38. Complete the standard multiplication algorithm for [tex]\(934 \times 396\)[/tex], including any "carried," or regrouped digits, if necessary.



Answer :

To multiply [tex]\(934\)[/tex] by [tex]\(396\)[/tex] using the standard multiplication algorithm, follow these steps. We will break down the multiplication into parts and then add the intermediate results.

### Step 1: Multiply 934 by 6 (the units place of 396)
First, we multiply each digit of 934 by 6:
[tex]\[ \begin{array}{c} \ \\\ 934 \times 6: \\\ \begin{array}{cccc} & 9 & 3 & 4 \\ \times & & & 6 \\ \hline & & 5 & 6 \\ & & 28 & 4 \\ \end{array} \][/tex]
Add the carries:
- [tex]\(4 \times 6 = 24\)[/tex] (write 4, carry 2)
- [tex]\(3 \times 6 = 18 + 2 = 20\)[/tex] (write 0, carry 2)
- [tex]\(9 \times 6 = 54 + 2 = 56\)[/tex] (write 56)

So the first intermediate result is:
[tex]\[ 5604 \][/tex]

### Step 2: Multiply 934 by 9 (the tens place of 396)
Next, we multiply each digit of 934 by 9 and shift one place to the left:
[tex]\[ \begin{array}{c} \ \\\ 934 \times 9: \\\ \begin{array}{ccccc} & 9 & 3 & 4 \\ \times & & 9 & & \\ \hline & & & 3 & 6 \\ & & 1 & 2 & 1 \\ \end{array} \][/tex]
Add the carries:
- [tex]\(4 \times 9 = 36\)[/tex] (write 6, carry 3)
- [tex]\(3 \times 9 = 27 + 3 = 30\)[/tex] (write 0, carry 3)
- [tex]\(9 \times 9 = 81 + 3 = 84\)[/tex] (write 84)

So the second intermediate result, shifted one place, is:
[tex]\[ 84060 \][/tex]

### Step 3: Multiply 934 by 3 (the hundreds place of 396)
Lastly, we multiply each digit of 934 by 3 and shift two places to the left:
[tex]\[ \begin{array}{c} \ \\\ 934 \times 3: \\\ \begin{array}{cccccc} & 9 & 3 & 4 \\ \times & 3 & & & \\ \hline & & & & 1 & 2 \\ & & & 1 & 0 & 2 \\ & & 2 & 7 & 0 \\ \end{array} \][/tex]
Add the carries:
- [tex]\(4 \times 3 = 12\)[/tex] (write 2, carry 1)
- [tex]\(3 \times 3 = 9 + 1 = 10\)[/tex] (write 0, carry 1)
- [tex]\(9 \times 3 = 27 + 1 = 28\)[/tex] (write 28)

So the third intermediate result, shifted two places, is:
[tex]\[ 280200 \][/tex]

### Step 4: Add the intermediate results
Now we add all the intermediate results together:
[tex]\[ \begin{array}{cccccc} & & & 5 & 6 & 0 & 4 \\ + & & & 8 & 4 & 0 & 6 & 0 \\ + & 2 & 8 & 0 & 2 & 0 & 0 \\ \hline = & 3 & 6 & 9 & 8 & 6 & 4 \\ \end{array} \][/tex]

Thus, the product of [tex]\(934\)[/tex] and [tex]\(396\)[/tex] is [tex]\(369,864\)[/tex].