Question:
Let $A=\left\{n \in \mathbf{N} \mid n^{2} \leq n+10,000\right\}, B=\{3 k+1 \mid k \in \mathbf{N}\}$ and $\mathrm{C}=\{2 \mathrm{k} \mid \mathrm{k} \in \mathrm{N}\}$, then the sum of all the elements
of the set $A \cap(B-C)$ is equal to ___________
Solution:
$\mathrm{B}-\mathrm{C} \equiv\{7,13,19, \ldots 97, \ldots .\}$
Now, $\mathrm{n}^{2}-\mathrm{n} \leq 100 \times 100$
$\Rightarrow \mathrm{n}(\mathrm{n}-1) \leq 100 \times 100$
$\Rightarrow \mathrm{A}=\{1,2, \ldots, 100\}$
So, $A \cap(B-C)=\{7,13,19, \ldots, 97\}$
Hence, sum $=\frac{16}{2}(7+97)=832$
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