You have been asked to select a positive integer N which is less than 1000 , such that it is either a multiple of 4, or a multiple of 6, or an odd multiple of 9. The number of such numbers is

In question it is asked to select a positive integer N which is less than 1000

So, basically we have to select from 1 to 999

Now, number of integers which is a multiple of 4 = $$\left[\dfrac{999}{4}\right]=249$$

number of integers which is a multiple of 6 = $$\left[\dfrac{999}{6}\right]=166$$

number of integers which is a multiple of 9 = $$\left[\dfrac{999}{9}\right]=111$$

So, the number of odd multiples of 9 will be = $$\left[\dfrac{111}{2}\right]+1=56$$ (we are adding 1 because the last number 999 is also an odd multiple of 9)

So, the number of possible integers are = $$249+166+56=471$$

But we also have to eliminate integers that are multiples of both 4 and 6.

Multiples of both 4 and 6 means multiple of 12 (as 12 is the L.C.M of 4 and 6)

So, number of integers which is a multiple of 12 = $$\left[\dfrac{999}{12}\right]=83$$

So, required number of integers = 471-83=388