Find the last digit of the exponent $x$.

Question

Let \begin{align} p&=396543857870745963499374527519378569849832249490600276007703072957912\cdots\\ &\phantom{=}8049490077183813353745228056691 \end{align}

This number is a 100-digit prime number and 2 is a primitive root modulo $p$. Let $x$ be the unique positive integer with $1 \leq x \leq p-1$ so that $2^x \equiv 5 \pmod{p}. $

What is the last digit of $x$?

Answer 1

We can work this out with a little help from any program capable of computing with large integers (I used Maple and Haskell (ghci)):

It turns out that $p-1=10q$ where $q$ is prime. Also, it turns out that $5^{5q}\equiv1\pmod p$, i.e., $2^{5qx}\equiv1\pmod p$. Since $2$ is a primitive root, this implies $5qx\equiv0\pmod{10q}$, and hence $x$ is even.

Edit the second: A bit more playing around (on a hunch) reveals that $2^{2q}\equiv5^{4q}\pmod{p}$, i.e., $2^{2q}\equiv2^{4qx}\pmod{p}$, and therefore $2q\equiv4qx\pmod{10q}$, i.e., $1\equiv2x\pmod5$. Since $x$ is even, we conclude $x\equiv8\pmod{10}$, i.e., the last digit of $x$ is $8$.