Finding the Unit Digit of 3 to the Power of 99

To find the unit digit of a number raised to a large power, we need to observe the pattern of the unit digits of the base number when raised to successive powers.

Let's look at the unit digits of the first few powers of 3:

• $\text{3}^1 = 3$ (Unit digit is 3)
• $\text{3}^2 = 9$ (Unit digit is 9)
• $\text{3}^3 = 27$ (Unit digit is 7)
• $\text{3}^4 = 81$ (Unit digit is 1)
• $\text{3}^5 = 243$ (Unit digit is 3)
• $\text{3}^6 = 729$ (Unit digit is 9)

We can see a repeating pattern in the unit digits: 3, 9, 7, 1. This cycle of unit digits has a length of 4.

The unit digit of $\text{3}^{\text{n}}$ depends on the remainder when the exponent 'n' is divided by 4. The position in the cycle (3, 9, 7, 1) corresponds to the remainder:

• If the remainder is 1, the unit digit is 3 (like $\text{3}^1$).
• If the remainder is 2, the unit digit is 9 (like $\text{3}^2$).
• If the remainder is 3, the unit digit is 7 (like $\text{3}^3$).
• If the remainder is 0 (or the exponent is a multiple of 4), the unit digit is 1 (like $\text{3}^4$).

In this problem, the exponent is 99. We need to find the remainder when 99 is divided by 4.

Divide 99 by 4:

\( \frac{99}{4} \)

We perform the division:

\( 99 = 4 \times 24 + 3 \)

The quotient is 24 and the remainder is 3.

Since the remainder is 3, the unit digit of $\text{3}^{99}$ will be the same as the unit digit of $\text{3}^3$, which is 7.

Therefore, the digit in the unit place of $\text{3}^{99}$ is 7.