Let $A = \{x : x^2 \leq x + 20000;\ x \in \mathbb{N}\}$, $B = \{x = 3k+2;\ k \in \mathbb{N}\}$, $C = \{x = 4k;\ k \in \mathbb{N}\}$. The sum of the elements of the set $A \cap (B - C)$ is
Step-by-Step Solution
Key Concept: $A$: naturals with $x^2-x\leq 20000$, so $x\leq 142$. $B-C$: elements of $B$ not in $C$; $B$: numbers $\equiv 2\pmod 3$; $C$: multiples of 4. Elements in $B\cap C$ are $\equiv 2\pmod 3$ and $\equiv 0\pmod 4$, i.e. $\equiv 8\pmod{12}$.
$A$: $x^2-x\leq 20000 \Rightarrow x\leq 142$. $B=\{2,5,8,11,...\}$ (assuming $k\geq 0$). $C=\{4,8,12,...\}$. $B-C$: elements $\equiv 2\pmod 3$ but not $\equiv 0\pmod 4$. In $A\cap B$: numbers $\{2,5,8,...,140\}$ minus those in $C$. Elements in $B\cap C$ up to 142: $\{8,20,32,...\}$, i.e. $\equiv 8\pmod{12}$. Sum of all in $A\cap B$ minus sum of $A\cap B\cap C$. Sum $= 2447$.
Correct Answer: B