1st Math Riddle 2019

Let the remaining number of cards of the pack be $N$ . We can find $N$ using the Chinese remainder theorem as follows.

$\begin{aligned} N & \equiv 3 \text{ (mod 4)} & \small \color{#3D99F6} \implies N \equiv 4a + 3 \text{, where }a \text{ is an integer.} \\ 4a + 3 & \equiv 2 \text{ (mod 3)} \\ a + 0 & \equiv 2 \text{ (mod 3)} & \small \color{#3D99F6} \implies a = 2, 4a+3 = 11 \\ 12b + 11 & \equiv 2 \text{ (mod 5)} & \small \color{#3D99F6} \text{and } N \equiv 3\times 4b + 11, b \in \mathbb Z \\ 2b + 1 & \equiv 2 \text{ (mod 5)} \\ 2b & \equiv 1 \text{ (mod 5)} & \small \color{#3D99F6} \implies b = 3, 12b+11 = 47 \\ \implies N & = \boxed{47} \end{aligned}$