__Product of Factors__

__Perfect square as a product of two factors__

In case of perfect square number we have odd number of factors i.e. the number of factors are odd hence in that case required number of ways in which we can write perfect square number as a product of its two factors are (n – 1)/2 if we do not include the square root of the number and

(n + 1 )/2 if we include the square root of the number. So number of ways to express a perfect square as product of two different factors (that means excluding its square root) is

½ {(p + 1)(q + I)(r + I) … – 1)}. And if we include the square root then required numbs 1/2 {(p+1)(q+1)(r+1) … +1}

**Let’s take one example to understand this. **

**Example 2**: **In how many ways you can express 36 as the product of two of its factors?**

**Solution :**

**Step 1**: Prime factorization of 36 i.e. we write 36 = 2^{2} x 3^{2}

**Step 2**: Number of factors of 36 will be (2+1)(2+1)=9 (i.e. factors are 1,2,3,4,6,9,12,18,36)

**Step 3**: Since we are asked total number of ways hence we include square root of 36 i.e. 6 as well. Thus number of ways you can express 36 as the product of two of its factors is (9+1)/2=5

__ __

__Product of two co-prime numbers __

To express the number as a product of co-prime factors we will use the following steps:

**Step 1:**Write Prime factorisation of given number i.e. convert the number in the form where p_{1} ,p_{2},p_{3}…..p_{n }are prime numbers and a,b,c….. are natural numbers as their respective powers.

**Step 2:** In the above step we have n prime numbers then the number of ways to express the number as the product of two co prime numbers =2^{n-1}

Because two primes are always co-prime and after we pick 1 prime the other prime can be picked in 2^{n-1}ways.Hence number of ways in which we can write given number as a product of two co prime factors =2^{n-1}

**Example 1: In how many ways you can write 315 as product of two of its co-prime factors. Solution: **

Step 1: Prime factorization of 315 i.e. we write 315 = 3

^{2}x 5

^{1}x 7

^{1}

Step2: In above step 3 prime numbers are used. Hence number of ways are 2

^{3-1 }= 4.Infact we can mention these cases as well 9×35, 5 x 63, 7 x 45, 15 x 21.

__Remember:__The number of numbers, which are less than N=**p**

^{a}q^{b}r^{c}(where p , q, r are prime numbers and the a,b,c are natural numbers as their respective powers)and are co-prime to N is given by N{(1-**1/p)(1-1/q**

**)(1-1/r**

**)}.**

__Product of all the factors__**To find product of all the factors we follow the following steps:**

Step1: Prime factorisation N= a

^{p}b

^{q}c

^{r}

Step2: Number of factors (say X)

Step3: Product of all the factors is given by = N

^{X/2}

**Example 2: Find the product of all the factors of 120 ?**

**Solution:**

**Step 1:**Prime factorisation : = 2

^{3}X5

^{1}X3

^{1}

**Step2:**Number of factors are ( 3+1)(1+1)(1+1)= 16

**Step 3:**Product of all the factors =( 2

^{3}X5

^{1}X3

^{1})

^{16/2}=(120)

^{8}

*Assignment:*

__Questions:__**1**: In how many ways can you express 216 as a product of two of its factors?

a) 4

b) 6

c) 8

d) 9

Answer: c

Solution:

Step 1: 216 = 2

^{3}3

^{3}

Step 2: Number of factors are (3+1)(3+1) = 4 x 4= 16

Step 3: So number of ways ½ (4)(4) = 8

**2**: Find the sum of factors of 216.

a. 950

b) 850

c) 600

d) 1000

Answer: c

Solution:

Step 1: 216 = 2^{3}3^{3}

Step 2: Sum of factors = (2^{0}+2^{1}+2^{2}+2^{3})(3^{0}+3^{1}+3^{2}+3^{3})= 15 x 40 = 600

**3**: Find the product of all the factors of 400

a. 80^{5}

b) 80^{4}

c) 80^{7}

d) 80^{6}

Answer: a

Solution:

Step 1: Prime factorisation of 400 = 2^{4}X5^{2}

Step2: Number of factors (4+1)(1+1)= 10

Step 3: Product of all the factors =( 2^{4}X5^{1})^{10/2} =(80)^{5}

**4**:**In how many ways you can write 200 as product of two of its co-prime factors. **

a) 1

b) 2

c) 4

d) 8

Answer: b

Step1: 200 = 2

^{3}5

^{2}

Hence number of ways are 2

^{2-1 }= 2.Infact we can mention these cases as well 8 x 25,

1 x 100.

**5**: In how many ways you can write 10890 as product of two of its co-prime factors.

a) 7

b) 3

c) 31

d) 15

**Answer: d Solution: **We will do the solution with step by step

Step1: first prime factorization i.e.2

^{1}x 3

^{2}x 5

^{1}x 11

^{2}

Step2: 3 prime number are used in this so the number of ways are 2

^{4}-1=15

This is w.r.t Q.5. The formula used is wrong. It should be 2^(n-1) according to my knowledge based on above given theory.

Hey, in the example of writing 315 as product of coprime factors, It’s written 15*21 but they aren’t Co prime. I think it should be 1*315 instead

Dear Prashanth, For Question.5 it is supposed to be 2^4-1 = 8. But the answer is mentioned as 16(2^4) – 1. Could you be more elaborative.

Please Solve this :

In how many ways 6400 can be written as a product of two perfect squares which are not co-primes?