Polynomial remainder-Nightmare

Let the polynomial be $f(x)=\sum_{k=0}^{99} a_k x^k$ so that $f\left(2^{999}\right)=\sum_{k=0}^{99} a_k 2^{999k}=13^{811986}\implies \sum_{k=0}^{99} a_k\equiv 1\pmod 7$ because $2^{999}=8^{333}\equiv 1\pmod 7$ . Also $4^{3987}=8^{2658}\equiv 1\pmod 7$ . So $f\left(4^{3987}\right)=\sum_{k=0}^{99} a_k 4^{3987k}\equiv \sum_{k=0}^{99} a_k\equiv \fbox{1}\pmod 7$ and note that the degree doesn't matter.

We use the well known fact that if $a, b, n$ are integers, $n|a-b$ then $n|f (a)-f (b)$ .

Note that $4^{3987}=(4^{3})^{1329}$ . Since $7|4^{3}-1, 7|4^{3987}-1.$ Also, $7|2^{3}-1 \implies 7|2^{999}-1$ . Thus both have the same remainder when divided by 7, and $13^{811986}, f(4^{3987})$ have the same remainder too. But $13^{811986}$ is congruent to $(-1)^{811986}=1$ mod 7 so the answer is 1.