p-adic valuations of some sums of multinomial coefficients

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that ${\nu }_p({\sum }_{k=0}^{n-1}\frac{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}{m^k})$ and ${\nu }_p({\sum }_{k=0}^{n-1}\left(\genfrac{}{}{0px}{}{n-1}{k}\right)(-1{)}^k\frac{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}{m^k})$ are at least ${\nu }_p(n)$, where ${\nu }_p(x)$ denotes the $p$-adic valuation of $x$. Furthermore, if $p>3$ then $$n^{-1}\sum _{k=0}^{n-1}\frac{\text{bi}2kk}{m^k}=\frac{\left(\genfrac{}{}{0px}{}{2n-1}{n-1}\right)}{4^{n-1}}(mod{p}^{{\nu }_p(m-4)})$$ and $$n^{-1}\sum _{k=0}^{n-1}\left(\genfrac{}{}{0px}{}{n-1}{k}\right)(-1{)}^k\frac{\left(\genfrac{}{}{0px}{}{2k}{k}\right)}{m^k}=\frac{C_{n-1}}{4^{n-1}}(mod{p}^{{\nu }_p(m-4)}),$$ where $C_k$ denotes the Catalan number $\left(\genfrac{}{}{0px}{}{2k}{k}\right)\mathrm{/}(k+1)$. This implies several conjectures of Guo and Zeng [GZ]. We also raise two conjectures, and prove that $n>1$ is a prime if and only if $$\sum _{k=0}^{n-1}multinomial(n-1)kk,\mathrm{.}\mathrm{.}\mathrm{.},k=0(modn),$$ where $multinomial{k}_1+\mathrm{.}\mathrm{.}\mathrm{.}+k_{n-1}{k}_1,\mathrm{.}\mathrm{.}\mathrm{.},k_{n-1}$ denotes the multinomial coefficient $(k_1+\mathrm{.}\mathrm{.}\mathrm{.}+k_{n-1})!\mathrm{/}(k_1!\mathrm{.}\mathrm{.}\mathrm{.}{k}_{n-1}!)$.