A simple modular arithmetic problem

Since $\gcd(5,177) = 1$ , we can use Euler's theorem . The Euler's totient function $\phi(117) = 117 \times \dfrac 23 \times \frac{12}{13} = 72$ . Then we have:

$\begin{aligned} 5^{5^{50}} & \equiv 5^{5^{20} \bmod \phi(117)} \text{ (mod 117)} \\ & \equiv 5^{5^{50} \bmod 72} \text{ (mod 117)} & \small \blue{\text{Again }\gcd(5,72)=1 \text{, Euler's theorem applies.}} \\ & \equiv 5^{5^{50 \bmod \phi(72)} \bmod 72} \text{ (mod 117)} & \small \blue{\text{and }\phi(72) = 72 \times \frac 12 \times \frac 23 = 24} \\ & \equiv 5^{5^{50 \bmod 24} \bmod 72} \text{ (mod 117)} \\ & \equiv 5^{5^2} \equiv 5^{25} \text{ (mod 117)} \\ & \equiv 5 \times 125^8 \text{ (mod 117)} \\ & \equiv 5(117+8)^8 \text{ (mod 117)} \\ & \equiv 5 (8^8) \equiv 5(8)(128^3) \text{ (mod 117)} \\ & \equiv 5(8)(117+11)^3 \equiv 5(8)(11^3) \text{ (mod 117)} \\ & \equiv 5(8)(11)(121) \equiv 5(8)(11)(4) \text{ (mod 117)} \\ & \equiv 8(220) \equiv 8(-14) \text{ (mod 117)} \\ & \equiv -112 \equiv \boxed 5 \text{ (mod 117)} \end{aligned}$

First of all, $117$ is not prime. In fact, $117=3^2\times 13$ .

We could use the Euler's Theorem to solve it.

The theorem says that given an integer $n$ , for any integer $a$ relatively prime to $n$ we have that $a^{\phi(n)} \equiv 1 (\text{mod n})$ , where $\phi(n)$ is the Euler's Totient Function .

We have that $\phi(117)=117\left(1-\frac{1}{3}\right)\left(1-\frac{1}{13}\right)=72$ .

Then $5^{72} \equiv 1 (\text{mod 117})$ , since $\text{gcd}(5,117)=1$ .

We can use the theorem again to find $5^{50} (\text{mod 72})$ . This is important because a consequence of this theorem is that $5$ to the power of any multiple of $72$ will give us a remainder of $1$ modulo $117$ .

We have that $72=2^3\times 3^2$ and then $\phi(72)=72\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=24$ .

Then $5^{24} \equiv 1 (\text{mod 72})$ , since $\text{gcd}(5,72)=1$ .

To find $5^{50} (\text{mod 72})$ we could use the fact that $5^{24}\cdot 5^{24} = 5^{48}\equiv 1 (\text{mod 72})$ .

Hence, $5^{48}\cdot 5^{2} = 5^{50}\equiv 25 (\text{mod 72})$ .

It means that $5^{50}-25$ is a multiple of $72$ .

This gives us that $5^{5^{50}-25}\equiv 1 (\text{mod 117})$ . Hence, $5^{5^{50}}\equiv 5^{25} (\text{mod 117})$ .

We can easily find $5^{25} (\text{mod 117})$ knowing that $5^{72} \cdot 5^{3} = 5^{75} = \left(5^{25}\right)^{3} \equiv 5^{3} (\text{mod 117})$ .

Finally, $5^{25} \equiv 5 (\text{mod 117})$ .

In the end we have that $5^{5^{50}}\equiv 5 (\text{mod 117})$ , and that is our answer.

If anyone has a simpler solution, I will appreciate to know!