Finding the Sum of Factors of 100

To find the sum of all the factors of 100, we first need to identify all the numbers that divide 100 evenly. These numbers are called the factors of 100.

Let's list the factors of 100:

• 1 (since 100 \(\div\) 1 = 100)
• 2 (since 100 \(\div\) 2 = 50)
• 4 (since 100 \(\div\) 4 = 25)
• 5 (since 100 \(\div\) 5 = 20)
• 10 (since 100 \(\div\) 10 = 10)
• 20 (since 100 \(\div\) 20 = 5)
• 25 (since 100 \(\div\) 25 = 4)
• 50 (since 100 \(\div\) 50 = 2)
• 100 (since 100 \(\div\) 100 = 1)

The factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100.

Now, we need to find the sum of these factors:

Sum = 1 + 2 + 4 + 5 + 10 + 20 + 25 + 50 + 100

Sum = 3 + 4 + 5 + 10 + 20 + 25 + 50 + 100

Sum = 7 + 5 + 10 + 20 + 25 + 50 + 100

Sum = 12 + 10 + 20 + 25 + 50 + 100

Sum = 22 + 20 + 25 + 50 + 100

Sum = 42 + 25 + 50 + 100

Sum = 67 + 50 + 100

Sum = 117 + 100

Sum = 217

So, the sum of all the factors of 100 is 217.

Alternative Method: Using Prime Factorization

A more systematic way to find the sum of factors, especially for larger numbers, is using prime factorization.

First, find the prime factorization of 100:

\(100 = 10 \times 10 = (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2\)

So, the prime factorization of 100 is \(2^2 \times 5^2\).

If a number N can be written as \(N = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}\), where \(p_1, p_2, \dots, p_k\) are distinct prime numbers and \(a_1, a_2, \dots, a_k\) are positive integers, then the sum of the factors of N is given by the formula:

Sum of factors \( = (1 + p_1 + p_1^2 + \dots + p_1^{a_1})(1 + p_2 + p_2^2 + \dots + p_2^{a_2}) \dots (1 + p_k + p_k^2 + \dots + p_k^{a_k})\)

Applying this formula to \(100 = 2^2 \times 5^2\):

• For the prime factor 2 with power 2, the sum of terms is \(1 + 2^1 + 2^2 = 1 + 2 + 4 = 7\).
• For the prime factor 5 with power 2, the sum of terms is \(1 + 5^1 + 5^2 = 1 + 5 + 25 = 31\).

The sum of all factors of 100 is the product of these sums:

Sum of factors of 100 \( = (1 + 2^1 + 2^2) \times (1 + 5^1 + 5^2)\)

Sum of factors of 100 \( = (1 + 2 + 4) \times (1 + 5 + 25)\)

Sum of factors of 100 \( = 7 \times 31\)

Sum of factors of 100 \( = 217\)

Both methods yield the same result. The sum of all the factors of 100 is 217.

Revision Table: Key Concepts for Factors and Sum of Factors

Additional Information: Related Number Theory Concepts

Understanding factors and their sums is fundamental in number theory. Here are a few related concepts:

• Number of Factors: If the prime factorization of N is \(p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}\), the total number of factors is given by \((a_1+1)(a_2+1)\dots(a_k+1)\). For 100 (\(2^2 \times 5^2\)), the number of factors is \((2+1)(2+1) = 3 \times 3 = 9\), which matches our list.
• Proper Factors: These are all factors of a number excluding the number itself. For 100, the proper factors are 1, 2, 4, 5, 10, 20, 25, 50.
• Sum of Proper Factors: This is the sum of all factors minus the number itself. For 100, the sum of proper factors is \(217 - 100 = 117\).
• Perfect Numbers: A perfect number is a positive integer that is equal to the sum of its proper positive divisors (sum of proper factors). For example, 6 is a perfect number because its proper divisors (1, 2, 3) sum to 6. 100 is not a perfect number as the sum of its proper factors (117) is not equal to 100.