A circle's circumference has 50 numbers written on it, each of which is either +1 or -1. We want to know the product of these numbers. What is the minimum number of questions needed to find out, if we can ask about the product of three consecutive numbers at a time?
Let the numbers be denoted in the order of the clock hands as $a_{1}, a_{2}, \ldots, a_{50}$ (so $a_{1}$ and $a_{50}$ are also adjacent).

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The positions of the three selected numbers - whose product we are asked for - are uniquely determined by the middle one. If this middle number is $a_{i}$, we denote the product by $b_{i}$. It is clear that we can ask at most 50 different questions, since more than 50 numbers cannot be in the middle. We claim that the product of the 50 numbers can be determined from these 50 questions. Indeed,

$$
\begin{gathered}
b_{1} \cdot b_{2} \cdot \ldots \cdot b_{50}= \\
=\left(a_{50} \cdot a_{1} \cdot a_{2}\right)\left(a_{1} \cdot a_{2} \cdot a_{3}\right) \cdot \ldots \cdot\left(a_{49} \cdot a_{50} \cdot a_{1}\right)= \\
=\left(a_{1} \cdot a_{2} \cdot \ldots \cdot a_{50}\right)^{3}
\end{gathered}
$$

Since the cube of a real number uniquely determines the number itself, the value of $a_{1} \cdot a_{2} \cdot a_{50}$ can also be determined.

On the other hand, we claim that the product cannot be determined from 49 or fewer questions. More precisely, we prove the following stronger statement: if the answer to the first 49 questions (whatever they may be) is that the product of the 3 numbers in question is +1, then it is still possible that the product of the 50 numbers is +1, and it is also possible that the product of the 50 numbers is -1.

The 49 questions can be asked about at most 49 of the values $b_{1}, b_{2}, \ldots, b_{49}$, so there is one for which the questioner does not know whether it is +1 or -1. We can assume that this is precisely $b_{50}$. Thus, to prove the statement formulated earlier, it is sufficient to write 50 +1s or -1s on the circumference of the circle in two different ways such that

(1) in one case the product of the written numbers is +1, and in the other case it is -1;

(2) in both cases the values of $b_{1}, b_{2}, \ldots, b_{49}$ (except for $b_{50}=a_{49} \cdot a_{50} \cdot a_{1}$) are +1.

In one of the writings, 50 +1s are placed on the circumference of the circle; the other is shown in the accompanying diagram: $a_{i}=+1$ if $i$ gives a remainder of 1 when divided by 3, and $a_{i}=-1$ otherwise.

In the first case, the product of the written numbers is +1; in the second case, we need to take the product of $1+48 / 3=17$ +1s and $50-17=33$ -1s, which is -1. In the first case, $b_{1}, b_{2}, \ldots, b_{49}$ are clearly +1, while in the second case, except for $b_{49}$, two of the three factors are -1 and one is +1, so $b_{1}=b_{2}=\ldots=b_{49}=+1$ in this case as well.

Therefore, at least 50 questions must be asked to determine the product.

Remark. From 50 questions, more can be inferred, namely, it can be determined for each number whether it is +1 or -1. Indeed, $a_{1} \cdot a_{2} \cdot a_{3}, a_{4} \cdot a_{5} \cdot a_{6}, \ldots, a_{46} \cdot a_{47} \cdot a_{48}$ are known, so their product is also known. The product $a_{1} \cdot \ldots \cdot a_{50}$ is also known, so the quotient of these, i.e., $a_{49} \cdot a_{50}$, can be determined. Given $a_{49} \cdot a_{50} \cdot a_{1}$, we can determine $a_{1}$ from it, and similarly, we can get all the numbers.