Consider the following product. \[ 23_{\text {five }} \cdot 33_{\text {five }} \] Determine the product using base 5 . \( 23_{\text {five }} \cdot 33_{\text {five }}=\square \) five

To multiply \( 23_{\text{five}} \) and \( 33_{\text{five}} \) in base 5, we first convert each number to base 10. \( 23_{\text{five}} = 2 \times 5^1 + 3 \times 5^0 = 10 + 3 = 13_{\text{ten}} \) \( 33_{\text{five}} = 3 \times 5^1 + 3 \times 5^0 = 15 + 3 = 18_{\text{ten}} \) Next, we multiply these two numbers together in base 10: \( 13 \cdot 18 = 234_{\text{ten}} \) Now, we convert \( 234_{\text{ten}} \) back to base 5. We repeatedly divide by 5 and keep track of the remainders: 1. \( 234 \div 5 = 46 \) remainder \( 4 \) 2. \( 46 \div 5 = 9 \) remainder \( 1 \) 3. \( 9 \div 5 = 1 \) remainder \( 4 \) 4. \( 1 \div 5 = 0 \) remainder \( 1 \) Reading the remainders from bottom to top, we find that \( 234_{\text{ten}} = 1414_{\text{five}} \). Therefore, the product is: \[ 23_{\text{five}} \cdot 33_{\text{five}} = 1414_{\text{five}} \]