Task 1. Determine the smallest natural number that is divisible by 7, and when divided by 2, 3, 4, 5, and 6, gives a remainder of 1.
Solution. If $x$ is the desired number, then $x-1$ is divisible by $2,3,4,5$ and 6, which means that $x-1=k \cdot \operatorname{LCM}(2,3,4,5,6)=60 k$, where $k$ is a natural number. Thus, $x=60 k+1$. The smallest number of the form $60 k+1$ that is divisible by 7 is obtained for $k=5$. The desired number is $x=60 \cdot 5+1=301$.