How to find the reminder when $982^{40167}$ is divided by 15? I probably have to use Euler's function but I'm not sure how.
 A: $$\phi(15) = 8$$ which means $$a^8 \equiv 1\pmod {15}$$
$$ 40167 = 7 + 5020 \times 8$$
And $$982 = 65 \times 15 + 7$$
$$982^{40167} \equiv 7^{7} \equiv 13 \pmod {15}$$
EDIT: one may figure out how to calculate $7^{7}$
The story is really simple:
$$7^2 \equiv 49 \equiv 4 \pmod{15} $$
Therefore, 
$$7^4 \equiv 4\times4 \equiv 1 \pmod{15} $$
$$7^7 \equiv 7^4 7^2 7 \equiv 1\times 4 \times 7 \equiv 13 \pmod {15}$$
A: One of the standard approaches is to find the remainder when divided by $3$ and $5$ respectively, then use Chinese Remainder Theorem.
$982^{40167}\equiv1^{40167}\equiv1(\text{mod }3)$
$982^{40167}\equiv2^{40167}\equiv2^3\equiv3(\text{mod }5)$ (by the cyclic property $2^4\equiv1(\text{mod }5)$)
Now by Chinese Remainder Theorem or simply some trial and error,
$982^{40167}\equiv13(\text{mod }15)$
A: One way to handle this is to look mod $3$ and mod $5$ and use the Chinese Remainder Theorem.
To simplify, we can reduce mod $3$ and mod $5$ and apply Little Fermat:
$$
982^{40167}\equiv1^{40167}\equiv1\pmod{3}
$$
and
$$
982^{40167}\equiv2^{40167}\equiv2^3\equiv3\pmod{5}
$$
The solution is $13\pmod{15}$.
A: One way is to say that $982\equiv 7\pmod{15}$, so $982^4\equiv 7^4\equiv (7^2)^2\equiv 4^2\equiv 1\pmod{15}$.
That means that $982^{40167}=982^{40164}\cdot 982^3\equiv (982^4)^{10033}\cdot 7^3\equiv 1\cdot 13=13\pmod{15}$.
A: $982\equiv 7\pmod{15}$ and $7^4\equiv 1\pmod{15}$ hence $$982^{40167}\equiv 7^{4\cdot10041+3}\equiv 1^{10041}\cdot7^3\equiv\color{red}{13}\pmod{15}$$
A: You can do it like that:
$982^{40167}\equiv7^{40167}\equiv343^{13389}\equiv13^{13389}\equiv2197^{4463}\equiv7^{4463}\equiv7*49^{2231}\equiv13*1^{1115}\equiv13(mod  15)$ 
