Subtract 17 (octal) from 2016 (octal). Show your work.

To find the number, we need to subtract 17_oct from 2016_oct. Let the unknown number be X_oct. The problem can be written as: X_oct + 17_oct = 2016_oct To find X_oct, we rearrange the equation: X_oct = 2016_oct - 17_oct Step 1: Set up the subtraction in octal. We will subtract digit by digit from right to left, remembering that we are in base 8. c c c c & 2 & 0 & 1 & 6_oct \\ & & & 1 & 7_oct \\ Step 2: Subtract the units column (8^0). We need to subtract 7 from 6. Since 6 < 7, we must borrow from the next column (the 8^1 place). The 1 in the 8^1 place becomes 0. The 6 in the units place becomes 6 + 8 = 14. Now, subtract: 14 - 7 = 7. c c c c & 2 & 0 & 1^0 & 6^14_oct \\ & & & 1 & 7_oct \\ & & & & 7 \\ Step 3: Subtract the 8^1 column. We need to subtract 1 from 0 (after borrowing). Since 0 < 1, we must borrow from the next column (the 8^2 place). The 0 in the 8^2 place cannot lend, so we borrow from the 2 in the 8^3 place. The 2 in the 8^3 place becomes 1. The 0 in the 8^2 place becomes 0 + 8 = 8. Now, the 8 in the 8^2 place lends 1 to the 8^1 place. The 8 in the 8^2 place becomes 7. The 0 in the 8^1 place becomes 0 + 8 = 8. Now, subtract: 8 - 1 = 7. c c c c & 2^1 & 0^7 & 0^8 & 6^14_oct \\ & & & 1 & 7_oct \\ & & & 7 & 7 \\ Step 4: Subtract the 8^2 column. The 0 in the 8^2 place became 8 and then 7 after lending. So, we have 7. Subtract 0 (since there's no digit in 17_oct for this place): 7 - 0 = 7. c c c c & 2^1 & 0^7 & 0^8 & 6^14_oct \\ & & & 1 & 7_oct \\ & & 7 & 7 & 7 \\ Step 5: Subtract the 8^3 column. The 2 in the 8^3 place became 1 after lending. So, we have 1. Subtract 0: 1 - 0 = 1. c c c c & 2^1 & 0^7 & 0^8 & 6^14_oct \\ & & & 1 & 7_oct \\ & 1 & 7 & 7 & 7_oct \\ The number is 1777_oct. The final answer is 1777_oct. 3 done, 2 left today. You're making progress.