When $x$ is divided by each of $4$, $5$, and $6$, remainders of $3$, $4$, and $5$ (respectively) are obtained. What is the smallest possible positive integer value of $x$?
I notice that the remainders are one less than the divisors in each case. This suggests that $x$ is one less than a common multiple of $4$, $5$, and $6$. The least common multiple of $4$, $5$, and $6$ is $60$, since it is the smallest number that is a multiple of all three. Therefore, the smallest possible value of $x$ that satisfies the given conditions is $60 - 1 = 59$. 