Make sure not to double count!

Let's call the set of multiples of $n$ as ${n_i}$ .

Let $\mathrm{X}$ be the set of numbers $1,2,3,...,6000$ .

Then what we want to find is actually " $\mathrm{A - {2_i} \cup {3_i} \cup {5_i}}$ "

We know that $n(A \cup B\cup C) = n(A)+n(B)+n(C) - n(A\cap B) - n(A\cap C) - n(B\cap C) + n(A\cap B\cap C)$

Here , $n(A)=n(2_i) = 3000$ ,

$n(B)=n(3_i) = 2000$ ,

$n(C)=n(5_i)=1200$ ,

$n(A\cap B) = n(6_i) = 1000$ ,

$n(B\cap C)=n(15_i) = 400$ ,

$n(A\cap C)=n(10_i)=600$

$n(A\cap B \cap C) = n(30_i) = 200$

Thus $n({2_i}\cup {3_i} \cup {5_i}) = 2000+3000+1200 - 1000 -400-600 +200 = 4400$

Hence the answer is $n(X) - n(A\cup B \cup C) = 6000-4400 = \boxed{\displaystyle{1600}}$

Out of 6000,
There are 6000/ 2= 3000 multiples of 2.
3000/3= 1000 multiples of 3 and 2000/5 = 400 multiples of 5. Therefore , 6000-(3000 +1000+400) = 1600 are not multiples of 2,3, or 5.

Of the numbers 1 to 6000:

3000 - multiples of 2, 6 (came from 2 x 3), 10 (came from 2 x 5) and 30 (came from 2 x 3 x 5) To obtain the multiples of 2 ONLY, subtract all the existing multiples of 6, 10, and 30 from 1 to 6000 from 2000 (which are 1000, 600 and 200, respectively). 3000 - (1000 + 600 + 200) = 1200 multiples of 2 only.

2000 - multiples of 3, 6 (came from 2 x 3), 15 (came from 3 x 5), and 30 (came from 2 x 3 x 5) To obtain the multiples of 3 ONLY, subtract all the existing multiples of 6, 15, and 30 from 1 to 6000 from 2000 (which are 1000, 400 and 200, respectively) 2000 - (1000 + 400 + 200) = 400 multiples of 3 only.

1200 - multiples of 5, 10 (came from 2 x 5), 15 (came from 3 x 5) and 30 (came from 2 x 3 x 5) To obtain the multiples of 5 ONLY, subtract all the existing multiples of 10, 15, and 30 from 1 to 6000 from 2000 (which are 600, 400 and 200, respectively). This gives 0 multiples of 5 only.

Add all the differences: 1200 + 400 = 1600