Subjects number systems

Octal Subtraction 74B088

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Search Solutions

Octal Subtraction 74B088


1. **Problem Statement:** Evaluate the following subtractions in base 8 (octal): a) $424_8 - 373_8$ b) $5641_8 - 4234_8$ c) $13523_8 - 6550_8$ d) $1134_8 - 762_8$ 2. **Important Note:** To subtract numbers in base 8, convert them to base 10, perform the subtraction, then convert the result back to base 8. --- ### a) $424_8 - 373_8$ - Convert to decimal: - $424_8 = 4 \times 8^2 + 2 \times 8^1 + 4 \times 8^0 = 4 \times 64 + 2 \times 8 + 4 = 256 + 16 + 4 = 276$ - $373_8 = 3 \times 64 + 7 \times 8 + 3 = 192 + 56 + 3 = 251$ - Subtract in decimal: $276 - 251 = 25$ - Convert back to octal: - $25 \div 8 = 3$ remainder $1$ - $3 \div 8 = 0$ remainder $3$ - So, $25_{10} = 31_8$ ### b) $5641_8 - 4234_8$ - Convert to decimal: - $5641_8 = 5 \times 8^3 + 6 \times 8^2 + 4 \times 8^1 + 1 = 5 \times 512 + 6 \times 64 + 4 \times 8 + 1 = 2560 + 384 + 32 + 1 = 2977$ - $4234_8 = 4 \times 512 + 2 \times 64 + 3 \times 8 + 4 = 2048 + 128 + 24 + 4 = 2204$ - Subtract in decimal: $2977 - 2204 = 773$ - Convert back to octal: - $773 \div 8 = 96$ remainder $5$ - $96 \div 8 = 12$ remainder $0$ - $12 \div 8 = 1$ remainder $4$ - $1 \div 8 = 0$ remainder $1$ - So, $773_{10} = 1405_8$ ### c) $13523_8 - 6550_8$ - Convert to decimal: - $13523_8 = 1 \times 8^4 + 3 \times 8^3 + 5 \times 8^2 + 2 \times 8 + 3 = 1 \times 4096 + 3 \times 512 + 5 \times 64 + 16 + 3 = 4096 + 1536 + 320 + 16 + 3 = 5971$ - $6550_8 = 6 \times 8^3 + 5 \times 8^2 + 5 \times 8 + 0 = 6 \times 512 + 5 \times 64 + 40 + 0 = 3072 + 320 + 40 = 3432$ - Subtract in decimal: $5971 - 3432 = 2539$ - Convert back to octal: - $2539 \div 8 = 317$ remainder $3$ - $317 \div 8 = 39$ remainder $5$ - $39 \div 8 = 4$ remainder $7$ - $4 \div 8 = 0$ remainder $4$ - So, $2539_{10} = 4753_8$ ### d) $1134_8 - 762_8$ - Convert to decimal: - $1134_8 = 1 \times 8^3 + 1 \times 8^2 + 3 \times 8 + 4 = 512 + 64 + 24 + 4 = 604$ - $762_8 = 7 \times 8^2 + 6 \times 8 + 2 = 448 + 48 + 2 = 498$ - Subtract in decimal: $604 - 498 = 106$ - Convert back to octal: - $106 \div 8 = 13$ remainder $2$ - $13 \div 8 = 1$ remainder $5$ - $1 \div 8 = 0$ remainder $1$ - So, $106_{10} = 152_8$ --- **Final answers:** a) $31_8$ b) $1405_8$ c) $4753_8$ d) $152_8$