How do I find a remainder using congruence?

Question

I'm asked to find the remainder of $305^{305}$ when it is divided by $42$.

My progress:

$305 \equiv 11 $ (mod 42) so $305^{305} -> 11^{305}$

$11^1 \equiv 11 $ (mod $42$)
$11^2 \equiv 121 \equiv 37$ (mod $42$)
$11^3 \equiv (11^2)(11) \equiv (37)(11) \equiv 29$ (mod $42$)

so this is where I'm having problems. It seems like I can't narrow my congruence to just $11^n \equiv 1$ (mod $42$) to state that $11^{nk} \equiv 1$ (mod $42$) where $k \in Z$ so I can easily substitute that into $11^{305} \equiv 11^{nk} + 11^j \equiv 1 + 11^j $ (mod $42$) where $j \in Z$, and complete the solution.

Answer 1

$305^{305}\equiv 11^{305}\equiv 121^{152}\times 11\equiv (-5)^{152}\times 11\equiv 625^{38}\times 11\\ \equiv (-5)^{38}\times 11\equiv 625^{9}\times 25\times 11\equiv(-5)^{9}\times 25\times 11\equiv(625)^{2}\times 25\times 25\times 11\\ -5\times -5 \times 11=23 \pmod {42}$

Answer 2

By Euler's generalization of Fermat'S little theorem, $x^{\varphi(42)}\equiv 1(\text{ mod } 42)$, where $\varphi$ is the Euler phi function. Hence anything to the $\varphi(42)= (2-1)(3-1)(7-1)=12$ is $1$.

Answer 3

Since Carmichael's function $\lambda(42)={\rm lcm}(6,2,1)=6$, and $\gcd(11,42)=1)$, we know that you will soon find for $k=6$ that $11^k\equiv 1 \bmod 42$. Just to check that:

$$11^3\equiv 29\bmod 42 \qquad \text{ so }\quad 11^6\equiv 29^2 \equiv 841 \equiv 1 \bmod 42$$

as expected. Then obviously $305 = 6\cdot50 +5$ and $11^{305}\equiv 11^5 \bmod 42$. Finally $$305^{305}\equiv 11^5 \equiv 37\cdot 29 \equiv -5\cdot-13 \equiv 65\equiv \fbox{23} \bmod 42$$

Answer 4

As $42=2\cdot3\cdot7$

$305\equiv1\pmod2\implies305^{305}\equiv1\equiv-1$

$305\equiv-1\pmod3\implies305^{305}\equiv(-1)^{305}\equiv-1$

$\implies305^{305}\equiv-1\pmod{\text{lcm}(2,3)}\ \ \ \ (1)$

$305\equiv-3\pmod7,305^{305}\equiv(-3)^{305}$

Now as $(-3)^3\equiv1$ and as $305\equiv2\pmod3$

$(-3)^{305}\equiv(-3)^2\equiv2\pmod7\ \ \ \ (2)$

Apply CRT on $(1),(2)$