To prove that ${2+i \choose i}\equiv k \mod n$ is not possible that $k=0,1,\ldots,n-1 \forall i\ge 0$ and $i \in \mathbb{Z}$ and $n$ is odd.
This is a problem from ISI 2014 written test in a little varied form. I could not attempt the problem at all. Please help. Thank you.
I think what you are asking is to show that if $n$ is odd, then for at least one $k\in\{0,1,\ldots,n-1\}$, there is no solution to ${2+i\choose i}\equiv k\bmod n$.
Well, ${2+i\choose i}={2+i\choose 2}$ and, since $n$ is odd, ${2+i\choose 2}\equiv {2+i+n\choose 2} \bmod n$. In other words, the sequence $\left({i+2\choose 2}\bmod n\right)_{i=0}^\infty$ is periodic with period $n$. So it's enough to consider the first $n$ terms, i.e., $$S=\left({2\choose 2}\bmod n,{3\choose 2}\bmod n,\ldots, {n+1\choose 2}\bmod n\right).$$ We need to show that there is at least one $k\in\{0,1,\ldots,n-1\}$ not appearing as an entry in $S$.
If $n=3$, $S=(1,0,0)$ and so $k=2$ is the required missing number $k$.
For $n\geq 5$, we have ${2\choose 2}\equiv 1\pmod n$ and ${n-1\choose 2}\equiv 1\bmod n$. Hence $1$ appears twice amongst the $n$ entries in $S$, and so at least one number in $\{0,1,\ldots,n-1\}$ does not appear in $S$.