Let's solve the multiplication [tex]\(934 \times 396\)[/tex] step-by-step using the standard multiplication algorithm.
### Step 1: Multiplying by the digit in the units place (6)
[tex]\[
\begin{array}{r}
934 \\
\times 6 \\
\hline
\end{array}
\][/tex]
1. [tex]\(4 \times 6 = 24\)[/tex]. Write down 4, carry over 2.
2. [tex]\(3 \times 6 = 18\)[/tex]. Adding the carried-over 2, we get [tex]\(18 + 2 = 20\)[/tex]. Write down 0, carry over 2.
3. [tex]\(9 \times 6 = 54\)[/tex]. Adding the carried-over 2, we get [tex]\(54 + 2 = 56\)[/tex].
So, the result of [tex]\(934 \times 6\)[/tex] is:
[tex]\[
5604
\][/tex]
### Step 2: Multiplying by the digit in the tens place (90)
First note that [tex]\(90 = 9 \times 10\)[/tex], so we multiply by 9 and then by 10:
[tex]\[
\begin{array}{r}
934 \\
\times 90 \\
\hline
\end{array}
\][/tex]
1. [tex]\(4 \times 9 = 36\)[/tex]. Write down 6, carry over 3.
2. [tex]\(3 \times 9 = 27\)[/tex]. Adding the carried-over 3, we get [tex]\(27 + 3 = 30\)[/tex]. Write down 0, carry over 3.
3. [tex]\(9 \times 9 = 81\)[/tex]. Adding the carried-over 3, we get [tex]\(81 + 3 = 84\)[/tex].
Then multiply the result by 10 (shift one place to the left):
[tex]\[
84060
\][/tex]
### Step 3: Multiplying by the digit in the hundreds place (300)
First note that [tex]\(300 = 3 \times 100\)[/tex], so we multiply by 3 and then by 100:
[tex]\[
\begin{array}{r}
934 \\
\times 300 \\
\hline
\end{array}
\][/tex]
1. [tex]\(4 \times 3 = 12\)[/tex]. Write down 2, carry over 1.
2. [tex]\(3 \times 3 = 9\)[/tex]. Adding the carried-over 1, we get [tex]\(9 + 1 = 10\)[/tex]. Write down 0, carry over 1.
3. [tex]\(9 \times 3 = 27\)[/tex]. Adding the carried-over 1, we get [tex]\(27 + 1 = 28\)[/tex].
Then multiply the result by 100 (shift two places to the left):
[tex]\[
280200
\][/tex]
### Step 4: Adding the results together
[tex]\[
\begin{array}{r}
& 5604 \\
+ & 84060 \\
+ & 280200 \\
\hline
& 369864 \\
\end{array}
\][/tex]
So, the final result of [tex]\(934 \times 396\)[/tex] is:
[tex]\[
369864
\][/tex]
