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

Question

ScanSolve AI Tutor · Accepted Answer

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}

Set up the subtraction in octal.
We will subtract digit by digit from right to left, remembering that we are in base 8.
$\begin{array}{c c c c} & 2 & 0 & 1 & 6_{oct} \\ - & & & 1 & 7_{oct} \\ \hline \end{array}$
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: $14 - 7 = 7$ .
$\begin{array}{c c c c} & 2 & 0 & \cancel{1}^0 & \cancel{6}^{14}_{oct} \\ - & & & 1 & 7_{oct} \\ \hline & & & & 7 \\ \end{array}$
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: $8 - 1 = 7$ .
$\begin{array}{c c c c} & \cancel{2}^1 & \cancel{0}^7 & \cancel{0}^8 & \cancel{6}^{14}_{oct} \\ - & & & 1 & 7_{oct} \\ \hline & & & 7 & 7 \\ \end{array}$
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): $7 - 0 = 7$ .
$\begin{array}{c c c c} & \cancel{2}^1 & \cancel{0}^7 & \cancel{0}^8 & \cancel{6}^{14}_{oct} \\ - & & & 1 & 7_{oct} \\ \hline & & 7 & 7 & 7 \\ \end{array}$
Subtract the $8^3$ column.
The 2 in the $8^3$ place became 1 after lending. So, we have 1.
Subtract 0: $1 - 0 = 1$ .
$\begin{array}{c c c c} & \cancel{2}^1 & \cancel{0}^7 & \cancel{0}^8 & \cancel{6}^{14}_{oct} \\ - & & & 1 & 7_{oct} \\ \hline & 1 & 7 & 7 & 7_{oct} \\ \end{array}$
The number is $1777_{oct}$ .

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

Answer

Step-by-step solution

Set up the subtraction in octal.

Subtract the units column ( $8^0$ ).

Subtract the $8^1$ column.

Subtract the $8^2$ column.

Subtract the $8^3$ column.

Try one yourself

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

Answer

Step-by-step solution

Set up the subtraction in octal.

Subtract the units column (808^080).

Subtract the 818^181 column.

Subtract the 828^282 column.

Subtract the 838^383 column.

Try one yourself

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Subtract the units column ( $8^0$ ).

Subtract the $8^1$ column.

Subtract the $8^2$ column.

Subtract the $8^3$ column.