how to calculate （1987^718） mod 60

Question

My attempt: $1987=60×33+7$，$\phi（60）=16$，so $7^{16}≡1 \bmod60$, $1987^{718} ≡ 7^{718} ≡ 7^{(16×44+14)} ≡ 7^{14} \bmod60$, then I have no idea how to solve it.

What do you think about it? Could you please show me?

Regards

Answer 1

Once you've reduced it to $7^{14} \pmod {60}$, note that $7^2 \equiv 49 \equiv -11$, and so $$7^4 \equiv (7^2)^2 \equiv (-11)^2 \equiv 121 \equiv 1 \pmod {60}.$$ So $7^{14} \equiv (7^4)^3 \cdot 7^2 =\equiv 1^3 \cdot 49 \equiv 49 \pmod{60}$.