Question

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

Answer 1

Answer: 3748

$X:1,6,11,\dots 10086$
$Y:9,16,23,\dots ,14128$
$X\cap Y:16,51,86,\dots$
Let $m=n(X\cap Y)$
$16+(m-1)\times 35\le 10086$
$m\le 288.71m=288$
$n(X\cup Y)=n(X)+n(Y)-n(X\cap Y)$
$=2018+2018=288=3748$