Understanding the Sum of Even Factors of (10^5): A Comprehensive Guide

When dealing with the task of finding the sum of even factors of a particular number, such as (10^5), a systematic and structured approach is essential. This article will provide a detailed explanation of the process, breaking down each step and explaining the underlying mathematical concepts. By the end of this guide, you will understand the method to calculate the sum of even factors of (10^5) and be able to apply these techniques to similar problems.

Step-by-Step Guide to Finding the Sum of Even Factors of (10^5)

Step 1: Prime Factorization

Prime factorization is the cornerstone of this calculation. We start by expressing (10^5) in its prime factors.

(10^5 2^5 times 5^5) (1)

Step 2: Total Number of Factors

The total number of factors of a number (n) with prime factorization (p_1^{n_1} times p_2^{n_2} times ldots times p_k^{n_k}) is given by the formula:

(n_1 1 times n_2 1 times ldots times n_k 1)

For (10^5 2^5 times 5^5):

(5 1 times 5 1 6 times 6 36) (2)

This means (10^5) has 36 factors in total.

Step 3: Identifying Even Factors

An even factor must contain at least one factor of 2. Therefore, we can express the even factors as:

(2^k times 5^m), where (k in {1, 2, 3, 4, 5}) and (m in {0, 1, 2, 3, 4, 5}).

Step 4: Counting the Even Factors

To count the even factors, we determine the number of combinations for (k) and (m) values:

There are 5 possible values for (k) (1 through 5). There are 6 possible values for (m) (0 through 5).

Thus, the total number of even factors is:

(5 times 6 30) (3)

Step 5: Sum of the Even Factors

Next, we calculate the sum of the even factors. We factor out the 2 from each even factor:

(2 times text{Sum of factors of } frac{10^5}{2} 2 times text{Sum of factors of } 5^5 times 2^4)

From equation (1), we know:

(frac{10^5}{2} 5^5 times 2^4)

Step 6: Calculating the Sum of Factors of (5^5 times 2^4)

Using the formula for the sum of factors:

(frac{p^{n} - 1}{p - 1})

We calculate the sum of factors for (5^5) and (2^4):

Sum of factors of (5^5): (frac{5^{5} - 1}{5 - 1} frac{5^6 - 1}{4} frac{15625 - 1}{4} frac{15624}{4} 3906) Sum of factors of (2^4): (frac{2^{5} - 1}{2 - 1} 31)

Therefore, the sum of factors of (5^5 times 2^4) is:

(3906 times 31 121116)

Step 7: Final Calculation

The final sum of the even factors is:

(2 times 121116 242232)

Thus, the sum of the even factors of (10^5) is:

(boxed{242232})

Additional Insights

To provide a broader understanding, let's also explore an alternate method to find the sum of even factors:

Alternate Method

First, we find the total sum of all factors of (10^5):

(2^0 times 5^0 2^1 times 5^0 2^2 times 5^0 ldots 2^5 times 5^5 246078)

Note that all odd factors do not include a power of 2 in their factorization. Therefore, the sum of the odd factors is:

(5^0 5^1 5^2 5^3 5^4 5^5 3906)

Subtracting the sum of the odd factors from the total sum of factors, we get the sum of even factors:

(246078 - 3906 242172)

Frequently Asked Questions

Q: How do you determine the number of factors of a number?

A: The number of factors of a number with prime factorization (p_1^{n_1} times p_2^{n_2} times ldots times p_k^{n_k}) is given by the product of ((n_1 1) times (n_2 1) times ldots times (n_k 1)).

Q: What is the formula for the sum of factors of a number?

A: The sum of factors of a number (n) with prime factorization (p_1^{n_1} times p_2^{n_2} times ldots times p_k^{n_k}) is given by:

(frac{p_1^{n_1 1} - 1}{p_1 - 1} times frac{p_2^{n_2 1} - 1}{p_2 - 1} times ldots times frac{p_k^{n_k 1} - 1}{p_k - 1})

Q: How can I practice similar problems?

A: To practice, try finding the sum of even factors for other numbers, such as (2^6 times 3^3) or (2^4 times 5^5 times 7^2). Use the steps outlined in this guide as a reference.