2. The least common multiple of seven natural numbers is 2012. Find the smallest possible sum of these numbers.
Solution: Let's factorize the number 2012 into prime factors $2012=2^{2} \cdot 503$.

One of the numbers, the LCM of which is 2012, must be divisible by $2^{2}=4$, and one (possibly the same) - by 503. If the same number is divisible by both 4 and 503, then it is divisible by 2012, and the sum of seven numbers is not less than 2018. If 4 and 503 are divisible by different numbers, then the sum of all seven numbers is not less than $4+503+1+1+1+1+1=512$. Since LCM(4,503,1,1,1,1,1)=2012, then 512 is the smallest possible value of the sum.

## Comments on Checking

The proof of minimality only considers the case where 4 and 503 are divisible by different numbers - 2 points.

An example of numbers with a sum of 512 is provided, but there is no proof of minimality - 3 points.

It is proven that the sum $\geq 512$, but no example of such numbers is given - 4 points.