3. Find the smallest natural number that, when divided by each of the numbers $2,3,4,5,6$ and 7, gives a remainder of 1.
Solution. If $n$ is the desired number, then $n-1$ is the smallest number divisible by $2,3,4,5,6$ and 7, i.e., $n-1$ is the LCM (least common multiple) of 2, 3, 4, 5, 6, and 7. According to this,

$$
n=2 \cdot 2 \cdot 3 \cdot 5 \cdot 7+1=421
$$