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 $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$.
Answer: 2310.

## Municipal Stage of the All-Russian Mathematics Olympiad 2018-2019 Academic Year

10th Grade

Grading Criteria

| Score | Points for |
| :---: | :---: | :--- |
| 7 | Complete solution, a correct algorithm of actions for Zhenya is provided, and all conditions are verified. |
| 6 | Complete solution, a correct algorithm of actions for Zhenya is provided, but it is not verified that the arrangement obtained by the algorithm satisfies all conditions. |
| 4 | An algorithm of actions for Zhenya is constructed, based on dividing the cells into pairs. Due to an unsuccessful choice of cell placement and (or) the sequence of their filling, the solution is not completed. The correct answer is provided or no answer is given. |
| 2 | An arrangement of numbers is provided that satisfies the condition of the problem, but the sequence of moves by the players is not described. |
| 1 | An algorithm of actions for Zhenya is constructed, based on dividing the cells into pairs. Due to an unsuccessful choice of cell placement and (or) the sequence of their filling, the end of the solution contains an error. An incorrect answer is provided (see 4 points). |
| 0 | Incorrect solution, or only the answer is written without reasoning, or the problem statement is misunderstood. |


| Problem 2 | Score | Points for |
| :---: | :---: | :---: |
|  | 7 | Fully correct solution. |
|  | 3 | The existence of two different roots is proven, but their location, as stated in the problem, is not indicated or justified. |
|  | 0 | Incorrect proof. In particular, an error in the choice of the initial statement and its consequences, or an error in working with inequalities. |


| Problem 3 | Score | Points for |
| :---: | :---: | :---: |
|  | 7 | Fully correct solution. Correct answer. |
|  | 5 | Correct solution containing an arithmetic error. The answer provided differs from the correct one due to an arithmetic error. |
|  | 4 | Generally correct solution containing an error in the application of trigonometric functions. The answer provided differs from the correct one due to an incorrect replacement of a trigonometric function with a cofunction and (or) an incorrect transition to a half-angle. |
|  | 3 | It is proven that the segments marked as \(x\) and \(y\) on the diagram are equal. |
|  | 0 | Incorrect solution, in particular due to a misunderstanding of the configuration of the tangents. |


| Score | Points for |
| :--- | :--- | :--- |
| Problem 4 | Fully correct solution. |
| 5 | Partially correct proof containing an arithmetic error. |
| 4 | Partially correct proof containing an error in counting the number of terms. |
|  | Partially correct proof containing an error in applying the formula for the sum of an arithmetic progression. |
| 1 | It is stated that the expression on the right is the sum of sums of arithmetic progressions. Further progress is absent. |
| 0 | Incorrect solution. |


|  | Score | Points for |
| :---: | :---: | :---: |
|  | 7 | Fully correct solution. |
|  | 6 | It is proven that among ten consecutive numbers, there is always a number that has a prime divisor different from 2, 3, 5, 7. The proven statement is applied to find the answer. The answer provided may be incorrect, differing from the correct one by replacing the divisor 11 with (one) another divisor. |
| Problem 5 | 5 | It is proven that among ten consecutive numbers, there is always a number that has a prime divisor different from 2, 3, 5, 7. It is indicated that this fact must be applied to find the answer. The application of the proven statement is not completed or is incorrect. Or it is proven that the sought number is greater than 210, but no answer is provided. |
|  | 3 | It is not proven that among ten consecutive numbers, there is always a number that has a prime divisor different from 2, 3, 5, 7, but it is indicated that this fact must be applied to find the answer. The correct answer may be provided. |
|  | 2 | Only divisibility by 2, 3, 5, 7 is considered, the question of the presence or absence of other prime divisors is not taken into account. |
|  | 1 | The correct answer is provided without justification. |
|  | 0 | Incorrect solution. Incorrect answer (see 6 points). |