9,10 $[\quad$ The Pigeonhole Principle (etc.).

The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take?
Example. Consider nine numbers equal to 91 and the number 182. Their sum is 1001.

Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \cdot 11 \cdot 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On the other hand, the smallest term in the sum (and thus the GCD) is no more than $1001: 10$, that is, no more than 101. It remains to note that 91 is the largest divisor of the number 1001 that satisfies this condition.

91. 

## [ LCM and GCD. Mutual simplicity ] Problem $\underline{65064}$ Topics: $\left.\begin{array}{l}{\left[\begin{array}{c}\text { Evenness and oddness }\end{array}\right]} \\ {[\text { Divisibility criteria by } 3 \text { and 9 }}\end{array}\right]$

Author: Emelyanova T.

Can the asterisks in the expression $\operatorname{LCM}\left(*,{ }^{*}, *\right)-\operatorname{LCM}(*, *, *)=2009$ be replaced by six consecutive natural numbers in some order so that the equality becomes true?

## Solution

Suppose such numbers were found. The least common multiple of several numbers is divisible by each of them and, consequently, by each of their divisors. Therefore, if there is an even number among the numbers from which the LCM is found, then the LCM will also be even. The difference between two even numbers is an even number, but 2009 is odd. Therefore, all even numbers must be in one LCM, that is, one LCM will be found from three consecutive even numbers, and the other from three consecutive odd numbers. Among three consecutive even or odd numbers, there is a multiple of 3. Therefore, both LCMs are multiples of 3. But 2009 is not divisible by 3. Contradiction.

## Answer

No.