I probably have to use Euler's function but I'm not sure how.
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You basically want to find 982^40167 (mod 15). You can replace 982 with its reduction mod 15 and 40167 with its reduction mod $\phi(15) = 8$ to simplify things greatly. – Lance Sackless Jun 26 '16 at 13:23
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Thank you guys. You make everything look so easy. – Lindow Jun 26 '16 at 13:51
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You might have a look at the post: How do I compute $a^b,\bmod c$ by hand? – Martin Sleziak Jun 26 '16 at 16:41
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See http://math.stackexchange.com/questions/1828655/remainder-when-dividing-133530-with-12348/1828673#1828673 – lab bhattacharjee Jun 28 '16 at 16:15
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$$\phi(15) = 8$$ 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,
$$7^4 \equiv 4\times4 \equiv 1 \pmod{15} $$
$$7^7 \equiv 7^4 7^2 7 \equiv 1\times 4 \times 7 \equiv 13 \pmod {15}$$
Zau
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2Or, $\ {\rm mod}\ 15!:,\ 7^7 \equiv \dfrac{7^8}{7}\equiv \dfrac{1}7\equiv \dfrac{-14}7\equiv -2\ \ $ – Bill Dubuque Jun 26 '16 at 16:53
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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,
$982^{40167}\equiv13(\text{mod }15)$
Brian Cheung
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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 $13\pmod{15}$.
robjohn
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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 $982^{40167}=982^{40164}\cdot 982^3\equiv (982^4)^{10033}\cdot 7^3\equiv 1\cdot 13=13\pmod{15}$.
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$982\equiv 7\pmod{15}$ and $7^4\equiv 1\pmod{15}$ hence $$982^{40167}\equiv 7^{4\cdot10041+3}\equiv 1^{10041}\cdot7^3\equiv\color{red}{13}\pmod{15}$$
Piquito
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You can do it like that: $982^{40167}\equiv7^{40167}\equiv343^{13389}\equiv13^{13389}\equiv2197^{4463}\equiv7^{4463}\equiv7*49^{2231}\equiv13*1^{1115}\equiv13(mod 15)$
janqo
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