How many integers 1 to 1000 are multiples of 3 or 9?

Question

I'm stumped about how to solve this, because every multiple of 9 is already a multiple of 3.

Here's what I have so far:

A = 999/3 = 333 B = 999/9 = 111

(A ∩ B) = multiples of 9 from 1 to 1,000, which equals 111

(A ∪ B) = A + B − (A ∩ B) = 333 + 111 − 111 = 333

Answer 1

This seems like a solid application of the inclusion/exclusion principle. You do have the correct answer. Is your apprehension as to how the inclusion/exclusion principle works in particular situations?

Answer 2

You can simply ignore the "or multiples of $9$" for the very reason you state. Inclusion exclusion plays no part. That shows up when you first add and then subtract $111$.

If the question asked about "multiples of $3$ or multiples of $5$" your argument would be necessary.