10.2. What is the greatest number of consecutive natural numbers, each of which has exactly four natural divisors (including 1 and the number itself)?
Answer: three numbers.

Solution. Suppose there are four consecutive numbers that satisfy the condition. Note that among four consecutive numbers, one is divisible by 4. Then, in the prime factorization of this number, there are at least two twos. If there is another prime divisor $p$ different from two, then the number of divisors of the number is at least six: $1, 2, 4, p, 2p, 4p$. If the factorization contains only twos, then for the number of divisors to be exactly four ($1, 2, 4, 8$), there must be exactly three twos. Thus, there is only one number divisible by 4 that has exactly four divisors - the number 8. Its neighbors (7 and 9) do not satisfy the condition, so the number of such numbers is no more than three.

Example of three consecutive numbers, each of which has exactly four natural divisors: $33, 34, 35$. There are other examples as well.

## Grading Criteria

+ A complete and well-reasoned solution is provided

± A generally correct but insufficiently justified solution is provided (without containing incorrect statements). For example, the correct answer and example are provided, it is noted that among four consecutive numbers one is divisible by 4, but the case $n=8$ is not mentioned

Ғ The correct answer and example are provided, but there is an error in the reasoning. For example, it is stated that any number divisible by four has more than four divisors

干 Only the correct answer and example are provided

- Only the answer "three numbers" is provided
- The problem is not solved or is solved incorrectly