Work Step by Step
We are trying to find a number $n$ such that division by $2$, $3$, $4$, $5$, or $6$ leaves a remainder of $1$. The solution to this problem is $1$ greater than the LCM of these numbers because adding $1$ to the LCM provides a number that leaves a remainder of $1$ when divided by the list above. The least common multiple of 2, 3, 4, 5, and 6 is the lowest number that is a multiple of these numbers, which is 60. Thus, we obtain: $=60+1=61$.