Question

how many numbers between 6000 and 9000 are co prime to 9000

Answer 1

This question is based on Euler function.
according to Euler's concept , if a number n is in such a way that $n=p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$
then, $\phi(n)=n\{1-\frac{1}{p_1}\}\{1-\frac{1}{p_2}\}.....\{1-\frac{1}{p_k}\}$
where $\phi(n)$ is the number of positive integers m, n with g.c.d(m,n) = 1

here, number of coprime upto 9000 :
n = 9000 = (2×5)³×3²
n = 2³ × 3² × 5³
$\phi(9000)=9000(1-1/2)(1-1/3)(1-1/5)$
= 9000 × 1/2 × 2/3 × 4/5
= 9000 × 8/30
= 2400

number of coprime upto 6000 :
n = 6000 = (2×5)³ × 2 × 3
n = 2⁴ × 3¹ × 5³
$\phi(6000)=6000(1-1/2)(1-1/3)(1-1/5)$
=6000 × 1/2 × 2/3 × 4/5
= 1600

hence numbers between 6000 to 9000 are coprime to 9000 = 2400 - 1600= 800

Answer 2

Answer:

800

Step-by-step explanation:

First of all taking the factors for 6000 and 9000 we have

 factors for 6000 = 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5

  factors for 9000 = 2 x 2 x 2 x 5 x 5 x 5 x 3 x 3

By taking hcf we get 2 x 3 x 5

Now applying Euler's rule that is

n(1 - 1/x1)(1 - 1/x2)-------------(1 - 1/xn)

9000(1 - 1/2)(1 - 1/3)(1 - 1/5) - 6000(1 - 1/2)(1 - 1/3)(1 - 1/5)

9000(1/2)(2/3)(4/5) - 6000(1/2)(2/3)(4/5)

9000(8/30) - 6000(8/30)

8/30(9000 - 6000)

8/30 x 3000 = 800

800 numbers are co prime to 9000