones. Therefore no configuration can produce $D>2$.

The maximal possible value is

$\boxed{2}.$