How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?

Question

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?

Answer 1

Your answer is incorrect. While it is true that there are positive odd integers less than or equal to $200$, what you have actually calculated with the expression $$\left\lfloor \frac{200}{2} \right\rfloor$$ is the number of even integers less than or equal to $200$. It just so happens that $200 - 100 = 100$. Similarly, there are $$\left\lfloor \frac{200}{2 \cdot 3} \right\rfloor = 33$$ positive integers less than or equal to $200$ that are divisible by both $2$ and $3$, $$\left\lfloor \frac{200}{2 \cdot 5} \right\rfloor = 20$$ positive integers less than or equal to $200$ that are divisible by both $2$ and $5$, and $$\left\lfloor \frac{200}{2 \cdot 3 \cdot 5} \right\rfloor = 6$$ positive integers less than or equal to $200$ that are divisible by $2$, $3$, and $5$.

However, we can work with these numbers.

Let $A$ be the set of positive odd integers less than or equal to $200$ which are odd; let $B$ be the set of positive integers less than or equal to $200$ which are multiples of $3$; let $C$ be the set of positive integers less than or equal to $200$ which are multiples of $5$. Then \begin{align*} |A| & = 200 - \left\lfloor \frac{200}{2} \right\rfloor = 100\\ |B| & = \left\lfloor \frac{200}{3} \right\rfloor = 66\\ |C| & = \left\lfloor \frac{200}{5} \right\rfloor = 40\\ |A \cap B| & = 66 - \left\lfloor \frac{200}{2 \cdot 3} \right\rfloor = 66 - 33 = 33\\ |A \cap C| & = 40 - \left\lfloor \frac{200}{2 \cdot 5} \right\rfloor = 40 - 20 = 20\\ |B \cap C| & = \left\lfloor \frac{200}{3 \cdot 5} \right\rfloor = 13\\ |A \cap B \cap C| & = 13 - \left\lfloor \frac{200}{2 \cdot 3 \cdot 5} \right\rfloor = 7 \end{align*} where we obtain $|A \cap B|$ by subtracting the number of even multiples of $3$ less than or equal to $200$ from the number of positive integer multiples of $3$ which are at most $200$, $|A \cap C|$ by subtracting the number of even multiples of $5$ less than or equal to $200$ from the number of positive integer multiples of $5$ which are at most $200$, and $|A \cap B \cap C|$ by subtracting the number of even multiples of $3$ and $5$ less than or equal to $200$ from the number of positive integer multiples of $3$ and $5$ less than or equal to $200$.

Hence, by the Inclusion-Exclusion Principle, the number of positive integers less than or equal to $200$ which are odd or divisible by $3$ or divisible by $5$ is $$100 + 66 + 40 - 33 - 20 - 13 + 7 = 147$$

Answer 2

In this simple case you don't really need the Inclusion/Exclusion principle $-$ you can do it 'by hand'.

First, count the even numbers between $1$ and $30$ that are divisible by $3$ or $5$: these are $6,10,12,18,20,24,$ and $30$. So there are $7$ of them. (Or you could use Inclusion/Exclusion on this part only: $\frac{30}{6}+\frac{30}{10}-\frac{30}{30}=7$.)

Next, note that this pattern repeats exactly every $30$ numbers; so between $1$ and $210$, there are $49$ of them. Subtract $204$ and $210$ to get $47$ between $1$ and $200$.

Now add the odd numbers to get $147$.

Answer 3

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).