5. From the odd natural numbers from 1 to 47, 12 fractions less than 1 were formed, using each number exactly once. The resulting fractions were divided into groups of equal values. What is the smallest number of groups that could have been obtained? (I. Rubanov)
Answer: 7. Solution: Evaluation. A fraction containing at least one of the prime numbers 17, 19, 23, 29, 31, 37, 41, 43, 47 cannot equal any of our other fractions, because already $17 \cdot 3 > 47$. Since there are no more than nine such "lonely" fractions, among the constructed fractions there are some that are not equal to them. Therefore, if there are at least six "lonely" fractions, there are no fewer than seven groups of equal fractions. Suppose there are five "lonely" fractions (obviously, there cannot be fewer). Replace one of the remaining fractions with an equivalent irreducible fraction $\mathrm{m} / \mathrm{n}$. Since all fractions are less than 1, then $n \geq 3$. The denominators of fractions equal to $m / n$ are odd, multiples of $n$, and do not exceed 47. If $n \geq 5$, then there are no more than five such denominators, and among the seven remaining fractions, there will be two that are not equal; the evaluation is proven. For $n=3$, there are eight odd numbers that are multiples of 3 and do not exceed 47: $3, 9, 15, 21, 27, 33, 39, 45$. But all the remaining fractions after reduction should give $1 / 3$, and in one set, fractions $1 / 3$ and $3 / 9$, as well as fractions $5 / 15$ and $15 / 45$, cannot coexist, so there are no more than six fractions equal to $1 / 3$. Thus, in this case, among the seven remaining fractions, there will be two that are not equal. Example when there are exactly 7 classes: $\{1 / 3, 5 / 15, 7 / 21, 9 / 27, 11 / 33, 13 / 39\}, \{25 / 41\}, \{35 / 43\}, \{45 / 47\}, \{17 / 19\}, \{23 / 29\}, \{31 / 37\}$.