To find the number of natural numbers less than 107 that are exactly divisible by 7, we can follow these steps:
### Step-by-Step Solution
1. **Identify the Range**:
We need to find natural numbers less than 107. The natural numbers start from 1, so we consider the range from 1 to 106.
2. **Determine the First and Last Natural Numbers Divisible by 7**:
- The first natural number divisible by 7 is 7 (since \(7 \times 1 = 7\)).
- To find the largest natural number less than 107 that is divisible by 7, we can divide 107 by 7 and take the floor of the result:
\[
\text{Largest integer } k \text{ such that } 7k < 107 \implies k = \left\lfloor \frac{107}{7} \right\rfloor = 15
\]
- Therefore, the largest number less than 107 that is divisible by 7 is:
\[
7 \times 15 = 105
\]
3. **List the Sequence of Natural Numbers Divisible by 7**:
The natural numbers divisible by 7 from 1 to 106 are:
\[
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105
\]
4. **Count the Numbers**:
We can see that these numbers form an arithmetic sequence where:
- First term \(a = 7\)
- Common difference \(d = 7\)
- Last term \(l = 105\)
We can use the formula for the \(n\)-th term of an arithmetic sequence:
\[
l = a + (n-1)d
\]
Substituting the known values:
\[
105 = 7 + (n-1) \cdot 7
\]
Simplifying this:
\[
105 - 7 = (n-1) \cdot 7 \implies 98 = (n-1) \cdot 7
\]
Dividing both sides by 7:
\[
n - 1 = \frac{98}{7} = 14 \implies n = 14 + 1 = 15
\]
5. **Conclusion**:
Therefore, the number of natural numbers less than 107 that are exactly divisible by 7 is **15**.
