Remainders one less than the divisor

Statement

Find the smallest positive integer that leaves remainders $2$, $3$, $4$ and $5$ when divided by $3$, $4$, $5$ and $6$ respectively.

Solution

In each case the remainder is one less than the divisor, i.e. the number is $1$ short of the next multiple. Hence $n+1$ is a common multiple of $3,4,5,6$: $n+1 = \operatorname{lcm}(3,4,5,6)\cdot k = 60k.$ The smallest positive value is obtained with $k=1$: $n = 60 - 1 = \boxed{59}.$