Question

Let $A=\left\{n \in N \mid n^{2} \leq n+10,000\right\}$, $B=\{3 k+1 \mid k \in N\}$ and $C=\{2 k \mid k \in N\}$, then the sum of all the elements of the set $A \cap(B-C)$ is equal to ___

Answer 1

$B-C \equiv\{7,13,19, \ldots 97, \ldots\}$
Now, $n^{2}-n \leq 100 \times 100$
$\Rightarrow n(n-1) \leq 100 \times 100$
$\Rightarrow A=\{1,2, \ldots \ldots, 100\}$
So, $A \cap(B-C)=\{7,13,19, \ldots \ldots, 97\}$
Hence, sum $=\frac{16}{2}(7+97)=832$