$12^{80}$ mod 143, given that $12^6 $≡ 1 mod 143

Question

Can someone explain this question? I fully understand calculating modulo but I don't get how it's calculated with power of and logical equivalence.

Answer 1

$12^{80}\equiv(12^6)^{13}\times12^2\equiv1^{13}\times12^2\equiv1\pmod {143} $

Also, alternatively:

If $ord _b a=m$ then $ord _b{a^u}=\frac{m}{(m,u)}$

$$ord_{143}12=2\implies ord_{143}12^{80}=\frac{2}{(2,80)}=1\implies 12^{80}\equiv 1\pmod{143} $$

Answer 2

$$12^{80} = 12^{(6)(13+2)} = (12^6)^{13}(12^2) = (12^6)^{13}(144)$$

Modulo $143$, this becomes:

$$((12^6)^{13})(144) \equiv (1^{13})(1) = 1\pmod{143}$$

Addendum: the "hint" is totally unnecessary.

$$12^{80} = 12^{(2)(40)} = 144^{40} \equiv 1^{40} = 1 \pmod{143}$$

Answer 3

$12^{80} = (12^{6})^{13} 12^{2} \equiv 1\times 144 \equiv 1$ (mod 143)

Answer 4

Hint: $12^{80} = (12^{6})^{13}\cdot 12^{2} $