3. Find the smallest natural number that when divided by 2 gives a remainder of 1, when divided by 3 gives a remainder of 2, when divided by 4 gives a remainder of 3, and when divided by 5 gives a remainder of 4.
Solution. Let the required number be $x$; then we have

$$
x=2 x_{1}+1=3 x_{2}+2=4 x_{3}+3=5 x_{4}+4
$$

For the number $x+1$, we will have

$$
x+1=2\left(x_{1}+1\right)=3\left(x_{2}+1\right)=4\left(x_{3}+1\right)=5\left(x_{4}+1\right)
$$

which means that $x+1$ is divisible by $2,3,4$ and 5. Therefore, the number $x+1$ is the least common multiple of the numbers $2,3,4$ and 5, i.e., $x+1=60$, from which we get $x=59$.