Question

there are multiple long - division problems. for example: \\(\

$$\begin{array}{r}\\square\\\\4\\enclose{longdiv}{24}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\5\\enclose{longdiv}{40}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\3\\enclose{longdiv}{18}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\4\\enclose{longdiv}{24}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\4\\enclose{longdiv}{20}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\4\\enclose{longdiv}{12}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\5\\enclose{longdiv}{45}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\3\\enclose{longdiv}{18}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\2\\enclose{longdiv}{24}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\2\\enclose{longdiv}{10}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\3\\enclose{longdiv}{18}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\5\\enclose{longdiv}{10}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\\\enclose{longdiv}{32}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\5\\enclose{longdiv}{20}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\3\\enclose{longdiv}{24}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\), \\(\

$$\begin{array}{r}\\square\\\\5\\enclose{longdiv}{45}\\\\ \\square\\ \\square\\\\\\hline\\end{array}$$

\\)

Answer:

To solve these long - division problems, we follow the basic long - division algorithm: divide the dividend by the divisor, find the quotient, multiply the quotient by the divisor to get the product, and then subtract this product from the dividend. Let's solve each problem one by one.

1. For \(

$$\begin{array}{r}4\enclose{longdiv}{24}\end{array}$$

\)

Step 1: Find the quotient

We know that \(4\times6 = 24\), so the quotient (the number in the green box) is \(6\).

Step 2: Find the product

Multiply the quotient (\(6\)) by the divisor (\(4\)): \(4\times6=24\) (this is the number in the blue boxes).

Step 3: Subtract

Subtract the product from the dividend: \(24 - 24=0\).

2. For \(

$$\begin{array}{r}5\enclose{longdiv}{40}\end{array}$$

\)

Step 1: Find the quotient

Since \(5\times8 = 40\), the quotient is \(8\).

Step 2: Find the product

Multiply the quotient (\(8\)) by the divisor (\(5\)): \(5\times8 = 40\) (blue boxes).

Step 3: Subtract

\(40-40 = 0\).

3. For \(

$$\begin{array}{r}3\enclose{longdiv}{18}\end{array}$$

\)

Step 1: Find the quotient

We know that \(3\times6=18\), so the quotient is \(6\).

Step 2: Find the product

Multiply the quotient (\(6\)) by the divisor (\(3\)): \(3\times6 = 18\) (blue boxes).

Step 3: Subtract

\(18 - 18=0\).

4. For \(

$$\begin{array}{r}4\enclose{longdiv}{24}\end{array}$$

\) (second occurrence)

Step 1: Find the quotient

As \(4\times6 = 24\), the quotient is \(6\).

Step 2: Find the product

\(4\times6 = 24\) (blue boxes).

Step 3: Subtract

\(24-24 = 0\).

5. For \(

$$\begin{array}{r}4\enclose{longdiv}{20}\end{array}$$

\)

Step 1: Find the quotient

Since \(4\times5=20\), the quotient is \(5\).

Step 2: Find the product

\(4\times5 = 20\) (blue boxes).

Step 3: Subtract

\(20 - 20=0\).

6. For \(

$$\begin{array}{r}4\enclose{longdiv}{12}\end{array}$$

\)

Step 1: Find the quotient

We know that \(4\times3 = 12\), so the quotient is \(3\).

Step 2: Find the product

\(4\times3=12\) (blue boxes).

Step 3: Subtract

\(12-12 = 0\).

7. For \(

$$\begin{array}{r}5\enclose{longdiv}{45}\end{array}$$

\)

Step 1: Find the quotient

Since \(5\times9 = 45\), the quotient is \(9\).

Step 2: Find the product

\(5\times9=45\) (blue boxes).

Step 3: Subtract

\(45 - 45=0\).

8. For \(

$$\begin{array}{r}3\enclose{longdiv}{18}\end{array}$$

\) (second occurrence)

Step 1: Find the quotient

We know that \(3\times6=18\), so the quotient is \(6\).

Step 2: Find the product

\(3\times6 = 18\) (blue boxes).

Step 3: Subtract

\(18-18 = 0\).

9. For \(

$$\begin{array}{r}2\enclose{longdiv}{24}\end{array}$$

\)

Step 1: Find the quotient

Since \(2\times12=24\), the quotient is \(12\).

Step 2: Find the product

\(2\times12 = 24\) (blue boxes).

Step 3: Subtract

\(24-24 = 0\).

10. For \(

$$\begin{array}{r}2\enclose{longdiv}{10}\end{array}$$

\)

Step 1: Find the quotient

We know that \(2\times5 = 10\), so the quotient is \(5\).

Step 2: Find the product

\(2\times5=10\) (blue boxes).

Step 3: Subtract

\(10 - 10=0\).

11. For \(

$$\begin{array}{r}3\enclose{longdiv}{18}\end{array}$$

\) (third occurrence)

Step 1: Find the quotient

Since \(3\times6=18\), the quotient is \(6\).

Step 2: Find the product

\(3\times6 = 18\) (blue boxes).

Step 3: Subtract

\(18-18 = 0\).

12. For \(

$$\begin{array}{r}5\enclose{longdiv}{10}\end{array}$$

\)

Step 1: Find the quotient

We know that \(5\times2=10\), so the quotient is \(2\).

Step 2: Find the product

\(5\times2 = 10\) (blue boxes).

Step 3: Subtract

\(10-10 = 0\).

13. For \(

$$\begin{array}{r}4\enclose{longdiv}{32}\end{array}$$

\) (assuming the divisor is \(4\) as the first digit is missing, but from the pattern it should be \(4\))

Step 1: Find the quotient

Since \(4\times8 = 32\), the quotient is \(8\).

Step 2: Find the product

\(4\times8=32\) (blue boxes).

Step 3: Subtract

\(32-32 = 0\).

14. For \(

$$\begin{array}{r}5\enclose{longdiv}{20}\end{array}$$

\)

Step 1: Find the quotient

We know that \(5\times4 = 20\), so the quotient is \(4\).

Step 2: Find the product

\(5\times4=20\) (blue boxes).

Step 3: Subtract

\(20 - 20=0\).

15. For \(

$$\begin{array}{r}3\enclose{longdiv}{24}\end{array}$$

\)

Step 1: Find the quotient

Since \(3\times8 = 24\), the quotient is \(8\).

Step 2: Find the product

\(3\times8=24\) (blue boxes).

Step 3: Subtract

\(24-24 = 0\).

16. For \(

$$\begin{array}{r}5\enclose{longdiv}{45}\end{array}$$

\) (second occurrence)

Step 1: Find the quotient

We know that \(5\times9=45\), so the quotient is \(9\).

Step 2: Find the product

\(5\times9 = 45\) (blue boxes).

Step 3: Subtract

\(45-45 = 0\).

The quotients (green boxes) and products (blue boxes) for each problem are as follows:

Division Problem	Quotient (Green Box)	Product (Blue Boxes)
\(5\enclose{longdiv}{40}\)	\(8\)	\(40\)
\(3\enclose{longdiv}{18}\)	\(6\)	\(18\)
\(4\enclose{longdiv}{24}\)	\(6\)	\(24\)
\(4\enclose{longdiv}{20}\)	\(5\)	\(20\)
\(4\enclose{longdiv}{12}\)	\(3\)	\(12\)
\(5\enclose{longdiv}{45}\)	\(9\)	\(45\)
\(3\enclose{longdiv}{18}\)	\(6\)	\(18\)
\(2\enclose{longdiv}{24}\)	\(12\)	\(24\)
\(2\enclose{longdiv}{10}\)	\(5\)	\(10\)
\(3\enclose{longdiv}{18}\)	\(6\)	\(18\)
\(5\enclose{longdiv}{10}\)	\(2\)	\(10\)
\(4\enclose{longdiv}{32}\)	\(8\)	\(32\)
\(5\enclose{longdiv}{20}\)	\(4\)	\(20\)
\(3\enclose{longdiv}{24}\)	\(8\)	\(24\)
\(5\enclose{longdiv}{45}\)	\(9\)	\(45\)