Bogdanov I.i.

Initially, there are 111 pieces of plasticine of the same mass on the table. In one operation, you can choose several groups (possibly one) with the same number of pieces and in each group, combine all the plasticine into one piece. What is the minimum number of operations required to get exactly 11 pieces, each of which has a different mass?

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Let the mass of one original piece be 1. If in the first operation in each group there are $k$ pieces, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to get 11 pieces of different masses in one operation.

We will show that the required result can be achieved in two operations. For the first operation, choose 37 groups of 2 pieces each; after the operation, there will be 37 pieces with masses of 1 and 2. For the second operation, choose 9 groups of 8 pieces each: in the $i$-th group $(1 \leq i \leq 9)$, there will be $i-1$ pieces of mass 2 and $9-i$ pieces of mass 1. Two pieces of masses 1 and 2 will remain unused, and from the $i$-th group, a piece of mass $9-i+2(i-1)=7+i$ will be obtained. Thus, 11 pieces with masses 1, 2, 8, 9, ..., 16 will be obtained.

## Answer

In two operations.

![](https://cdn.mathpix.com/cropped/2024_05_06_b41405b02b6e888d68cag-01.jpg?height=110&width=1258&top_left_y=2401&top_left_x=1)
Complexity: 3
Classes: 5,6
Arrange the numbers $1,2,3, \ldots, 9$ in the circles so that the sum of the numbers on each side of the triangle equals ..... 17.

![](https://cdn.mathpix.com/cropped/2024_05_06_b41405b02b6e888d68cag-02.jpg?height=391&width=418&top_left_y=-1&top_left_x=820)

## Answer

See the diagram

![](https://cdn.mathpix.com/cropped/2024_05_06_b41405b02b6e888d68cag-02.jpg?height=368&width=414&top_left_y=565&top_left_x=822)

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