division of a three-digit number by a one-digit number

1. Understand the Problem: The problem asks us to divide the number 567 by 9. This is a division problem.

1. Set up the Division: We can set this up as a long division problem. We want to find out how many times 9 fits into 567.

$$ \begin{array}{r} 9 \overline{)567} \end{array} $$

1. Divide the First Digit (or first few digits): Look at the first digit of the dividend (5). Since 5 is less than 9, we look at the first two digits (56).

1. Determine the First Digit of the Quotient: How many times does 9 go into 56? We can list multiples of 9: 9, 18, 27, 36, 45, 54, 63. The largest multiple of 9 that is less than or equal to 56 is 54. So, 9 goes into 56 six times ($6 imes 9 = 54$). Write 6 above the 6 in 567.

$$ \begin{array}{r} 6 \ 9 \overline{)567} \end{array} $$

1. Multiply and Subtract: Multiply the digit we just placed in the quotient (6) by the divisor (9): $6 imes 9 = 54$. Write 54 below 56 and subtract.

$$ \begin{array}{r} 6 \ 9 \overline{)567} \ \underline{-54} \ \phantom{00}2 \end{array} $$

1. Bring Down the Next Digit: Bring down the next digit from the dividend (7) next to the remainder (2) to form the new number 27.

$$ \begin{array}{r} 6 \ 9 \overline{)567} \ \underline{-54} \ \phantom{00}27 \end{array} $$

1. Determine the Next Digit of the Quotient: How many times does 9 go into 27? We know from our multiples of 9 that $3 imes 9 = 27$. So, 9 goes into 27 three times. Write 3 above the 7 in 567.

$$ \begin{array}{r} 63 \ 9 \overline{)567} \ \underline{-54} \ \phantom{00}27 \end{array} $$

1. Multiply and Subtract Again: Multiply the new digit in the quotient (3) by the divisor (9): $3 imes 9 = 27$. Write 27 below 27 and subtract.

$$ \begin{array}{r} 63 \ 9 \overline{)567} \ \underline{-54} \ \phantom{00}27 \ \underline{\phantom{00}-27} \ \phantom{000}0 \end{array} $$

1. Check for Remainder: The remainder is 0. This means 567 is perfectly divisible by 9.

1. State the Final Answer: The quotient is 63.