Multiples Problem Question:
Anna writes the first 1000 positive integers. She then circles the even ones with a green pen.
Bob circles the multiples of three in red. Cindy circles the multiples of five in blue. How
many numbers are circled exactly twice?
My Solution:
1000 / 6 + 1000/ 15 + 1000/10 - 1000/ 30 = 299. However the correct answer is 233. What did I do wrong?
 A: Let $A_2,A_3,A_5,A_6,A_{10},A_{15},A_{30}$ represent the sets of integers between $1$ and $1000$ (inclusive) which are multiples of $2,3,5,\dots$ respectively.
You are tasked with calculating $|(A_6\cup A_{10}\cup A_{15})\setminus A_{30}|$
Notice that $A_6 = \{6,12,18,24,\color{red}{30},36,\dots,\color{red}{60},\dots\}$
$A_{10}=\{10,20,\color{red}{30},40,50,\color{red}{60},\dots\}$
$A_{15}=\{15,\color{red}{30},45,\color{red}{60},\dots\}$
and $A_{30}=\{30,60,90,\dots\}$
You correctly counted $|A_6|=\lfloor\frac{1000}{6}\rfloor=166$,$|A_{10}|=\lfloor\frac{1000}{10}\rfloor=100$,$|A_{15}|=\lfloor\frac{1000}{15}\rfloor=66$
By adding these together however, you will have let the numbers $30,60,90,\dots$, count in your summation multiple times.  In fact, you will have counted $30$ three times, $60$ three times, etc... once in $A_6$, once in $A_{10}$, and once in $A_{30}$.
We didn't want to count them any times.  To correct the count, we must subtract three times the size of $A_{30}$, giving 
$$\left\lfloor\frac{1000}{6}\right\rfloor+\left\lfloor\frac{1000}{10}\right\rfloor+\left\lfloor\frac{1000}{15}\right\rfloor-3\left\lfloor\frac{1000}{30}\right\rfloor=233$$
A: We use your strategy. There are $\left\lfloor \frac{1000}{6}\right\rfloor$ numbers in our interval that are divisible by $6$. We can get similar expressions for the number of numbers  divisible by $15$, and for the number of numbers divisible by $10$. If we add up, we have counted three times the numbers divisible by $30$. But we must count these $0$ times. So we must subtract $3\left\lfloor \frac{1000}{30}\right\rfloor$. 
