Remainder 1 when dividing by 3, 4, 5 and 6

Statement

What is the smallest integer greater than $1$ that leaves remainder $1$ when divided by $3$, $4$, $5$ and $6$?

Solution

If $n$ leaves remainder $1$ for each of those divisors, then $n-1$ is a common multiple of $3,4,5,6$, hence a multiple of their LCM: $\operatorname{lcm}(3,4,5,6) = 60.$ So $n-1 = 60,120,\dots$ and the smallest $n>1$ is $n = 60 + 1 = \boxed{61}.$