The number of lying between 1 and 200 which are divisible by either of 2,3 or 5 is:

Options

A. 146
B. 145
C. 158
D. none the these

Answer

To find the number of integers between 1 and 200 that are divisible by either 2, 3, or 5, we can use the principle of Inclusion-Exclusion.

### Step 1: Count the multiples of each number
1. **Count multiples of 2**:
- The multiples of 2 between 1 and 200 are: 2, 4, 6, ..., 200.
- This is an arithmetic sequence where the first term \( a = 2 \) and the last term \( l = 200 \) with a common difference \( d = 2 \).
- The number of terms \( n \) can be found using the formula:
\[ n = \frac{l - a}{d} + 1 = \frac{200 - 2}{2} + 1 = 100 \]

2. **Count multiples of 3**:
- The multiples of 3 between 1 and 200 are: 3, 6, 9, ..., 198.
- Here, \( a = 3 \), \( l = 198 \), and \( d = 3 \).
- The number of terms \( n \) is:
\[ n = \frac{198 - 3}{3} + 1 = 66 \]

3. **Count multiples of 5**:
- The multiples of 5 between 1 and 200 are: 5, 10, 15, ..., 200.
- Here, \( a = 5 \), \( l = 200 \), and \( d = 5 \).
- The number of terms \( n \) is:
\[ n = \frac{200 - 5}{5} + 1 = 40 \]

### Step 2: Count the overlaps (multiples of combinations)
1. **Count multiples of 6 (2 and 3)**:
- The multiples of 6 between 1 and 200 are: 6, 12, 18, ..., 198.
- Here, \( a = 6 \), \( l = 198 \), and \( d = 6 \).
- The number of terms \( n \) is:
\[ n = \frac{198 - 6}{6} + 1 = 33 \]

2. **Count multiples of 10 (2 and 5)**:
- The multiples of 10 between 1 and 200 are: 10, 20, 30, ..., 200.
- Here, \( a = 10 \), \( l = 200 \), and \( d = 10 \).
- The number of terms \( n \) is:
\[ n = \frac{200 - 10}{10} + 1 = 20 \]

3. **Count multiples of 15 (3 and 5)**:
- The multiples of 15 between 1 and 200 are: 15, 30, 45, ..., 195.
- Here, \( a = 15 \), \( l = 195 \), and \( d = 15 \).
- The number of terms \( n \) is:
\[ n = \frac{195 - 15}{15} + 1 = 13 \]

4. **Count multiples of 30 (2, 3, and 5)**:
- The multiples of 30 between 1 and 200 are: 30, 60, 90, ..., 180.
- Here, \( a = 30 \), \( l = 180 \), and \( d = 30 \).
- The number of terms \( n \) is:
\[ n = \frac{180 - 30}{30} + 1 = 7 \]

### Step 3: Apply Inclusion-Exclusion Principle
Using the Inclusion-Exclusion principle, we can find the total count:
\[ N = (N_2 + N_3 + N_5) - (N_{2,3} + N_{2,5} + N_{3,5}) + N_{2,3,5} \]
Substituting the values we found:
\[ N = (100 + 66 + 40) - (33 + 20 + 13) + 7 \]
Calculating this gives:
\[ N = 206 - 66 + 7 = 147 \]

### Final Answer
The number of integers between 1 and 200 that are divisible by either 2, 3, or 5 is **147**.

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