RMO 2003 ; Part - 2

Since $n\in\Bbb{Z_{\gt 7}}$ , its base $7$ representation contains more than or equal to $2$ digits. More precisely, consider the base $7$ representation of $n$ as follows:

$n=\sum_{k=0}^q n_k7^k~\textrm{where }7^q\leq n\lt 7^{q+1}$

where $q$ is some non-negative integer $\geq 1$ depending on the value of $n$ satisfying the inequality (bounding) stated above.

Also, it's obvious by definition of base $7$ expansion that all the $n_i$ 's are non-negative integers depending on the value of $n$ which satisfy $0\leq n_i\leq 6$ (this idea is used while evaluating the floor expression).

Also, we have $\overline{7}_{10}=\overline{10}_7=\overline{\underbrace{000\cdots}_{(q-1)\textrm{ digits}}10}~_{7}$ .

Claim: The given expression in the problem is divisible by $7$ .

Proof: We use Lucas' Theorem noting the fact that $7$ is a prime.

$\binom n7\equiv \binom{n_0}{0}\cdot\binom{n_1}{1}\cdot\prod_{k=2}^q\binom{n_k}{0}\equiv 1\cdot n_1\cdot\prod_{k=2}^q 1\equiv n_1\pmod7$

Then again, using the base $7$ expansion of $n$ , we easily have,

$\begin{aligned}\left\lfloor\frac n7\right\rfloor=\left\lfloor\left(\sum_{k=2}^q n_k7^{k-1}\right)+n_1+\frac{n_0}{7}\right\rfloor=\left(\sum_{k=2}^q n_k7^{k-1}\right)+n_1+\left\lfloor\frac{n_0}{7}\right\rfloor\\ \implies\left\lfloor\frac n7\right\rfloor=\left(\sum_{k=2}^q n_k7^{k-1}\right)+n_1+0\equiv 0+n_1+0\equiv n_1\pmod7\end{aligned}$

Hence, we conclude that $\forall~n\in\Bbb{Z_{\gt 7}}$ , we have,

$\binom n7-\left\lfloor\frac n7\right\rfloor\equiv n_1-n_1\equiv 0\pmod{7}\iff 7\mid\binom n7-\left\lfloor\frac n7\right\rfloor$

For disproving the other options, it's easy to show a counter-example like $n=8$ which $\in\Bbb{Z_{\gt 7}}$ easily disproves all the other three options.

Does anyone have an elementary solution ? I am unable to find one.