6 divisors, 6 remainders, 1 answer

Relevant wiki: Chinese Remainder Theorem

Let the required integer be $N$ . We can find $N$ using the Chinese remainder theorem as follows:

$\begin{aligned} N & \equiv 1 \text{ (mod 2)} & \small \color{#3D99F6} \implies N \equiv 2k+1 \text{, where }k \in \mathbb Z \\ \implies 2k+1 & \equiv 1 \text{ (mod 3)} \\ 2k & \equiv 0 \text{ (mod 3)} & \small \color{#3D99F6} \implies k = 0 \implies N \equiv 2\times 3 m + 2k + 1 \\ \implies 6m+1 & \equiv 2 \text{ (mod 5)} \\ m & \equiv 1 \text{ (mod 5)} & \small \color{#3D99F6} \implies m = 1 \implies N \equiv 6\times 5 n + 6m+1 \\ \implies 30n+7 & \equiv 4 \text{ (mod 7)} \\ 2n & \equiv 4 \text{ (mod 7)} & \small \color{#3D99F6} \implies n = 2 \implies N \equiv 30\times 7 p + 30n+7 \\ \implies 210p + 67 & \equiv 3 \text{ (mod 11)} \\ p + 1 & \equiv 3 \text{ (mod 11)} \\ p & \equiv 2 \text{ (mod 11)} & \small \color{#3D99F6} \implies p = 2 \end{aligned}$

Therefore, $N \equiv 210 \times 2 + 67 = \boxed{487}$