Long Division with a Remainder

36 R 4

Explanation

Problem: \( 832 \div 23 \)

Step 1: Divide the first two digits.
How many times does 23 go into 83? \( 23 \times 3 = 69 \), \( 23 \times 4 = 92 \) (too high). So, it goes 3 times.

\[ \begin{array}{r} \mathbf{3}\phantom{0} \\ \hline 23 \big) 8\ 3\ 2 \\ \underline{-6\ 9}\phantom{0} \\ 1\ 4\phantom{0} \end{array} \]

• Divide: \(83 \div 23 \approx 3\)
• Multiply: \(23 \times 3 = 69\)
• Subtract: \(83 - 69 = 14\)

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Step 2: Bring down the last digit (2).
Now we have 142. How many times does 23 go into 142? \( 23 \times 6 = 138 \), \( 23 \times 7 = 161 \) (too high). So, it goes 6 times.

\[ \begin{array}{r} 3\ \mathbf{6} \\ \hline 23 \big) 8\ 3\ 2 \\ \underline{-6\ 9}\phantom{0} \\ 1\ 4\ 2 \\ \underline{-1\ 3\ 8} \\ \mathbf{4} \end{array} \]

• Bring down: 2 comes down to join 14.
• Divide: \(142 \div 23 \approx 6\)
• Multiply: \(23 \times 6 = 138\)
• Subtract: \(142 - 138 = 4\)

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Final Result
The quotient is 36 and the remainder is 4.

\[ 832 \div 23 = 36 \text{ R } 4 \]

Or as a mixed number:

\[ 36 \frac{4}{23} \]

Why this works:
We broke the number into pieces:
\(23000 \div 23 = 1000\)
\(0 \div 23 = 0\)
\(690 \div 23 = 30\)
\(138 \div 23 = 6\)
Plus the leftover \(4\).

Adding the partial quotients: \(1000 + 0 + 30 + 6 = 1036\) with remainder \(4\).

\(832 \div 23 = 36\) R 4. Check: \((36 \times 23) + 4 = 828 + 4 = 832\) ✓