Multiplication worksheet for practicing multiplying numbers by 100.

Problem Description:

The task is to solve a series of multiplication problems where one of the factors is always 100. The goal is to find the products for each pair of numbers.

Key Concept:

When multiplying any number by 100, you simply add two zeros to the end of that number. For example:
- \( 79 \times 100 = 7900 \)
- \( 100 \times 48 = 4800 \)

This rule simplifies the calculations significantly.

Solution:

#### Left Column:
1. \( 79 \times 100 = 7900 \)
2. \( 100 \times 41 = 4100 \)
3. \( 100 \times 32 = 3200 \)
4. \( 40 \times 100 = 4000 \)
5. \( 77 \times 100 = 7700 \)
6. \( 82 \times 100 = 8200 \)
7. \( 100 \times 75 = 7500 \)
8. \( 56 \times 100 = 5600 \)
9. \( 80 \times 100 = 8000 \)
10. \( 86 \times 100 = 8600 \)

#### Right Column:
1. \( 100 \times 48 = 4800 \)
2. \( 100 \times 35 = 3500 \)
3. \( 31 \times 100 = 3100 \)
4. \( 100 \times 72 = 7200 \)
5. \( 100 \times 62 = 6200 \)
6. \( 100 \times 83 = 8300 \)
7. \( 100 \times 58 = 5800 \)
8. \( 100 \times 19 = 1900 \)
9. \( 100 \times 11 = 1100 \)
10. \( 30 \times 100 = 3000 \)

Final Answer:

\[
\boxed{
\begin{array}{ll}
79 \times 100 = 7900 & 100 \times 48 = 4800 \\
100 \times 41 = 4100 & 100 \times 35 = 3500 \\
100 \times 32 = 3200 & 31 \times 100 = 3100 \\
40 \times 100 = 4000 & 100 \times 72 = 7200 \\
77 \times 100 = 7700 & 100 \times 62 = 6200 \\
82 \times 100 = 8200 & 100 \times 83 = 8300 \\
100 \times 75 = 7500 & 100 \times 58 = 5800 \\
56 \times 100 = 5600 & 100 \times 19 = 1900 \\
80 \times 100 = 8000 & 100 \times 11 = 1100 \\
86 \times 100 = 8600 & 30 \times 100 = 3000 \\
\end{array}
}
\]

Explanation:

Each problem involves multiplying a number by 100. By adding two zeros to the end of the number being multiplied, we obtain the product quickly and efficiently. This method works because multiplying by 100 shifts the decimal point two places to the right, effectively appending two zeros.

This approach ensures accuracy and saves time, making it ideal for solving these types of problems.