Calculating $-6^{20} \cdot 5^{70} \pmod {131}$

Question

Using Fermat's little theorem:

$(-6^{20} \cdot 5^{70})^{130}\equiv 1 \pmod{131}$

$-6^{2600} \cdot 5^{9100} \equiv 1 \pmod{131}$

I'm a bit stuck on where to go after this since I haven't really dealt with these sort of situations before however I feel like the answer is pretty simple to find from this step. All i'm asking for is a point in the right direction.

Possible duplicate of How do I compute $a^b,\bmod c$ by hand? — Martin Sleziak, Jun 14 '18 at 13:31

score 3 · Accepted Answer · answered Jun 13 '18 at 14:40

3

Use the fact that $5^3 \equiv - 6 \pmod{131}$. Then you have

$$-6^{20}\cdot 5^{70} \equiv (5^3)^{20}\cdot 5^{70} \equiv 5^{130} \equiv 1 \pmod{131}$$

answered Jun 13 '18 at 14:40

Stefan4024

35,843

Calculating $-6^{20} \cdot 5^{70} \pmod {131}$

1 Answers1