Problem 107 - Fundamentals of Mathematics

Question

Compute each product using the distributive property. \( 15 \cdot 105 \)

Step-by-Step Solution

Answer

The product is 1575.

Step 1: Decompose 105

Recognize that 105 can be decomposed into two simpler numbers: 100 and 5. So, the expression becomes \( 15 \times (100 + 5) \).

Step 2: Apply Distributive Property

Use the distributive property \( a(b + c) = ab + ac \). So, \( 15 \times (100 + 5) = 15 \times 100 + 15 \times 5 \).

Step 3: Compute Each Product Separately

Calculate each separate multiplication: \( 15 \times 100 = 1500 \) and \( 15 \times 5 = 75 \).

Step 4: Add the Partial Products

Add the results from Step 3: \( 1500 + 75 = 1575 \).

Step 5: Final Result

The product of \( 15 \cdot 105 \) using the distributive property is \( 1575 \).

Key Concepts

Understanding Multiplication

Multiplication is one of the basic arithmetic operations, which is essentially repeated addition. When we talk about multiplying two numbers, say \( 15 \) and \( 105 \), it means adding \( 15 \) together with itself \( 105 \) times. While this approach is straightforward, it can be time-consuming with larger numbers.

In such cases, using different mathematical properties, like the distributive property, can make multiplication more efficient. The objective is to simplify the process and reduce opportunity for errors, especially when dealing with more complex numbers.

• *Commutative Property*: States that the order of multiplication doesn't affect the product (i.e., \( a \times b = b \times a \)).
• *Associative Property*: Allows grouping of numbers being multiplied without changing the product (i.e., \[(a \times b) \times c = a \times (b \times c)\]).
• *Distributive Property*: Combines both addition and multiplication, allowing a number to distribute over a sum or difference (i.e., \( a(b + c) = ab + ac \)).

Partial Products Approach

In the partial products approach, we break a complex multiplication into simpler calculations that are easier to manage. Let's consider \( 15 \times 105 \) as an example.

Instead of doing the entire multiplication at once, we decompose one of the numbers, in this case \( 105 \), into parts that are easier to multiply. Here, \( 105 \) can be rewritten as \( 100 + 5 \), making it convenient to apply the distributive property. The expression becomes \( 15 \times (100 + 5) \). This transformation simplifies the arithmetic process.

By multiplying \( 15 \) separately with \( 100 \) and \( 5 \) (termed as each partial product), we cater to simpler operations:

• \( 15 \times 100 = 1500 \)
• \( 15 \times 5 = 75 \)

Applying the distributive property in this way divides the problem into smaller, more manageable steps, leading us to the next concept of arithmetic decomposition.

Arithmetic Decomposition in Detail

Arithmetic decomposition is a strategy used to simplify calculations by breaking numbers into smaller components. This method is particularly effective in making complex multiplications more straightforward.

For example, with \( 105 \) in the equation \( 15 \times 105 \), we decompose \( 105 \) into \( 100 + 5 \). The core idea is to work with simpler numbers that are easier to multiply.

Here's why it's practical and beneficial:

• Simplified calculation steps reduce potential errors.
• Each multiplication operation becomes more straightforward, as seen in the example: \( 15 \times 100 \) is simply adding a zero to \( 1500 \), and \( 15 \times 5 \) is basic multiplication giving \( 75 \).
• Combining partial results, as in \( 1500 + 75 = 1575 \), is the final step to get the complete product.

This breakdown not only helps in understanding the distributive property but also enhances mental math skills by encouraging flexible thinking with numbers.