How many four digit numbers have this property? How many four digit numbers are multiples of 3 but not the multiples of 2 or 5 or multiples of 2 but not the multiples of 3 or 5 or multiples of 5 but not the multiples of 2 or 3?
 A: You should prove it using the Inclusion–exclusion principle:
Let's calculate how many 4 digits number are multiples of 2 but not multiples of 3 and 5.
There are $\frac{9998-1000}{2} + 1=4500$ 4 digits numbers that are multiples of 2
There are $\frac{9996-1002}{6} + 1=1500$ 4 digits numbers that are multiples of 2 and 3
There are $\frac{9990-1000}{10}+ 1=900$ 4 digits numbers that are multiples of 2 and 5
There are $\frac{9990-1020}{30} + 1=300$ 4 digits numbers that are multiples of 2, 3 and 5
There fore there are $4500 - 1500 - 900 + 300 = 2400$ 4 digits number are multiples of 2 but not multiples of 3 and 5.
Do the same for 3 and 5.
A: There are $9000$ four-digit numbers. Half of them, i.e. $4500$ are divisible by $2$. Of these, again, a third ($4500/3 = 1500$) are divisible by $3$, and a fifth ($4500/5 = 900$) are divisible by $5$, so we subtract each of those to get $2100$. However, the numbers that are divisible by both three and five at the same time (i.e. divisible by $15$, there are $300$ of those) were subtracted twice when we only wanted to subtract them once, so we need to counter that by adding $300$ back. Therefore there are $2400$ four-digit numbers that are divisible by $2$, and at the same time not divisible by neither $3$ nor $5$. (This way of counting, by subtracting each of them and adding back what was double-counted, is called the inclusion-exclusion principle, and it some times makes seemingly difficult counting easy.)
The two other cases are done exactly the same way to get $1200$ for $3$ and $600$ for $5$. Add all those together to get the final answer of $4200$.
