Given 25 containers, their capacities are $1, 2, \ldots, 25$ liters. In how many ways can ten of them be chosen such that any amount of 1 liter of water can be measured using any two of the selected containers? (An unlimited amount of water is available, and water can be transferred from one container to another.)





Translating the text into English while preserving the original line breaks and format, the result is as above.
The task clearly consists of two almost independent parts. First, we show that if \(a\) and \(b\) are positive integers, then 1 liter of water can be measured precisely using an \(a\) liter capacity container \(A\) and a \(b\) liter capacity container \(B\) if and only if the greatest common divisor of \(a\) and \(b\) is 1. After this, the question is how many ways can ten numbers be chosen from the set \(\{1, 2, \ldots, 25\}\) such that any two of them are relatively prime.

i) If in a given state container \(A\) contains \(a_1\) liters and container \(B\) contains \(b_1\) liters of water, then in the next step we can empty or fill any of the containers to the brim, or start pouring the contents of one container into the other until the latter is full or the former is empty. After any step, at least one of the containers is either empty or full.

Measuring 1 liter of water under these conditions means that after a finite sequence of the above steps, one of the containers contains 1 liter of water. (The other is either full or empty, which is not significant.)

The condition \(\gcd(a, b) = 1\) is clearly necessary, since if \(d\) is a common divisor of \(a\) and \(b\), then the amount of water in the initially empty containers is clearly divisible by \(d\), and this property remains after the execution of any of the above steps.

On the other hand, if \(\gcd(a, b) = 1\), then 1 liter of water can be measured. If \(\gcd(a, b) = 1\) and \(a = b = 1\), then the statement is trivial, so we can assume, for example, that \(b > 1\). The proof now relies on a well-known arithmetic statement:

If \(\gcd(a, b) = 1\) and \(b > 1\), then the congruence

\[
a x \equiv 1 \quad (\bmod b)
\]

has a solution. Indeed, if \(\gcd(a, b) = 1\), then two multiples of \(a\), \(t_1 a\) and \(t_2 a\), give the same remainder when divided by \(b\) if and only if \(b \mid a t_1 - a t_2 = a(t_1 - t_2)\), or, because \(\gcd(a, b) = 1\), \(b \mid t_1 - t_2\). Therefore, if we take the multiples of \(a\) that give different remainders when divided by \(b\), namely \(1, 2, \ldots, b\), then these also give different remainders when divided by \(b\). Hence, every remainder appears exactly once, so there is a multiple that gives a remainder of 1: \(k a = q b + 1\). In the case \(b > 1\), 1 liter of water can also be measured starting from the solution of the congruence \(b y \equiv 1 (\bmod a)\).

2. Since under the given conditions the congruence

\[
a x \equiv b_1 \quad (\bmod b)
\]

is solvable for any \(0 \leq b_1 < b\), any integer \(1 \leq c \leq a + b\) can also be measured with the two containers.

3. It follows from the solution that for any positive integers \(a\) and \(b\), \(c\) liters of water can be measured precisely if and only if \(c \leq a + b\) and \(c\) is divisible by the greatest common divisor of \(a\) and \(b\). From this, it can easily be shown by induction that the same is true for more than two containers.
4. An interesting geometric interpretation can be given to the above proof and the theorem used. If the containers contain \(0 \leq a_1 \leq a\) and \(0 \leq b_1 \leq b\) liters of water, then the state of the system can be represented by the point \(P(a_1, b_1)\), and the process itself can be visualized by the trajectory of this point (Figure 1). During the process, the point \(P\) is a point of the rectangle \(O A T B\), and the diagram shows an intermediate state when the content of one of the containers changes. Horizontal and vertical movements represent the emptying or filling of the respective containers, while the "diagonal" movement represents the transfer of content from one container to the other. The point \(P\) starts from the origin and, during its journey, moves to the boundary of the rectangle in one of the allowed directions, then continues its journey.

The states that can actually be reached are thus selected from the integer-coordinate points on the perimeter of the rectangle \(A O T B\). The statement means that if \(\gcd(a, b) = 1\), there is a trajectory that passes through the point \(C_A(1,0)\) or \(C_B(0,1)\). The second diagram shows the solution for \(a = 7\) and \(b = 5\).

If you like, the "billiard" on the diagram "solves" the congruence

\[
7 x \equiv 1 \quad (\bmod 5)
\]

The value of \(x\) is the number of reflections on the \(T A\) wall, and the value of \(q\) is the number of reflections on the \(T B\) wall. Here, \(x = 3\) and \(q = 4\), and indeed: \(7 \cdot 3 = 4 \cdot 5 + 1\). The point \(P\) reaches the point \(C_A(1,0)\) after a total of \(7 + 5\) reflections, which is 13 transfers. The third diagram shows the solution of the symmetric congruence

\[
5 y \equiv 1 \quad (\bmod 7)
\]

on the "billiard table"; here \(y = 3\) and \(q = 2\), and after 8 reflections, 9 transfers, we reach the point \(C_B(0,1)\).

On both diagrams, it can be verified that if we continue the trajectory, it passes through all integer-coordinate points on the perimeter of the rectangle \(O A T B\) and closes at the point \(O\), thus illustrating the statements made in the second remark.

ii) Let's now address the question of the problem. First, note that more than ten numbers cannot be chosen from the set \(\{1, 2, \ldots, 25\}\) such that any two of them are relatively prime. In the above set, there are 9 prime numbers: \(2(4), 3(2), 5(2), 7(1), 11(1), 13(1), 17(1), 19(1), 23(1)\). The number in parentheses after each prime is the largest exponent with which the prime appears in the set. If two numbers are relatively prime, then their prime factors must not have any common elements, so we need to select 10 subsets from the 9-element set of prime factors, such that no two subsets have a common element. This is only possible if one of them is the empty set - the number 1 is among the selected numbers - and the other 9 are the 9 one-element subsets. Therefore, the 10-element sets that meet the conditions are of the form

\[
\left\{1, 2^a, 3^b, 5^c, 7, 11, 13, 17, 19, 23\right\}
\]

where \(1 \leq a \leq 4\), \(1 \leq b \leq 2\), and \(1 \leq c \leq 2\). This gives a total of \(4 \cdot 2 \cdot 2 = 16\) possibilities, so there are 16 ways to choose 10 numbers from the first 25 positive integers such that any two of them are relatively prime.
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