To find the unit digit of $(137^{13})^{47}$, we can simplify this problem by focusing only on the unit digits. Here’s a step-by-step solution:

1. Identify the Unit Digit of the Base: The unit digit of $137$ is $7$. So, we only need to consider the powers of $7$.

2. Determine the Pattern in Powers of $7$: The unit digits of powers of $7$ follow a repeating cycle. Let's examine the first few powers:

   • $7^1 = 7$, so the unit digit is $7$.
   • $7^2 = 49$, so the unit digit is $9$.
   • $7^3 = 343$, so the unit digit is $3$.
   • $7^4 = 2401$, so the unit digit is $1$.

   Thus, the unit digits of powers of $7$ cycle every 4 terms: $7, 9, 3, 1$.

3. Reduce the Power Modulo 4: We now need to find the unit digit of $7^{13 \times 47}$. Simplify $13 \times 47$ modulo 4, since the cycle length is 4.

   $13 \times 47 = 611$

   Now, find $611 \mod 4$: $611 \div 4 = 152\text{ remainder }3$ Therefore, $611 \equiv 3 \mod 4$.

4. Determine the Unit Digit: Since $611 \equiv 3 \mod 4$, the unit digit of $7^{611}$ will be the same as the unit digit of $7^3$, which is $3$.

Answer: The unit digit of $(137^{13})^{47}$ is 3.

Would you like more details or have any questions?

Here are some related questions for practice:

1. What is the unit digit of $27^{53}$?
2. How would the unit digit change if the exponent was even?
3. Can you explain why cycles occur in powers of integers?
4. What if the base number was $1377$ instead of $137$?
5. How do you find unit digits of expressions like $(125^5)^{10}$?

Tip: Finding unit digits often involves identifying cycles in the powers of single-digit bases.