WakeUp wrote: Show that the sequence \[\binom{2002}{2002},\binom{2003}{2002},\binom{2004}{2002},\ldots \]considred modulo $2002$, is periodic. Let $k=2002$ and $p \mid k$ a prime divisor of $k$. We will find the period $a_p$ of the sequence $\binom{n}{k}, \binom{n+1}{k},\dots$ $\pmod{p}$. In our case $k$ is squarefree, so the period of our sequence $\pmod{k}$ would be $\prod a_i$. Let $s_p(n)$ denote the sum of digits of $n$ in base $p$. Consider the congruence \[ \binom{n+a}{k} \equiv \binom{n}{k} \pmod{p} \]for some $a \in \mathbb{Z}^+$. If $\binom{n}{k} \equiv 0 \pmod{p}$ and $\binom{n+a}{k} \equiv 0 \pmod{p}$, the congruence will be true. Otherwise we must have at least one of the following: $v_p(\binom{n}{k})=0$ or $v_p(\binom{n}{k})=0$. Legendre's formula: \[ v_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor = \frac{n-s_p(n)}{p-1} \]\[ v_p(\binom{n}{k}) = v_p(\frac{n!}{k!(n-k)!}) = \frac{s_p(k)+s_p(n-k)-s_p(n)}{p-1} \]Hence it corresponds to at least one of $s_p(n)=s_p(n-k)+s_p(k), \: s_p(n+a)=s_p(n-k+a)+s_p(k)$ being true. Choose $a=p^b, b\in \mathbb{Z}^+$, where $p^b>k$. Lemma 1: If one equation is true, then the other one is, too. Proof: Note that by adding $a$ we increase the $b+1$-th digit of the number by $1$ in base $p$. Assume that exactly one of these two holds, say $s_p(n)=s_p(n-k)+s_p(k)$. It means there's no carrying when addong numbers $n-k$ and $k$. But this means that there would be no carrying when adding $n+a-k$ and $k$, as the digits of $k$ are in the first $b$ positions. So the second equality must also be true, a contradiction. $\square$ Therefore we obtain $v_p(\binom{n+a}{k})=v_p(\binom{n}{k})=0$. Equation (1) rewrites as \[ \binom{n}{k} \frac{(n+1)\dots(n+a)}{(n-k+1)\dots(n-k+a)} \equiv \binom{n}{k} \pmod{p} \]Upon dividing by $\binom{n}{k} \not\equiv 0 \pmod{p}$, we obtain \[ \frac{(n+1)\dots(n+a)}{(n-k+1)\dots(n-k+a)} \equiv 1 \pmod{p} \]Note that we must also have $v_p(\frac{(n+1)\dots(n+a)}{(n-k+1)\dots(n-k+a)}) =0$. Let $y=v_p((n+1)\dots(n+a))=v_p((n-k+1)\dots(n-k+a))$. Both numerator and denominator consist of the product of $a=p^b$ consecutive integers, giving $p^b$ different residues $\pmod{p^b}$. Writing each number in the numerator and in the denominator as $u=p^v \cdot w$, $w \not\equiv 0 \pmod{p}$, we get the same set of $w's$ in them. Denote the product of each such set by $Y \not\equiv 0 \pmod{p}$ and rewrite (2): \[ \frac{p^y}{p^y} \cdot \frac{Y}{Y} \equiv 1 \pmod{p} \]\[ \Leftrightarrow p^y \equiv p^y \pmod{p^{y+1}} \]The last one is obviously true, so we intial congruence is also true. Therefore we may take $a_p=a$, and so we are done. $\blacksquare$