Let A = nN : n^2 n+10000, B=3k+1 : kN and C=2k : kN.

Question

Let $A = \{n\in\mathbb{N} : n^2\leq n+10000\}$, $B=\{3k+1 : k\in\mathbb{N}\}$ and $C=\{2k : k\in\mathbb{N}\}$. Then the sum of all the elements of the set $A\cap(B-C)$ is equal to:

Answer: 832

Accepted Answer

$n^2\leq n+10000 \iff n^2-n\leq10000 \iff n(n-1)\leq100	imes100$, giving $A=\{1,2,\ldots,100\}$.

$B = \{4,7,10,13,\ldots\}$ (numbers $\equiv1\pmod3$), $C=\{2,4,6,\ldots\}$ (even numbers).

$B-C$ = numbers of the form $3k+1$ that are odd: $\{7,13,19,\ldots\}$ (an AP with common difference 6, since every other term of $B$ is odd).

$A\cap(B-C) = \{7,13,19,\ldots,97\}$, an AP with $d=6$.

Number of terms: $\dfrac{97-7}{6}+1=16$.

Sum: $\dfrac{16}{2}(7+97) = 8	imes104 = 832$

Solutions · 1