Find the least positive integer $n$ such that when $3^n$ is written in base $143$, its two right-most digits in base $143$ are $01$.
Note that the given condition is equivalent to $3^n \equiv 1 \pmod{143^2}$ and $143=11\cdot 13$. Because $\gcd(11^2, 13^2) = 1$, the desired condition is equivalent to $3^n \equiv 1 \pmod{121}$ and $3^n \equiv 1 \pmod{169}$.
If $3^n \equiv 1 \pmod{121}$, one can see the sequence $1, 3, 9, 27, 81, 1, 3, 9...$ so $5|n$.
Now if $3^n \equiv 1 \pmod{169}$, it is harder. But we do observe that $3^3 \equiv 1 \pmod{13}$, therefore $3^3 = 13a + 1$ for some integer $a$. So our goal is to find the first number $p_1$ such that $(13a+1)^ {p_1} \equiv 1 \pmod{169}$. Then, $p_1 \equiv 0 \pmod{13},$ which follows from the binomial theorem. It is not difficult to see that the smallest $p_1=13$, so ultimately $3^{39} \equiv 1 \pmod{169}$. Therefore, $39|n$.
The first $n$ satisfying both criteria is thus $5\cdot 39=\boxed{195}$.
-expiLnCalc