<s>[INST] How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$? How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?
\begin{align*}
 A_1 & = \left\lfloor{\frac{200}{3}} \right\rfloor = 66 && \text{(divisible by $3$)}\\
 A_2 & = \left\lfloor{\frac{200}{5}} \right\rfloor = 40 && \text{(divisible by $5$)}\\
 A_3 & = \left\lfloor{\frac{200}{2}} \right\rfloor = 100 && \text{(odd)}\\
| A_1 \cap A_2 | & = \left\lfloor{\frac{200}{3 \cdot 5}}\right\rfloor = 13\\
| A_1 \cap A_3 | & = \left\lfloor{\frac{200}{3 \cdot 2}}\right\rfloor = 33\\
| A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2}} \right\rfloor= 20\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2 \cdot 3}}\right\rfloor = 6
\end{align*}
Therefore, by the principle exclusion inclusion theorem
 $= 66 + 40 + 100- (13 + 33 + 20) + 6 = 146$
Is this logically right?
 [/INST] There is a simplier way to count the number.
First let separate the even and odd numbers. There are $100$ odd numbers between $1$ and $200$. All of them are part of the answer. We will continue only with even number. We will use the inclusion-exclusion principle, as you did.
How many even number between $2$ and $200$ are divisible by $3$? Every third number is divisible by $3$ ( divisible by $6$, actually).
$$\left\lfloor\frac{200}6\right\rfloor=33$$
How many even number between $2$ and $200$ are divisible by $5$? Every fifth number is divisible by $5$ ( divisible by $10$, actually).
$$\left\lfloor\frac{200}{10}\right\rfloor=20$$
We counted twice the even numbers that where divisible by $15$. How many even number between $2$ and $200$ are divisible by $15$? Every fifteenth number is divisible by $15$ ( divisible by $30$, actually).
$$\left\lfloor\frac{200}{30}\right\rfloor=6$$
Final answer
$$100+33+20-6=147$$
My answer is one more than yours. You have a mistake in your last intersection. Your formula
$$|A_1\cap A_2\cap A_3|=\left\lfloor\frac{200}{2\cdot3\cdot5}\right\rfloor$$
counts how many are divisible by $30$, but we need how many have a remainder of $15$ when divided by $30$. And there is one more. It could calculated like this.
$$|A_1\cap A_2\cap A_3|=\left\lfloor\frac{200-15}{2\cdot3\cdot5}\right\rfloor+1$$
Substract $15$ to every numbers and forget the negative ones. We now have $185$ numbers. How many are divisible by $30$? We add one to count the number $15$ (which became $0$ and wasn't accounted for).
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