China Mathematical Competition (Complementary Test)

Let $m=k+t\times l\times (k!)$, where $t$ is any positive integer. To prove that $\left(\genfrac{}{}{0ex}{}{m}{k}\right)$ and $l$ are relatively prime, we only need to prove that for any prime factor $p$ of $l$, $p\nmid \left(\genfrac{}{}{0ex}{}{m}{k}\right)$.

If $p\nmid k!$, we have $$\begin{array}{rl}k!\left(\genfrac{}{}{0ex}{}{m}{k}\right)& =\prod _{i=1}^k(m-k+i)\\ & =\prod _{i=1}^k[i+tl(k!)]\\ & =\prod _{i=1}^{k}i\equiv k!(\mathrm{mod}p).\end{array}$$ Therefore, $p\nmid \left(\genfrac{}{}{0ex}{}{m}{k}\right)$.

If $p\mid k!$, there exists integer $\alpha \ge 1$ such that $p^{\alpha }\mid k!$ but $p^{\alpha +1}\nmid k!$. Then $p^{\alpha +1}\mid l(k!)$. We have $$\begin{array}{rl}k!\left(\genfrac{}{}{0ex}{}{m}{k}\right)& =\prod _{i=1}^k(m-k+i)\\ & =\prod _{i=1}^k[i+tl(k!)]\\ & =\prod _{i=1}^{k}i\equiv k!(\mathrm{mod}{p}^{\alpha +1}).\end{array}$$ Therefore, $p^{\alpha }\mid k!\left(\genfrac{}{}{0ex}{}{m}{k}\right)$ and $p^{\alpha +1}\nmid k!\left(\genfrac{}{}{0ex}{}{m}{k}\right)$. Since $p^{\alpha }\mid k!$, we get $p\nmid \left(\genfrac{}{}{0ex}{}{m}{k}\right)$. The proof is completed.

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Alternative solution.

Let $m=k+t\times l\times (k!{)}^z$, where $t$ is any positive integer. The following proof steps are similar to those in Solution I, and are omitted.