If two consecutive numbers are removed from first $n$ natural numbers, and the arithmetic mean of remaining numbers is$ \frac {105}{4}$, find $n $.
Let $k $ and $k+1$ be two removed numbers, then we have :
$$ 1+2+3+\cdots +n = 2k+1+\frac{105n-210}{4}. $$
Rearranging I got quadratic as below
$$ 2n^2-103n+206-8k=0.$$
So:
$$ 4n=103 \pm \sqrt {8961+64k}. $$
How do I proceed from here?
Best Answer
Hint:
After getting $2n^2-103n+206=8k$, notice that $1\leq k\leq n-1$
So $$8\leq 2n^2-103n+206\leq 8(n-1)$$ Then you get only few possible values of n.