Number System: Factors and Coprimes

Formula 6: The number of ways of writing a number N as a product of two co-prime numbers = 2n−1 where 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⎞⎠⎟

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

                                                

1. P is the product of all the factors of 15552.  If P = 12N×M, where M is not a multiple of 12, then find the value of M.  [M and N are positive Integers]
Ans: 15552 = 26×35 
Product of all the factors of 15552 = (26×35)(6+1).(5+1)2=2126×3105 
⇒(22)63×363×342 = 1263×342 
So M = 342

2. Let M be the set of all the distinct factors of the number N=65×52×10,Which are perfect squares.  Find the product of the elements contained in the set M. 
N = 65×52×10 = 26×35×53
Even powers of 2 available: 20,22,24,26
Even powers of 3 available: 30,32,34
Even powers of 5 available: 50,52
Therefore number of factors of the number N that are perfect squares = 4 x 3 x 2 = 24
Product of the elements contained in M 
= 2(2+4+6)×6×3(2+4)×8×52×12=272×348×524