<s>[INST] $3^{1234}$ 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?
 [/INST] 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.
$\quad  \displaystyle \large 3^{1234} \equiv (3^4)^{308} \cdot 3^2 \equiv$
$\quad  \displaystyle \large  9(1-20)^{308} \equiv$
$\quad  \displaystyle \large 9(1-308 \cdot 20^1 + \Large 0  \large ) \equiv \quad \quad \normalsize \; (20^k \text{ usually has residue } 0)$
$\quad  \displaystyle \large 9 \cdot 41 \equiv 69 \pmod{100}$
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