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So, I've figured out that: $12 \equiv 2 \mod 10$

And hence,

$$ 12^{157} \equiv 2^{157} \mod 10$$

I thought of expanding out $2^{157} = 2^{156} \cdot 2 = 4^{78} \cdot 2 = 16^{39} \cdot 2$, then I could write: $$ 12^{157} \equiv 2^{157} \equiv 16^{39} \cdot 2 \equiv 6^{39} \cdot 2 \equiv 36^{19}\cdot 12\equiv 6^{19} \cdot 2 \equiv 36^9 \cdot 12 \equiv 6^9 \cdot 2 \equiv 36^4 \cdot 12 \equiv 6^4 \cdot 2 \equiv 72 \cdot 36 \equiv 2 \cdot 6 \equiv 2 $$

This is quite a long process. I found that you could use fermats little theorem to speed up the case when the number we taking mod with is prime, however are there any theorems/methods which can be used to speed up calculations when we take a composite numbers?

Theo Bendit
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tryst with freedom
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1 Answers1

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You could use Fermat`s Little Theorem and Chinese Remainder Theorem together.

Note that

$12^{157} \equiv 0 \pmod{2} \Rightarrow 12^{157}=2k$

$12^4 \equiv 1 \pmod{5} \Rightarrow 12^{157} \equiv 2 \pmod{5}$

$2k \equiv 2\pmod{5} \Rightarrow k \equiv 1 \pmod{5}$

$k \equiv 1 \pmod{5} \Rightarrow k=5x+1 $ so $12^{157}=2k=10x+2 \equiv 2 \pmod{10}$

Nevzat Eren Akkaya
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