- Set up the division \(258 \div 6\) by choosing a large multiple of 6 to subtract, such as \(40 \times 6 = 240\).
- Subtract 240 from 258 to leave a remainder of 18.
- Identify the next multiple of 6, which is \(3 \times 6 = 18\), and subtract it from 18 to leave 0.
- Add the partial quotients 40 and 3 to get the final quotient, \(\boxed{43}\).
Explanation
- Problem Analysis and Strategy
We need to divide \(258\) by \(6\) using the partial quotients method. In this method, we repeatedly subtract convenient multiples of the divisor (\(6\)) from the dividend (\(258\)) until we reach zero. We keep track of the multipliers (the partial quotients) used at each step, and their sum will provide our final result. The dividend is \(258\) and the divisor is \(6\).
- First Step of Division
We begin by choosing a large, manageable multiple of \(6\) that is less than \(258\). A convenient choice is \(6 \times 40 = 240\). Subtracting \(240\) from \(258\) leaves us with a remainder of \(258 - 240 = 18\). We record the partial quotient as \(40\).
- Second Step of Division
Now we divide the remaining amount, \(18\), by \(6\). We know that \(6 \times 3 = 18\). Subtracting \(18\) from \(18\) leaves us with a remainder of \(0\). We record the partial quotient as \(3\).
- Calculating Final Result
To find the final quotient, we sum the partial quotients used: \(40 + 3 = 43\). Thus, \(258 \div 6 = 43\). Represented in the requested tabular format:
\[ \begin{array}{r@{\quad}l} 6 \overline{) 258} & \ -240 & 40 \times 6 \\ 18 & \ -18 & 3 \times 6 \\ 0 & 43 \ \end{array} \]
- Final Answer
The final result of the division \(258 \div 6\) using the partial quotients method is \(\boxed{43}\).
Examples
The partial quotients method is a practical way to manage large division problems by breaking them down into simpler, recognizable parts. Imagine you are a baker who has baked 258 cookies and needs to pack them into boxes that hold 6 cookies each. By using this method, you can quickly determine how many full boxes you will have (43 boxes) without needing to perform the entire long division process all at once. This approach builds number sense and makes complex division intuitive.