The question goes as such: The number $N = {}^{20}C_{7}-{}^{20}C_{8}+{}^{20}C_{9}-{}^{20}C_{10}+.....-{}^{20}C_{20}$ is divisibly by:
(A)$3$ $ $ (B)$4$ $ $ (C)$7$ $ $ (D)$19$
There are multiple correct answers.
My attempt:
Let $\binom{n}{r}$ indicate ${}^{n}C_{r}$
${}^{20}C_{0}+{}^{20}C_{2}+{}^{20}C_{4}+{}^{20}C_{8}+.....+{}^{20}C_{20} = 2^{19}$ (1)
${}^{20}C_{1}+{}^{20}C_{3}+{}^{20}C_{5}+{}^{20}C_{7}+.....+{}^{20}C_{19} = 2^{19}$ (2)
$\therefore$ (2) - (1): $-{}^{20}C_{0}+{}^{20}C_{1}-{}^{20}C_{2}+{}^{20}C_{3}+.....-{}^{20}C_{20} = 0$
$\therefore$ (2) - (1): ${}^{20}C_{7}-{}^{20}C_{8}+{}^{20}C_{9}-{}^{20}C_{10}+.....-{}^{20}C_{20} = {}^{20}C_{0}-{}^{20}C_{1}+{}^{20}C_{2}-{}^{20}C_{3}+{}^{20}C_{4}-{}^{20}C_{5}+{}^{20}C_{6}$
If we can find $${}^{20}C_{0}-{}^{20}C_{1}+{}^{20}C_{2}-{}^{20}C_{3}+{}^{20}C_{4}-{}^{20}C_{5}+{}^{20}C_{6}$$ then then the problem is solved, but finding this seemingly small sum is a Herculean task.
We can write the sum as: $$S = ({}^{20}C_{0}+{}^{20}C_{2}+{}^{20}C_{4}+{}^{20}C_{6})-({}^{20}C_{1}+{}^{20}C_{3}+{}^{20}C_{5})$$
I tried searching online but I couldn't find any easy ways to find this sum other than manually finding each ${}^{n}C_{r}$, which is painstaking and lengthy for larger $r$ without a calculator (which is how the problem is intended to be solved, by the author).
This led me to have another thought: Could we find the value of $\sum_{r=0}^n \binom{k}{2r}$? This would allow us to find the value of $S_{1} = {}^{20}C_{0}+{}^{20}C_{2}+{}^{20}C_{4}+{}^{20}C_{6}$, and with a similar logic we should also be able to find $S_{2} = {}^{20}C_{1}+{}^{20}C_{3}+{}^{20}C_{5}$.
But I have no clue how to proceed. Any input on how to solve the problem (without manually calculating the remaining $nC_{r}$, and either finding the sum $S_{1}$ or $S$ directly would be much appreciated.
PS: A similar problem was asked in a competitive exam in India (called JEE) in which students have to solve each problem within 4 minutes. I also intend to solve this within 4 minutes and simulate that environment, which is why I'm trying to expedite this final calculation (as it would probably take longer than 4 minutes to calculate and add all the ${}^{n}C_{r}$ not withstanding the time taken to deduce this sum), which might appear "nit-picky" to users who feel the problem is already solved.
