Find $a$ such that $83359654581 ... 10843035531 ^ a \equiv 1 \pmod{32003240081}$ 
Find $a$ such that
$83359654581036155008716649031639683153293510843035531 ^ a \equiv 1 \pmod{32003240081}$

I am quite familiar with sort of equation like find m such that m^d = c (mod n), but I think it don't help me in this problem since it relative exponent.
I have read this, but in my problem, the number is so big.
 A: Let $n = 83359654581036155008716649031639683153293510843035531$.
Let $m = 32003240081$. It has the prime factorization $m = pq$ where $p = 160009, q = 200009$.
One can verify $\gcd(n,p) = \gcd(n,q) = 1$. By Fermat's little theorem, we have
$$
\begin{cases}
n^{p-1} \equiv 1 \pmod p\\
n^{q-1} \equiv 1 \pmod q
\end{cases}
\quad\implies\quad
\begin{cases}
n^\ell \equiv 1 \pmod p\\
n^\ell \equiv 1 \pmod q
\end{cases}
\quad\text{ where }\quad
\ell = {\tt lcm}(p-1,q-1)
$$
Together with Chinese Remainder theorem, this leads to
$$n^\ell \equiv 1 \pmod {pq}$$
Since we only want an $a$, not the smallest one, $\ell = 
{\tt lcm}(p-1,q-1) = 4000360008$ is a solution.
A: Suppose $n$ and $k$ are relatively prime. Then $n^{\varphi(k)}\equiv 1 \bmod k$. The same is true if we change $\varphi(k)$ for any multiple of $\varphi(k)$. For example $k!$.
We therefore have that $n^{k!}\equiv 1 \bmod k$ whenever $k$ and $n$ are relatively prime.
So all you have to do is check that those numbers are relatively prime, you can do that really fast with the euclidean algorithm (on a computer of course).
