Large Exponent Modular Arithmetic $6^{1939}\equiv x\mod22$

Question

I haven't seen this particular question, and many other questions I've looked ad required a use of a theorem that I don't think applies here.

Just a bit confused as to where to start.

Thanks.

Answer 1

Try to find a cycle in the sequence of normalized representers of $6^k$ for $k=1,2,3,\ldots$. You will see that $6^{11} \equiv 6 \mod 22$. Now you can reduce your work by exploiting this property: $6^{1331+608} \equiv 6^{11^3}6^{608} \equiv 6^{{{11}^{11}}^{11}}6^{608} \equiv 6^{{{11}^{11}}}6^{608}\equiv 6^{{{11}}}6^{608}\equiv 6\cdot 6^{608} \mod 22$. Now proceed on $6^{608}$ and so on.

Answer 2

1. Do it mod $2$.
2. Do it mod $11$ using Fermat's little theorem.
3. Put them together.

Answer 3

Carmichael's function gives $\lambda(22)= {\rm lcm}(\lambda(11),\lambda(2)) = \color{blue}{10}$ (the same as $\phi(22)$ on this occasion) as the value which all exponential cycles will divide, so $6^{1939}\equiv 6^9 \bmod 22$.

Then calculate.

Answer 4

In fact you can summarize the process since $\varphi(22)=10$. Then the exponent mod totient $\Rightarrow 1939\equiv 9 \mod \varphi(22)$ (the last digit of the exponent).

Then the answer is given evaluating: $6^9 \equiv 2 \pmod{22}$. In other cases is better to factorize the dividend and modulus, then use the modular arithmetic properties and put all together.

Answer 5

As $(6,22)=2$

let us find $6^{1939-1}\pmod{11}$

As $(6,11)=1$ and $\lambda(11)=\phi(11)=10,1938\equiv8\pmod{10},$

$6^{1939-1}\equiv6^8\pmod{11}$

Now $6^2\equiv3\pmod{11},6^8=(6^2)^4\equiv3^4\equiv4\pmod{11}$

$\implies6^{1938}\equiv4\pmod{11}$

$\implies6^{1938+1}\equiv4\cdot6\pmod{11\cdot6}\equiv ?\pmod{22}$

Answer 6

$6^{\large 1939}\!\bmod 22\, =\, 2\,\underbrace{(3\cdot 6^{\large 1938}\!\bmod{11})}_{\dfrac{3}{6^{\large 2}}\ \equiv\ \dfrac{1}{12}\ \equiv\ \dfrac{1}{1}} =\, 2,\, $ using $\ 6^{\large 10}\equiv 1\pmod{11}\,$ by Fermat