Remainder when $^{40}C_{12}$ is divided by $7$.

Question

I was solving a binomial summation problem, and I got $^{40}C_{12}$ as the answer. Now, the question demands to find the remainder when it is divided by $7$. $40!$ can be divided $5$ times by $7$ (using a famous GIF trick.) Unfortunately, $28!$ and $12!$ can be divided $4$ and $1$ time respectively by $7$, and hence the answer is clearly not zero. I am hence unable to find a way to deduce the remainder (without using a calculator to calculate $^{40}C_{12}$).

Answer 1

We could do the binomial expansion of $(x+1)^{40}$ in the seven-element field, by exploiting that $(x+1)^7=x^7+1$, so $$ (x+1)^{40}=((x+1)^7)^5(x+1)^5=(x^7+1)^5(x+1)^5 $$ The power $x^{12}$ can only be obtained as $x^7$ from the first factor, where the coefficient is $\binom{5}{1}=5$ and $x^5$ from the second, where the coefficient is $1$. Therefore $$ \binom{40}{12}\equiv5\pmod{7} $$

Answer 2

\begin{align*} \binom{40}{12} &= \frac {(40)\cdots (29)} {(12)\cdots (1)} \\[6pt] &= \Bigl(\frac{40}{12}{\cdots}\frac{36}{8}\Bigr) {\,\cdot\,} 5 {\,\cdot\,} \Bigl(\frac{34}{6}{\cdots}\frac{29}{1}\Bigr) \\[6pt] &\equiv \bigl(1{\cdots}1\bigr) {\,\cdot\,} 5 {\,\cdot\,} \bigl(1{\cdots}1\bigr) \;(\text{mod}\;7)&&\text{[since each fraction in the above line} \\[0pt] &&&\;\text{reduces to $1$, mod $7$]} \\[4pt] &\equiv 5 \;(\text{mod}\;7) \\[4pt] \end{align*}