Let $X$ be the set consisting of the first $2018$ terms of the arithmetic progression $1, 6, 11, \ldots,$ and $Y$ be the set consisting of the first $2018$ terms of the arithmetic progression $9, 16, 23, \ldots .$ Then, the number of elements in the set $X \cup Y$ is ______.

$\text{Given: } X = \{\text{first } 2018 \text{ terms of } 1, 6, 11, \ldots\}$ $Y = \{\text{first } 2018 \text{ terms of } 9, 16, 23, \ldots\}$ $\text{To find: } |X \cup Y|$ $\text{Plan:}\quad 1.\ \text{write general terms of both APs}$ $2.\ \text{form the congruence for a common term}$ $3.\ \text{count them} \to |X\cup Y| = |X|+|Y|-|X\cap Y|$ $X:\ a = 1,\ d = 6-1 = 5$ $Y:\ a = 9,\ d = 16-9 = 7$ $x_n = 1 + (n-1)\cdot 5 = 5n - 4$ $y_n = 9 + (n-1)\cdot 7 = 7n + 2$ $\text{Common term } n \text{ must satisfy:}$ $n \equiv 1 \pmod{5}, \quad n \equiv 2 \pmod{7}$ $n = 1 + 5t \quad (\text{so } n \equiv 1 \pmod 5)$ $1 + 5t \equiv 2 \pmod{7} \Rightarrow 5t \equiv 1 \pmod{7}$ $\text{Inverse of } 5 \pmod{7} \text{ is } 3 \quad (5\cdot 3 = 15 \equiv 1)$ $t \equiv 3 \pmod{7}$ $t = 3 + 7k \quad (k \in \mathbb{Z})$ $n = 1 + 5(3 + 7k) = 16 + 35k$ $x_{2018} = 1 + 2017\cdot 5 = 10086$ $y_{2018} = 9 + 2017\cdot 7 = 14128$ $16 + 35k \le 10086$ $35k \le 10070 \Rightarrow k \le 287.71$ $k = 287:\ 16 + 35\cdot 287 = 10061 \le 10086\ \checkmark$ $k = 288:\ 16 + 35\cdot 288 = 10096 > 10086\ \times$ $k = 0, 1, 2, \ldots, 287$ $\text{Number of common terms} = 288$ $|X \cup Y| = |X| + |Y| - |X \cap Y|$ $= 2018 + 2018 - 288$ $= 4036 - 288$ $\boxed{|X \cup Y| = 3748}$