1. The problem asks us to add two numbers written in base eight (octal). 2. The numbers given are $742_8$ and $347_8$. 3. First, convert each octal number to decimal for easier addition. 4. Convert $742_8$ to decimal: $$742_8 = 7 \times 8^2 + 4 \times 8^1 + 2 \times 8^0 = 7 \times 64 + 4 \times 8 + 2 = 448 + 32 + 2 = 482_{10}$$ 5. Convert $347_8$ to decimal: $$347_8 = 3 \times 8^2 + 4 \times 8^1 + 7 \times 8^0 = 3 \times 64 + 4 \times 8 + 7 = 192 + 32 + 7 = 231_{10}$$ 6. Add the decimal equivalents: $$482 + 231 = 713_{10}$$ 7. Convert the result back to base eight by repeated division: - $713 \div 8 = 89$ remainder $1$ - $89 \div 8 = 11$ remainder $1$ - $11 \div 8 = 1$ remainder $3$ - $1 \div 8 = 0$ remainder $1$ Reading remainders from bottom to top gives $1131_8$. 8. Therefore, the sum in base eight is: $$742_8 + 347_8 = 1131_8$$