How many three digit numbers are divisible by 7 ?

Answer

To find how many three-digit numbers are divisible by 7, we can follow these steps:

### Step 1: Identify the smallest three-digit number
The smallest three-digit number is 100.

### Step 2: Find the smallest three-digit number divisible by 7
To find the smallest three-digit number that is divisible by 7, we divide 100 by 7 and round up to the nearest whole number:
\[ 100 \div 7 \approx 14.2857 \]
Taking the integer part, we have 14. Now, we multiply 14 by 7 to find the largest multiple of 7 that is less than 100:
\[ 14 \times 7 = 98 \]
Since 98 is not a three-digit number, we take the next whole number, which is 15:
\[ 15 \times 7 = 105 \]
Thus, the smallest three-digit number divisible by 7 is **105**.

### Step 3: Identify the largest three-digit number
The largest three-digit number is 999.

### Step 4: Find the largest three-digit number divisible by 7
To find the largest three-digit number that is divisible by 7, we divide 999 by 7:
\[ 999 \div 7 \approx 142.7143 \]
Taking the integer part, we have 142. Now, we multiply 142 by 7:
\[ 142 \times 7 = 994 \]
Thus, the largest three-digit number divisible by 7 is **994**.

### Step 5: Use the formula for the number of terms in an arithmetic progression
Now we have the first term \( a = 105 \) and the last term \( l = 994 \), with a common difference \( d = 7 \).

The formula for the \( n \)-th term of an arithmetic progression is given by:
\[ l = a + (n - 1) \cdot d \]
Substituting the known values:
\[ 994 = 105 + (n - 1) \cdot 7 \]
Now, simplify and solve for \( n \):
\[ 994 - 105 = (n - 1) \cdot 7 \]
\[ 889 = (n - 1) \cdot 7 \]
\[ n - 1 = \frac{889}{7} \]
\[ n - 1 = 127 \]
\[ n = 127 + 1 = 128 \]

### Conclusion
There are **128 three-digit numbers** that are divisible by 7.

---