What is $65^{3903}$ $(\text{mod }81)$?

Question

How do you find $65^{3903}$ $(\text{mod }81)$? I found that $65^{3903} \equiv 2 \ \text{ (mod } 3)$ with Fermat's Little Theorem. Is that the right first step? I don't know how I would apply the Chinese remainder theorem here.

Answer 1

$65^{3903}\equiv(-16)^{3903}\equiv-2^{4\times3903}\equiv-2^6\equiv-64\equiv17\bmod81$

since $2^{54}\equiv81$ by Euler's theorem (a generalization of Fermat's, for composite numbers),

and $4\times3903=54\times289+6$.

The Chinese remainder theorem doesn't apply,

since $81$ doesn't have non-trivial factors that are relatively prime.

Answer 2

We can't apply FLT since $81$ is not prime.

We can proceed by steps as follows

$$65\equiv -16 \implies \color{blue}{65^2\equiv 256 \equiv 13} \implies 65^4 \equiv7\mod 81$$

$$\implies 65^6 \equiv 91 \equiv10\implies 65^7 \equiv-160 \equiv2 \implies \color{blue}{65^{56}\equiv 13 \mod 81}$$

therefore

$$65^{54}\equiv 1 \mod 81 \implies 65^{3903} \equiv 65^{15} \equiv 17 \mod 81$$

Answer 3

Apply $a^{\phi(n)} = 1 \pmod n$ by Fermat's little theorem, as generalised by Euler, provided $\gcd(a,n)=1$, which is indeed the case for $a=65$ and $n=81$. Of course $\phi(81)=3^4(1-\frac13)= 3^3\times 2 = 54$, so $$65^{54} = 1 \pmod{81}$$

Now $3903 = 15 + 72\times54$ so

$$65^{3903} = 65^{15} \pmod{81}$$

Now apply some squaring/multiplying to see that this equals $17$.

Answer 4

You probably need to ramp up Fermat's Theorem to Euler's Theorem. You have $\phi(81) = 54$, and $3903 \equiv 15 \pmod{54}$, so your problem reduces to $65^{15} \pmod{81}$.

Second, you might notice that $65 \equiv -16 \equiv -2^4 \pmod{81}$, so you have $-2^{4\cdot15} \equiv -2^{60} \equiv -2^6 \equiv -64 \equiv 17 \pmod{81}$ since $60 \equiv 6 \pmod{54}.$

Anyway, Euler's Theorem.