How many of the integers between $100$ and $200$ are divisible by $3$ or divisible by $2$ but not by $5$?
is the range of integer 200-100+1=101 or 100?
$A_5$ number that is divisible by 5
\begin{align*}
A_1 & = \left\lfloor{\frac{101}{3}} \right\rfloor = 33 && \text{(divisible by $3$)}\\
A_2 & = \left\lfloor{\frac{101}{2}} \right\rfloor = 50 && \text{(divisible by $2$)}\\
\\
| A_1 \cap A_5 | & = \left\lfloor{\frac{101}{3 \cdot 5}}\right\rfloor = 6\\
| A_2 \cap A_5 | & = \left\lfloor{\frac{101}{2 \cdot 5}}\right\rfloor = 10\\
\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{101}{2 \cdot 3 \cdot 5}}\right\rfloor = 3
\end{align*}
Therefore, by the principle exclusion inclusion theorem
$= 50 + 33 – (6 + 10) + 3 =70 $
Is this right?
Im trying to count one by one such as, and enumerate
for $100-130$ , number that divisible by 2,3 but not 5
$\{102,104,106,111,108,112,114,116,118,122,123,124,126,128\}=14$ number
$14*3=42 $ number $ (100-190)$
for
$191-200= ${ $192,194,196,198\}=4$ number
$42+4=46 $
i used program to check:
102 104 106 108 111 112 114 116 117 118 122 123 124 126 128 129 132 134 136 138 141 142 144 146 147 148 152 153 154 156 158 159 162 164 166 168 171 172 174 176 177 178 182 183 184 186 188 189 192 194 196 198
total=52
which one is right?
Best Answer
Your notation is not consistent. It looks like you are using both $A_3$ and $A_5$ to denote numbers divisible by $5$.
Let $A$ denote the set of positive integers satisfying $100 \leq n \leq 200$ which are divisible by $2$; let $B$ denote the set of positive integers satisfying $100 \leq n \leq 200$ which are divisible by $3$; let $C$ denote the set of positive integers satisfying $100 \leq n \leq 200$ which are divisible by $5$. We wish to find $$|(A \cup B) - C| = |A| + |B| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$ You took a shortcut that does not always work. The number of multiples of an integer within $101$ consecutive integers is not always the same as it is for the first $101$ positive integers. In particular, there are $50$ positive even integers between $1$ and $101$ inclusive but $51$ positive even integers between $100$ and $200$ inclusive since both $100$ and $200$ are even. To address this issue, we subtract the number of multiples that are at most $99$ from the number of multiples that are at most $200$. \begin{align*} |A| & = \left\lfloor \frac{200}{2} \right\rfloor - \left\lfloor \frac{99}{2} \right\rfloor = 100 - 49 = 51\\ |B| & = \left\lfloor \frac{200}{3} \right\rfloor - \left\lfloor \frac{99}{3} \right\rfloor = 66 - 33 = 33\\ |A \cap B| & = \left\lfloor \frac{200}{2 \cdot 3} \right\rfloor - \left\lfloor \frac{99}{2 \cdot 3} \right\rfloor = 33 - 16 = 17\\ |A \cap C| & = \left\lfloor \frac{200}{2 \cdot 5} \right\rfloor - \left\lfloor \frac{99}{2 \cdot 5} \right\rfloor = 20 - 9 = 11\\ |B \cap C| & = \left\lfloor \frac{200}{3 \cdot 5} \right\rfloor - \left\lfloor \frac{99}{3 \cdot 5} \right\rfloor = 13 - 6 = 7\\ |A \cap B \cap C| & = \left\lfloor \frac{200}{2 \cdot 3 \cdot 5} \right\rfloor - \left\lfloor \frac{99}{2 \cdot 3 \cdot 5} \right\rfloor = 6 - 3 = 3 \end{align*} Hence, the number of positive integers between $100$ and $200$ which are divisible by $2$ or $3$ but not by $5$ is $$51 + 33 - 17 - 11 - 7 + 3 = 52$$