**FACTORS**

In this article, we will take the concept of factors further and learn about the sum of factors, and sum of even and odd factors. So, let’s get started.

** ****Sum of Factors**:

Let us take an example to understand how to calculate the sum of factors.

**Example:** Find sum of factors of 12.

**Solution:**

**Method 1:**

**Method 2:**

We can write above process as follows:

We can generalize the process and formula :

**Sum Of Factors: **

To find sum of factors of any number we will follow the following steps:

**Step 1: **Write Prime factorization of given number i.e. convert the number in the form a^{p}b^{q}c^{r} where a,b,c are prime numbers and the p,q,r are natural numbers as their respective powers.

**Step 2 **: Sum of all the factors [(a^{p+1}-1)/(a-1) x (b^{q+1}-1)/(b-1) x (c^{r+1}-1)/(c-1)………]

**Sum of Even Factors of a Number:**

For this purpose we will exclude 2^{0}

**Example: **Calculate sum of even factors of 72.

Solution:

For even factors

**Sum of Odd Factors of a Number**:

In this case only power of 2 which will be included is 2^{0}

**Example: **Calculate sum of odd factors of 72.

**Solution: **

To get through the above concepts thoroughly, solve the exercise below:

**EXERCISE:**

**Question 1. ** Find the sum of all the factors of 600.

(1) 1860

(2) 3200

(3) 930

(4) 630

### Answer and Explanation

**Solution: Option 1**

**Step1:** Prime factorization of 600 i.e. we write 600= 2^{3}3^{1}5^{2}

Factor must have

2^{(0 or 1 or 2 or 3)}

3^{(0 or 1)}

5^{(0 or 1 or 2)}

**Step 2:** Hence Sum of all the factors will be obtained by

(2^{0}+2^{1}+2^{2}+2^{3})(3^{0}+3^{1})(5^{0 }+5^{1}+5^{2})

**Step 3: **To add terms inside the bracket we make use of G.P. to get

{(2^{3+1}-1)/(2-1)x(3^{1+1}-1)/(3-1)x(5^{1+2}-1)/(5-1)} =1860

Hence sum of all the factors of 600 is 1860.

**Question 2.** Find the sum of all the even factors of 600.

(1) 80

(2) 326

(3) 1736

(4) 124

### Answer and Explanation

**Solution: Option 3**

**Step1:** Prime factorization of 600 i.e. we write 600= 2^{3}3^{1}5^{2}

Factor must have

2^{(0 or 1 or 2 or 3)}

3^{(0 or 1)}

5^{(0 or 1 or 2)}

**Step 2:** Hence Sum of even factors will be obtained by

(2^{1}+2^{2}+2^{3})(3^{0}+3^{1})(5^{0 }+5^{1}+5^{2})=14*4*31=1736

**Question 3. **Find the sum of all the odds factors of 600.

(1) 80

(2) 326

(3) 1736

(4) 124

### Answer and Explanation

**Solution: Option 4**

**Step1:** Prime factorization of 600 i.e. we write 600= 2^{3}3^{1}5^{2}

Factor must have

2^{(0 or 1 or 2 or 3)}

3^{(0 or 1)}

5^{(0 or 1 or 2)}

**Step 2:** Hence Sum of odd factors will be obtained by

(2^{0})(3^{0}+3^{1})(5^{0 }+5^{1}+5^{2})= 1*4*31=124