7.3. In a bag for a bingo game, there are 90 barrels with numbers from 1 to 90. What is the minimum number of barrels that need to be drawn at random from the bag to guarantee getting a barrel with a number divisible by 3 or 5 (or both 3 and 5)?
Answer: 49. Solution. We will call numbers that are divisible by 3 or 5 desirable. The number of numbers from 1 to 90 that are divisible by 3 is 30 ( =90: 3 ), and those divisible by 5 is 18 ( =90:5). If we add $30+18=48$, then all numbers that are divisible only by 3 and only by 5 will be counted once, and numbers that are divisible by both 3 and 5, i.e., divisible by 15, will be counted twice. The number of such (doubly counted) numbers is $6=90: 15$. Therefore, by subtracting $48-6$, we get 42 desirable numbers. The remaining 48 undesirable numbers ( =90 - 42) could have appeared on the first 48 drawn tokens, but then the 49th drawn token will definitely be desirable.