$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?
 A: Here is some information to achieve an alternative solution.
$3^4\equiv1\bmod4$.  $\;3^{20}\equiv1\bmod25,$ by Euler's theorem.  Therefore, $3^{20}\equiv1\bmod100$.
Therefore, $3^{1234}\equiv3^{14}\equiv9^7=(10-1)^7\equiv7\times10-1\bmod100$.
A: You can directly find the last $2$ digits of $3^{1234}$ by writing it as $3^{1232}\cdot 9=(81)^{308}\cdot 9$.
Last two digits of $(81)^{308}=(80+1)^{308}\equiv 41 \pmod{100} $. Hence $41\cdot 9 \equiv 69 \pmod{100}$.
A: The Carmichael function value for $100$, $\lambda(100)=\text{lcm}(\lambda(25),\lambda(4)) = \text{lcm}(20,2) = 20$ gives us $1234 \equiv 14 \bmod \lambda(100) \implies 3^{1234} \equiv 3^{14} \bmod 100$.
Probably the easiest way to calculate $3^{14} \bmod 100$ is as $81^3 \cdot 9$ discarding higher digits, so $81^2 \equiv 61$ and $81\cdot 61 \equiv 41$ leading to $3^{1234} \equiv 3^{14} \equiv 69 \bmod100$.
Starting alternatively from your observation that $3^{15} \equiv 7 \bmod 100$, we could find $3^{14} \equiv 7\cdot 3^{-1} \equiv 7\cdot 67 \equiv 469 \equiv 69 \bmod 100$.
A: To get things going you can calculate the inverse of $[3]_{100}$ and then try relating it to powers of $[3]_{100}$, keeping in mind that $10^2 \equiv 0 \pmod{100}$.
In $\text{modulo-}100$ you know that the inverse has the form $[10n +7]_{100}$ and solving you can write out
$\tag 1 3 \cdot 67 \equiv 1  \pmod{100}$
We can form a relation right away, suspending further ${[3^n]_{100}}$ calculations,
$\quad  3^3 \equiv 67 - 40$
and so
$\tag 2  3^4 \equiv 1 - 20 \pmod{100}$
We can now use the binomial theorem.
$\quad  \displaystyle \large 3^{1234} \equiv (3^4)^{308} \cdot 3^2 \equiv$
$\quad  \displaystyle \large  9(1-20)^{308} \equiv$
$\quad  \displaystyle \large 9(1-308 \cdot 20^1 + \Large 0  \large ) \equiv \quad \quad \normalsize \; (20^k \text{ usually has residue } 0)$
$\quad  \displaystyle \large 9 \cdot 41 \equiv 69 \pmod{100}$
