To find how many of the first 100 natural numbers are divisible by 2, 3, or 5, we can use the principle of inclusion-exclusion. Here’s a step-by-step solution:
### Step 1: Count the multiples of each number
1. **Count multiples of 2**:
- The multiples of 2 up to 100 are: 2, 4, 6, ..., 100.
- This is an arithmetic series where the first term \(a = 2\) and the last term \(l = 100\) with a common difference \(d = 2\).
- The number of terms \(n\) can be calculated using the formula for the nth term of an arithmetic series:
\[
n = \frac{l - a}{d} + 1 = \frac{100 - 2}{2} + 1 = 50
\]
- So, there are 50 multiples of 2.
2. **Count multiples of 3**:
- The multiples of 3 up to 100 are: 3, 6, 9, ..., 99.
- Here, \(a = 3\), \(l = 99\), and \(d = 3\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{99 - 3}{3} + 1 = 33
\]
- So, there are 33 multiples of 3.
3. **Count multiples of 5**:
- The multiples of 5 up to 100 are: 5, 10, 15, ..., 100.
- Here, \(a = 5\), \(l = 100\), and \(d = 5\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{100 - 5}{5} + 1 = 20
\]
- So, there are 20 multiples of 5.
### Step 2: Count the multiples of combinations of the numbers
1. **Count multiples of 6 (2 and 3)**:
- The multiples of 6 up to 100 are: 6, 12, 18, ..., 96.
- Here, \(a = 6\), \(l = 96\), and \(d = 6\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{96 - 6}{6} + 1 = 16
\]
- So, there are 16 multiples of 6.
2. **Count multiples of 10 (2 and 5)**:
- The multiples of 10 up to 100 are: 10, 20, 30, ..., 100.
- Here, \(a = 10\), \(l = 100\), and \(d = 10\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{100 - 10}{10} + 1 = 10
\]
- So, there are 10 multiples of 10.
3. **Count multiples of 15 (3 and 5)**:
- The multiples of 15 up to 100 are: 15, 30, 45, ..., 90.
- Here, \(a = 15\), \(l = 90\), and \(d = 15\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{90 - 15}{15} + 1 = 6
\]
- So, there are 6 multiples of 15.
4. **Count multiples of 30 (2, 3, and 5)**:
- The multiples of 30 up to 100 are: 30, 60, 90.
- Here, \(a = 30\), \(l = 90\), and \(d = 30\).
- The number of terms \(n\) is:
\[
n = \frac{l - a}{d} + 1 = \frac{90 - 30}{30} + 1 = 3
\]
- So, there are 3 multiples of 30.
### Step 3: Apply the principle of inclusion-exclusion
Using the inclusion-exclusion principle, we can find the total count of numbers divisible by 2, 3, or 5:
\[
N(2 \cup 3 \cup 5) = N(2) + N(3) + N(5) - N(6) - N(10) - N(15) + N(30)
\]
Substituting the values:
\[
N(2 \cup 3 \cup 5) = 50 + 33 + 20 - 16 - 10 - 6 + 3
\]
Calculating this gives:
\[
N(2 \cup 3 \cup 5) = 103 - 32 + 3 = 74
\]
### Final Answer
Thus, the number of natural numbers from 1 to 100 that are divisible by 2, 3, or 5 is **74**.
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