Kozhevnikov P.A.

Petya and Vasya were given identical sets of $N$ weights, in which the masses of any two weights differ by no more than 1.25 times. Petya managed to divide all the weights of his set into 10 equal mass groups, and Vasya managed to divide all the weights of his set into 11 equal mass groups. Find the smallest possible value of $N$.

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Example. Let the set consist of 20 weights of 50 g and 30 weights of 40 g. In this case, Petya can divide all the weights into ten groups, each containing two weights of 50 g and three weights of 40 g; and Vasya can divide all the weights into five groups, each containing four weights of 50 g, and six groups, each containing five weights of 40 g. Thus, the value $N=50$ is possible.

Estimate. Suppose $N<50$, and Petya and Vasya managed to arrange the weights into groups as required. Then in one of Petya's groups, there are no more than four weights, and in one of Vasya's groups, there are also no more than four weights.

Lemma. Suppose there are two groups of weights of equal total mass, consisting of $k$ and $l$ weights, where $k<l$. Then $k \geq 4$, and in the case $k=4$ we have

$l=5$, and in each of the groups, the weights are of equal mass.

Proof. Let $m$ be the mass of the lightest weight in the group of $l$ weights, then the mass of the second group is at least $l m \geq(k+1) m$, and the mass of the first group is at most $1.25 k m$, from which $k+1 \leq 1.25 k$ and $k \geq 4$. If $k=4$, then in the above estimates, the inequalities turn into equalities, so

$l=k+1=5$, in the group of five weights, all weights are of the same mass $m=4 x$, and in the group of four weights, all weights are of mass $5 x$.

Consider two cases.

1) Among Petya's groups, there are two groups with a different number of weights. Then, from the lemma, it follows that in several of Petya's groups, there are four weights of mass $5 x$, and in the remaining Petya's groups, there are five weights of mass $4 x$. The total weight of all weights is then $200 x$, which is impossible, since 200 is not divisible by 11, and the total mass in each of Vasya's groups must be an integer multiple of $x$.
2) In ten of Petya's groups, the number of weights is equal. Then $N$ is a multiple of 10. If among Vasya's groups, there are two groups with a different number of weights, then from the lemma, it follows that in each of Vasya's groups, there are at least four weights, from which $N \leq 44 \Rightarrow N \geq 50$ (since $N$ is a multiple of 10). Contradiction.

If in 11 of Vasya's groups, the number of weights is equal, then $N$ is a multiple of 11 $\Rightarrow N$ is a multiple of 110 $\Rightarrow N \geq 110$. Contradiction.

## Answer

50 weights.