![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-25.jpg?height=225&width=1123&top_left_y=2563&top_left_x=-1)

There are seven glasses of water: the first glass is half full, the second is one-third full, the third is one-fourth full, the fourth is $1 / 5$ full, the fifth is $1 / 8$ full, the sixth is $1 / 9$ full, and the seventh is $1 / 10$ full. It is allowed to pour all the water from one glass into another or to pour water from one glass into another until it is completely full. Can any glass end up being filled a) to $1 / 12$; b) to $1 / 6$ after several pourings?
a) If the capacity of a glass is considered to be 1, then the first three glasses together contain $1 \frac{1}{12}$ of water. Pour all the water from the second glass into the first, and then from the third glass into the first until the first glass is full. After this, the third glass will contain $1 / 12$.

b) We will prove by induction on the number of pourings that the amount of water in a non-empty glass after the pourings is either 1 or a fractional part of the sum of some of the numbers $1 / 2, 1 / 3, 1 / 4, 1 / 5, 1 / 8, 1 / 9, 1 / 10$, with different glasses containing non-repeating numbers in their sums.

The base case is obvious.

Inductive step. Suppose the amount of water in glasses $A$ and $B$ is $a$ and $b$ respectively. If all the water from glass $A$ is poured into glass $B$, the new amounts are 0 and $a+b$. If glass $B$ is filled to the top from $A$, the new amounts are $a+b-1$ and 1. It is now clear that the statement remains true.

Let $1 / 6 = \{a_{1} + \ldots + a_{k}\}$, where $a_{i}$ are some of the numbers $1 / 2, 1 / 3, 1 / 4, 1 / 5, 1 / 8, 1 / 9, 1 / 10$. Then this sum does not contain the numbers $1 / 4$, $1 / 5$, $1 / 8$, $1 / 9$, $1 / 10$. Indeed, if at least one of the numbers $1 / 5$ or $1 / 10$, one of the numbers $1 / 4$ or $1 / 8$, or the number $1 / 9$ is present, the denominator of the resulting fraction is divisible by 5, 4, or 9, respectively. However, the number 6 is not divisible by any of these numbers. The numbers $1 / 2$ and $1 / 3$ cannot be added to form a number with a fractional part of $1 / 6$.

## Answer

a) It can; b) it cannot.