Question

A pack contains $n$ cards numbered from $1$ to $n$. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$. If the smaller of the numbers on the removed cards is k, then $k-20$ is

Answer 1

Clearly, $1+2+3+\dots +n-2\le 1224\le 3+4+\dots n$

$\Rightarrow \frac{(n-2)(n-1)}{2}\le 1224\le \frac{(n-2)}{2}(3+n)$

$\Rightarrow n^2-3n-2446\le 0$ and

$n^2+n-2454\ge 0$

$\Rightarrow 49<n<51\Rightarrow n=50$

$\therefore \frac{n(n+1)}{2}-(2k+1)=1224\Rightarrow k=25\Rightarrow k-20=5$