Congruence 10 modulo 7

Question

I'm reading some proofs for divisibility rules and see that $10 \equiv -1 \pmod{11}$, then $10^2 \equiv 1 \pmod{11}$ and etc., it's clear for me, but I can't understand the situation with modulo 7. If we have $10 \equiv 3 \pmod{7}$ how can we get the rest of the values? They are $10^2 \equiv 2 \pmod{7}$, $10^3 \equiv -1 \pmod{7}$ and etc.

I'm sorry if it's a stupid question, but I want to figure out this very well.

Answer 1

As a start, we use the rule that $$a \equiv b \pmod{n} \Rightarrow a^k \equiv b^k \pmod{n},\forall k\in \mathbb{N} \tag{1}$$ Hint $$10 \equiv 3 \pmod{7} \Rightarrow 10^2 \equiv 3^2 \pmod{7} \Rightarrow \\ 10^2 \equiv 9\equiv 7+2\equiv 2 \pmod{7}$$ or $$10^3 \equiv 3^3 \pmod{7} \Rightarrow \\ 10^3 \equiv 27\equiv 28-1\equiv 4\cdot7-1\equiv -1 \pmod{7}$$

Answer 2

• $10\equiv3\pmod7\implies10^2\equiv3^2\equiv2\pmod7$;
• $10^2\equiv2\pmod7\implies10^3=10^2\times10\equiv2\times3=6\pmod7$;
• $10^3\equiv6\pmod7\implies10^4=10^3\times10\equiv6\times3\equiv4\pmod7$

and so on …