<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] 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 $41\cdot 9 \equiv 69 \pmod{100}$.
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