Find the remainder of $S = \sum_{i=0}^{99} (n+i)^6 + 2^{2^{2558}} + 1$ divided by $100$. If $n$ is a positive integer, find the remainder of the following number divided by $100$:
$$S = \sum_{i=0}^{99} (n+i)^6 + 2^{2^{2558}} + 1$$
I've wrote a program using logarithmic exponentiation and big integers to find that $2^{2^{2558}} \equiv 16 \pmod{100}$. But I don't know what can I do with $\sum_{i=0}^{99} (n+i)^6$. Can you help me continue the problem, please? Thanks!
 A: If you know how to calculate the least universal exponent or Carmichael function $\lambda$, you know that $\lambda(100)=20$ and $\lambda(20)=4$ and thus
$2^{\large 2^{2558}}\equiv 2^{\large (2^{2558} \bmod 20)}\equiv 2^{\large (2^{2558 \bmod 4} \bmod 20)}\equiv 2^{\large (2^2\bmod 20)}\equiv 2^4 \equiv 16 \bmod 100$
Note that the $(n+i)$ term will take on every residue value $\bmod 100$ exactly once, no matter what the value of $n$ is. So there is only one answer, rather than a function based on $n$, and we can simply consider
$$S\equiv \left (\sum_{i=0}^{99} i^6 + 16 + 1 \right )\bmod 100$$
By reviewing the expansion it's clear that $i^6\equiv (50-i)^6\equiv (50+i)^6 \equiv (100-i)^6\bmod 100$ . Then 
$$S\equiv\left( 4\sum_{i=1}^{24} i^6  + 2\cdot 25^6 + 17 \right )\bmod 100$$
Clearly $25^k\equiv 25 \bmod 100$ for $k>0$. Now there's probably something smart to do with that remaining sum, but I can't see it it immediately. It's pretty clear to me from considering primitive roots that the sum of all sixth powers $\bmod 100$ will be the same as the sum of all squares, but a little hard to prove briefly. Nevertheless I enlisted the help of a spreadsheet to show
$$\sum_{i=1}^{24} i^6 \equiv 0 \bmod 100$$
Thus 
$$S\equiv\left( 4\cdot 0  + 2\cdot 25 + 17 \right ) \equiv 67 \bmod 100$$
