We will prove several congruences modulo a power of a prime such as $$
\sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv

{lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$
is odd} -{2^{n+1}+4\over n6^n} B_{p-n}({1\over 3}) &\pmod{p} &{if $n$ is even}.
$$ where $n$ is a positive integer and $p$ is prime such that $p>\max(n+1,3)$.

Let p be a prime and let a be a positive integer. In this paper we determine
$\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and
$\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a,
where m is any integer not divisible by p. For example, we show that if
$p\not=2,5$ then
$$\sum_{k=1}^{p-1}(-1)^k\frac{\binom{2k}k}k=-5\frac{F_{p-(\frac p5)}}p (mod
p),$$ where F_n is the n-th Fibonacci number and (-) is the Jacobi symbol. We
also prove that if p>3 then $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}k={8/9}
p^2B_{p-3} (mod p^3),$$ where B_n denotes the n-th Bernoulli number.