Now answer the following questions. Evaluate: \( 141_{\text {six }}+233_{\text {six }}-102_{\text {six }} \) A. \( 412_{\text {six }} \) B. \( 405_{\text {six }} \) C. \( \quad 312_{\text {six }} \) D. \( 305_{\text {six }} \)

To evaluate \( 141_{\text{six}} + 233_{\text{six}} - 102_{\text{six}} \), we can first convert all numbers from base six to base ten, perform the operations, and then convert the result back to base six. 1. Convert \( 141_{\text{six}} \) to base ten: \[ 1 \cdot 6^2 + 4 \cdot 6^1 + 1 \cdot 6^0 = 36 + 24 + 1 = 61 \] 2. Convert \( 233_{\text{six}} \) to base ten: \[ 2 \cdot 6^2 + 3 \cdot 6^1 + 3 \cdot 6^0 = 72 + 18 + 3 = 93 \] 3. Convert \( 102_{\text{six}} \) to base ten: \[ 1 \cdot 6^2 + 0 \cdot 6^1 + 2 \cdot 6^0 = 36 + 0 + 2 = 38 \] Now, perform the calculation in base ten: \[ 61 + 93 - 38 = 116 \] Next, we convert \( 116 \) back to base six. We can do this by dividing by six and keeping track of the remainders: - \( 116 \div 6 = 19\) remainder \(2\) - \( 19 \div 6 = 3\) remainder \(1\) - \( 3 \div 6 = 0\) remainder \(3\) Reading the remainders from last to first gives us: \( 312_{\text{six}} \) Thus, the answer is: C. \( \quad 312_{\text{six}} \)