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\newcommand{\leg}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
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\begin{document}


\begin{abstract}
We prove several congruences modulo a power of a prime, such as
\[
\sum_{0<k_1<\cdots<k_{n}<p}\leg{p-k_{n}}{3}
\frac{(-1)^{k_{n}}}{k_1\cdots k_{n}}\equiv
\begin{cases}\displaystyle
-\frac{2^{n+1}+2}{6^{n+1}}\,p\,B_{p-n-1}\left({1/3}\right)
\pmod{p^2}, &\text{if $n$ is odd,}
\\[.8em]
\displaystyle
-\frac{2^{n+1}+4}{n6^n}\,B_{p-n}\left({1/3}\right)
\pmod{p}, &\text{if $n$ is even,}
\end{cases}
\]
where $n$ is a positive integer and $p$ is a prime such that $p>\max(n+1,3)$.
\end{abstract}

\end{document}
