Binomial coefficient modularity conjecture

Question

I am not sure if I have seen this before:

$$\dbinom{n}{p}\equiv\big\lfloor{\dfrac{n}{p}}\big\rfloor\mod p$$

where $p$ is prime and $\lfloor x\rfloor$ is the floor function.

So, for example, $\dbinom{34}{7}\equiv\lfloor\dfrac{34}{7}\rfloor=4\mod 7$.

Answer 1

Your statement is a consequence of Lucas' theorem for binomial coefficients, which says that if $k$ is a positive integer and $0\leq a_i, b_i\leq p-1$ are integers for $i=0,1,\ldots,k$, then $$ {b_0+b_1 p+\ldots+b_k p^k\choose a_0+a_1p+\ldots+a_k p^k}\equiv {b_0\choose a_0}\cdots {b_k\choose a_k}\mod p. $$ If we write $n$ in base $p$ as $n=b_0+b_1p+\ldots +b_kp^k$, then $$ {n\choose p}\equiv {b_0\choose 0}{b_1\choose 1}{b_2\choose 0}\cdots {b_k\choose 0}= b_1\equiv \left\lfloor\frac{n}{p}\right\rfloor\mod p. $$