# Lucas' Theorem and Pascal's Triangle

I have a general question about Lucas' Theorem. Lucas' Theorem says the following:

Theorem (Lucas' Theorem) Let $p$ be a prime number. Write $n$ and $k$ in base $p$: $n = a_0 + a_{1}+a_{2}p^{2} + \cdots + a_{d}p^{d}$ and $k = b_0+b_{1}p+b_{2}p^{2}+ \cdots +b_{d}p^{d}$ where $0 \leq a_i,b_i \leq p-1$. Then $$\ \binom{n}{k} \equiv \prod_{i=0}^{d} \binom{a_i}{b_i} \pmod{p}$$

Another way of viewing this is that in $\mathbb{Z}_p$, $\binom{n}{k}$ is a product of binomial coefficients.

Question. How does this theorem correspond to the geometric interpretation of Pascal's Triangle? Namely, for any entry in Pascal's triangle which is odd mark a $\text{x}$. Else leave it blank. So we basically get a fractal like structure. Is Lucas' Theorem just a statement of a way to get extra rows of Pascal's Triangle given that we know $2^n$ rows?

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Where is $2^n$ coming from ? –  Sasha Aug 30 '11 at 21:30
$2^n$ is the total number of subsets of an $n$-element set. –  Damien Aug 30 '11 at 21:36
When $p = 2$, Lucas' Theorem states that ${n \choose m} \equiv 0 \pmod{2}$ if and only if in the binary expansion of $n = (\overline{ \cdots n_1n_0})_2$ and $m = (\overline{ \cdots m_1m_0})_2$, some binary digit (say the $d$th binary digit) satisfies $n_d = 0, m_d = 1$. In particular, this shows that ${2^k + n \choose m} \equiv {2^k + n \choose 2^k + m} \equiv {n \choose m} \pmod{2}$ when $2^k > n > m$, which describes the recursive nature of the Sierpinski triangle that can be found in Pascal's triangle by highlighting the odd elements (which it seems is the geometric interpretation you referred to). However, this is a fairly restricted application of Lucas' Theorem and can be seen by easier means.