Number System: Factors and Coprimes

Number system is a very important chapter and you can find good number of questions in various competitive exams.  Important formulas are the following.
A number can be written in its prime factorization format.  For example 100= 2² x 5²

Formula 1: The number of factors of a number N=ap.bq.cr... = (p+1).(q+1).(r+1)...

Example: Find the number of factors of 100.  
Ans: We know that 100 = 22×52
So number of factors of 100 = (2 +1 ).(2 +1) =  9.
Infact the factors are 1, 2, 4, 5, 10, 20, 25, 50, 100

Formula 2: The sum of factors of a number N=ap.bq.cr... can be written as ap+1−1a−1×bq+1−1b−1×cr+1−1c−1...
Example: Find the sum of the factors of 72
Ans: 72 can be written as 23×32. 

Sum of all the factors of 72 = 23+1−12−1×32+1−13−1 = 15 x 13= 195

Formula 3: The number of ways of writing a number as a product of two number =12×[(p+1).(q+1).(r+1)...] (if the number is not a perfect square)

If the number is a perfect square then two conditions arise:

1.  The number of ways of writing a number as a product of two distinct numbers =12×[(p+1).(q+1).(r+1)...−1]
2.  The number of ways of writing a number as a product of two numbers and those numbers need not be distinct= 12×[(p+1).(q+1).(r+1)...+1]


Example: Find the number of ways of writing 140 as a product of two factors
Ans: The prime factorization of 140 = 22×5×7
So number of ways of writing 140 as a product of two factors = 12×[(p+1).(q+1).(r+1)...]  = 12×[(2+1).(1+1).(1+1).]=6

Example: Find the number of ways of writing 144 as a product of two factors subjected to the following conditions a. Both factors should be different b. Both factors need not be different.
Ans: The prime factorization of 144 = 24×32
a. If both factors are different, then total ways = 12×[(p+1).(q+1).(r+1)...−1] = 12×[(4+1).(2+1)−1] = 7

If both factors need not be different, then total ways = 12×[(p+1).(q+1).(r+1)...+1] =12×[(4+1).(2+1)+1] = 8 

Formula 4: The number of co-primes of a number  N=ϕ(N)=ap.bq.cr... can be written as N×(1−1a)×(1−1b)×(1−1c)...
Example: Find the number of co-primes to 144 which are less than that of it
Ans: The prime factorization of 144 = 24×32
The number of co-primes which are less than that of 144 = 144×(1−12)×(1−13)= 48
Formula 5: The sum of co-primes of a number N= ϕ(N)×N2

Example: Find the sum of all the co-primes of 144
Ans: The sum of co-primes of the 144 = ϕ(N)×N2 = 48 x 1442 = 3456 
Formula 6: The number of ways of writing a number N as a product of two co-prime numbers = 2n−1where n=the number of prime factors of a number.

Example: Find the number of ways of writing 60 as a product of two co - primes
Ans: The prime factorization of 60 = 22×3×5
The number of ways of writing 60 as a product of two co - primes = 23−1 = 4


Formula 7: Product of all the factors of N = N
(Number of factors2
) =

N
((p+1).(q+1).(r+1)....2
)N

Example: Find the product of all the factors of 50
Ans: Prime factorization of 50 = 2×52 
Product of all the factors of 50 = 50(1+1).(2+1)2=503