In a set of first 123 natural numbers find the number of integers, which are divisible by either 2 or 3 or 5.

Options

A. 90
B. 31
C. 33
D. 11

Answer

To solve the problem of finding the number of integers in the first 123 natural numbers that are divisible by either 2, 3, or 5, we can use the principle of Inclusion-Exclusion.

### Step-by-Step Solution:

1. **Count the numbers divisible by 2:**
- The numbers divisible by 2 in the range of 1 to 123 are: 2, 4, 6, ..., 122.
- This forms an arithmetic sequence where:
- First term (a) = 2
- Last term (l) = 122
- Common difference (d) = 2
- The number of terms (n) in this sequence can be found using the formula:
\[ n = \frac{l - a}{d} + 1 \]
- Plugging in the values:
\[ n = \frac{122 - 2}{2} + 1 = \frac{120}{2} + 1 = 60 + 1 = 61 \]
- Therefore, there are **61 numbers divisible by 2**.

2. **Count the numbers divisible by 3:**
- The numbers divisible by 3 in the range of 1 to 123 are: 3, 6, 9, ..., 123.
- This forms an arithmetic sequence where:
- First term (a) = 3
- Last term (l) = 123
- Common difference (d) = 3
- Using the same formula for n:
\[ n = \frac{l - a}{d} + 1 \]
- Plugging in the values:
\[ n = \frac{123 - 3}{3} + 1 = \frac{120}{3} + 1 = 40 + 1 = 41 \]
- Therefore, there are **41 numbers divisible by 3**.

3. **Count the numbers divisible by 5:**
- The numbers divisible by 5 in the range of 1 to 123 are: 5, 10, 15, ..., 120.
- This forms an arithmetic sequence where:
- First term (a) = 5
- Last term (l) = 120
- Common difference (d) = 5
- Using the same formula for n:
\[ n = \frac{l - a}{d} + 1 \]
- Plugging in the values:
\[ n = \frac{120 - 5}{5} + 1 = \frac{115}{5} + 1 = 23 + 1 = 24 \]
- Therefore, there are **24 numbers divisible by 5**.

4. **Count the numbers divisible by combinations of 2, 3, and 5:**
- **Divisible by both 2 and 3 (i.e., 6):**
- Numbers: 6, 12, 18, ..., 120.
- First term (a) = 6, Last term (l) = 120, d = 6.
\[ n = \frac{120 - 6}{6} + 1 = \frac{114}{6} + 1 = 19 + 1 = 20 \]
- **Divisible by both 2 and 5 (i.e., 10):**
- Numbers: 10, 20, 30, ..., 120.
- First term (a) = 10, Last term (l) = 120, d = 10.
\[ n = \frac{120 - 10}{10} + 1 = \frac{110}{10} + 1 = 11 + 1 = 12 \]
- **Divisible by both 3 and 5 (i.e., 15):**
- Numbers: 15, 30, 45, ..., 120.
- First term (a) = 15, Last term (l) = 120, d = 15.
\[ n = \frac{120 - 15}{15} + 1 = \frac{105}{15} + 1 = 7 + 1 = 8 \]
- **Divisible by 2, 3, and 5 (i.e., 30):**
- Numbers: 30, 60, 90, 120.
- First term (a) = 30, Last term (l) = 120, d = 30.
\[ n = \frac{120 - 30}{30} + 1 = \frac{90}{30} + 1 = 3 + 1 = 4 \]

5. **Applying Inclusion-Exclusion Principle:**
- Total = (Count of 2) + (Count of 3) + (Count of 5) - (Count of 2 and 3) - (Count of 2 and 5) - (Count of 3 and 5) + (Count of 2, 3, and 5)
\[ \text{Total} = 61 + 41 + 24 - 20 - 12 - 8 + 4 = 90 \]

### Final Answer:
Thus, the number of integers in the first 123 natural numbers that are divisible by either 2, 3, or 5 is **90**.