8. (40 points) In a mental arithmetic competition, several teams participated. Each team received one natural number and was supposed to find the largest and the smallest, but not equal to 1, odd divisors of this number. All teams solved their tasks correctly. When reviewing the results, it turned out that any initial number obtained by any team can be represented in the form $15 M + 11 m$, where $M$ is the largest, and $m$ is the smallest odd divisors. How many teams participated in the competition and for which numbers did the teams find the divisors?
Answer: 4 teams; 528, 880, 1232, 1936.

Solution: It is easy to see that $M$ is divisible by $m$, and $\frac{15 M+11 m}{M}$ is a power of two. Since $m \leqslant M$,

$$
15<\frac{15 M+11 m}{M} \leqslant \frac{15 M+11 M}{M}=26
$$

Among the integers in the interval $(15,26]$, only 16 is a power of two. Therefore, $15 M+11 m=16 M$, from which $M=11 m$, and the original number is $176 m$. Since $m$ is its smallest odd divisor, and 176 is divisible by $11, m$ can only take the values $3,5,7,11$. In this case, $176 m$ will be respectively $528,880,1232,1936$. Verification shows that the obtained values satisfy the condition.