Question

how many ways can 1200 be expressed as a product of two co primes

Answer 1

Coprime factors have disjoint set of prime factors . I mean, if P is a set . And number of elements = n ∴ number of subsets = 2ⁿ . Then, possible number of coprime number = 2ⁿ/2 = 2ⁿ⁻¹

Here, 1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 2⁴ × 3¹ × 5²
∴ set P = { 2, 3 , 5} , n = 3
Then, number of possible ways as a product of two coprimes = 2ⁿ⁻¹
= 2³⁻¹ = 2² = 4
Hence, number of ways = 4
e.g., {2⁴, 3.5² } , {2⁴.3, 5²} ,{2⁴.5², 3} , {2⁴.3.5², 1 }

Answer 2

Solution :-

Prime factors of 1200 = 2*2*2*2*3*5*5

= 2⁴ × 3¹ × 5² 

The number of factors of ways writing a number N as a product of two co-primes = 2ⁿ⁻¹ , where N = the number of prime factors of a number.

Number of prime factors of 1200 are 2, 3 and 5

So, total number of prime factors of 1200 = 3

Therefore, total number of ways of writing 1200 as a product of 2 co-primes

= 2ⁿ⁻¹

= 2³⁻¹

= 2²

= 4

So, 1200 can be written in 4 ways as a product of two co-primes.

Answer.