5. We took ten consecutive natural numbers greater than 1, multiplied them, found all the prime divisors of the resulting number, and multiplied these prime divisors (taking each exactly once). What is the smallest number that could have resulted? Fully justify your answer.

Solution. We will prove that among ten consecutive numbers, there must be a number that has a prime divisor different from $2, 3, 5, 7$. Let's count how many of the ten consecutive numbers can have the specified four divisors. Out of ten numbers, exactly five are divisible by 2, and at least one of these five numbers is divisible by 3. (This fact can be taken for granted in the work, but let's recall one possible proof. The remainders from dividing six consecutive numbers by 6 form a set from 0 to 5, i.e., there is a number with a remainder of 0, which is divisible by both 2 and 3.) Further, the set of ten numbers contains at least three and no more than four numbers that are divisible by 3. If there are three such numbers, at least one of these three is divisible by 2 and has already been counted. If there are four such numbers, they have the form $3k, 3k+3, 3k+6, 3k+9$. Depending on the parity of $k$, either $3k, 3k+6$ or $3k+3, 3k+9$ are divisible by 2. Thus, when considering numbers divisible by 3, we can add no more than two new numbers to the five even numbers already counted. Ten consecutive numbers contain exactly two numbers that are divisible by 5, one of which is even and has already been counted. Therefore, divisibility by 5 adds no more than one number to the seven already counted. Among ten consecutive numbers, no more than two are divisible by 7, one of which is even and has already been counted. Divisibility by 7 adds no more than one number to the eight already counted. Thus, among ten consecutive numbers, no more than nine have at least one of the divisors $2, 3, 5, 7$, i.e., there is a number with a larger prime divisor and not divisible by any of the first four prime numbers. Therefore, in the product of ten numbers, there will be at least one divisor different from $2, 3, 5, 7$, so the product of all prime divisors will be greater than 210. The smallest possible product is $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$.