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 = 2² x 3²
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 p₁ ,p₂,p₃…..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-1ways.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  = 3² x 5¹ x 7¹
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^pb^qc^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 : = 2³X5¹X3¹
Step2: Number of factors are ( 3+1)(1+1)(1+1)= 16
Step 3: Product of all the factors =( 2³X5¹X3¹)^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 = 2³3³
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 = 2³3³
Step 2: Sum of factors = (2⁰+2¹+2²+2³)(3⁰+3¹+3²+3³)= 15 x 40 = 600

3: Find the product of all the factors of 400
a. 80⁵
b) 80⁴
c) 80⁷
d) 80⁶
Answer: a
Solution:
Step 1: Prime factorisation of 400 = 2⁴X5²
Step2: Number of factors (4+1)(1+1)= 10
Step 3: Product of all the factors =( 2⁴X5¹)^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 = 2³5²
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.2¹ x 3² x 5¹x 11²
Step2: 3 prime number are used in this so the number of ways are 2⁴-1=15