Finding the Last Digit of $\text{8}^{\text{25}}$

To find the last digit of a large power like $\text{8}^{\text{25}}$, we can look at the pattern of the last digits of the first few powers of 8.

Let's calculate the first few powers of 8 and observe their last digits:

• $\text{8}^{\text{1}} = 8$. The last digit is 8.
• $\text{8}^{\text{2}} = 8 \times 8 = 64$. The last digit is 4.
• $\text{8}^{\text{3}} = 64 \times 8 = 512$. The last digit is 2.
• $\text{8}^{\text{4}} = 512 \times 8 = 4096$. The last digit is 6.
• $\text{8}^{\text{5}} = 4096 \times 8 = 32768$. The last digit is 8.

We can see a repeating pattern in the last digits: 8, 4, 2, 6, 8, ...

The pattern of the last digits of powers of 8 is 8, 4, 2, 6. This sequence repeats every 4 powers. This is called the cyclicity of the last digit.

The cyclicity of the last digit of powers of 8 is 4.

To find the last digit of $\text{8}^{\text{25}}$, we need to determine where the 25th power falls in this repeating cycle. We can do this by dividing the exponent, which is 25, by the length of the cycle, which is 4.

Divide 25 by 4:

$\text{25} \div \text{4} = \text{6}$ with a remainder of $\text{1}$.

The remainder tells us which position in the cycle the last digit will be. If the remainder is 1, it's the first digit in the cycle. If the remainder is 2, it's the second digit, and so on. A remainder of 0 (or 4 in a cycle of 4) corresponds to the last digit in the cycle.

In this case, the remainder is 1.

The pattern is (1st: 8, 2nd: 4, 3rd: 2, 4th: 6).

Since the remainder is 1, the last digit of $\text{8}^{\text{25}}$ will be the first digit in the cycle.

The first digit in the cycle (8, 4, 2, 6) is 8.

Therefore, the last digit of $\text{8}^{\text{25}}$ is 8.