Remainders Revisited #2

$65^{37^{80}}\equiv 65^1 \pmod{3}$ by Fermat. This holds modulo 13 as well so $65^{37^{80}}\equiv 65\equiv \boxed{26}\pmod{39}$ .

$\text{}65$ $\equiv$ $\text{}26~{\pmod{39}}$

Being lambda of $39 = 12$ $37 \equiv 1 \pmod{12}$ so $65^{37^{80}} \equiv 65 \pmod{39}$ and $65 \equiv 26 \pmod{39}$ . Thus the solution is $\boxed{26}$