JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
Let $A=\{n\in\mathbf{N}\mid n^2\le n+10{,}000\}$, $B=\{3k+1\mid k\in\mathbf{N}\}$ and $C=\{2k\mid k\in\mathbf{N}\}$, then the sum of all the elements of the set $A\cap(B-C)$ is equal to ________.
Sets, Relations and Function
Answer$832$
Solution
$n^2\le n+10000$ gives $n\le100$ ($100^2=10000\le10100$, $101^2=10201>10101$), so $A=\{1,\ldots,100\}$. $B-C$ = elements $3k+1$ that are odd, i.e. $k$ even, $k=2m$: $6m+1$, $m\ge1$. Within $A$: $7,13,\ldots,97$ ($m=1,\ldots,16$). Sum $=6\cdot\frac{16\cdot17}{2}+16=816+16=832$.