What? Last digit of $\pi$ ?

The power cycle of $3^n$ is the following when $n$ is of the form $(\green{4}\orange{k} + 1), (\green{4}\orange{k} + 2), (\green{4}\orange{k}+3), (\green{4}\orange{k})$ :

$\begin{array}{c}&3 &\rightarrow &9 &\rightarrow &7 &\rightarrow &1 \\ &(\green{4}\orange{k}+1) &&(\green{4}\orange{k}+2) &&(\green{4}\orange{k}+3) &&(\green{4}\orange{k}) \end{array}$

Then, let $n$ be $14^{159}$ , then clearly

$\begin{aligned} 14^{159} &= \green{4} \times \orange{(14^{157} \times 7^2)} \\ &= \green{4}\orange{k} \end{aligned}$

Hence, the ending digit of $3^{14^{159}}$ will be:

$\boxed{1}$

We need to find $3^{14^{159}} \bmod 10$ .

Solution 1: By binomial expansion

$\begin{aligned} 3^{14^{159}} & \equiv (3^2)^{2^{158}\times 7^{159}} \text{ (mod 10)} \\ & \equiv 9^{2^{158}\times 7^{159}} \text{ (mod 10)} \\ & \equiv (10-1)^{2^{158}\times 7^{159}} \text{ (mod 10)} \\ & \equiv (-1)^\blue{2^{158}\times 7^{159}} \text{ (mod 10)} & \small \blue{\text{The exponent }2^{158}\times 7^{159} \text{ is even}} \\ & \equiv \boxed 1 \text{ (mod 10)} \end{aligned}$

Solution 2: By Euler's theorem . Since $\gcd(3,10) = 1$ , we can apply Euler's theorem as follows. Note that Euler's totient function $\phi(10) = 10 \times \frac 12 \times \frac 45 = 4$ .

$\begin{aligned} 3^{14^{159}} & \equiv 3^{14^{159} \bmod \phi(10)} \text{ (mod 10)} \\ & \equiv 3^{14^{159} \bmod 4} \text{ (mod 10)} \\ & \equiv 3^0 \text{ (mod 10)} \\ & \equiv \boxed 1 \text{ (mod 10)} \end{aligned}$

$3^{14^{159}}=3^{14\times 14^{158}}=(4782969)^{14^{158}}$ . Last digit of $9^{2n}$ is $1$ for any non-negative integer $n$ . So the required last digit is $\boxed 1$ .

Of course, one way to do this problem is to actually multiply everything (and break your calculator in the process), however, the easier way is to figure out the pattern in the last digits of the exponents.

You can break this problem down into two main parts.

First, find the last digit of $3^{14}$ .

The last digits of powers of $3$ go in a cycle of four digits: $9$ , $7$ , $1$ , and $3$ .

$3^2 = \textbf{9}, 3^3 = 2\textbf{7}, 3^4 = 8\textbf{1}, 3^5 = 24\textbf{3}, ...$

$14$ divided by $4$ is $3$ with the remainder $2$ , which means that the last digit of $3^{14}$ will be $7$ .

Now, to find the answer to the actual problem.

You don't have to find the last digits of the powers of $3^{14}$ because is essentially the same as finding the last digits of powers of $7$ .

The last digits of powers of $7$ also go in a cycle of four digits: $9$ , $3$ , $1$ , and $7$ .

$7^2 = 4\textbf{9}, 7^3 = 34\textbf{3}, 7^4 = 240\textbf{1}, 7^5 = 1680\textbf{7}, ...$

$159$ divided by $4$ is $39$ with the remainder $3$ , which means that the last digit of $\displaystyle (3^{14})^{159}$ will be $\boxed{1}$ .

Last digit of 14 = 4 so 3^4 = 8 1 Last digit of 159 = 9 So 1^9 = 1 is the last number