Find the sum of the all the three digit numbers, which leave the remainder 2 when divided by 5.

Answer

To find the sum of all three-digit numbers that leave a remainder of 2 when divided by 5, we can follow these steps:

### Step 1: Identify the first and last three-digit numbers that leave a remainder of 2 when divided by 5.
- The smallest three-digit number is 100. The first three-digit number that leaves a remainder of 2 when divided by 5 is 102 (since 102 mod 5 = 2).
- The largest three-digit number is 999. The largest three-digit number that leaves a remainder of 2 when divided by 5 is 997 (since 997 mod 5 = 2).

### Step 2: Identify the sequence of numbers.
- The sequence of three-digit numbers that leave a remainder of 2 when divided by 5 is: 102, 107, 112, ..., 997.
- This is an arithmetic progression (AP) where the first term \( a = 102 \) and the common difference \( d = 5 \).

### Step 3: Find the number of terms in the sequence.
- The general formula for the \( n \)-th term of an AP is given by:
\[ a_n = a + (n-1)d \]
- Setting \( a_n = 997 \) (the last term), we can solve for \( n \):
\[ 997 = 102 + (n-1) \cdot 5 \]
\[ 997 - 102 = (n-1) \cdot 5 \]
\[ 895 = (n-1) \cdot 5 \]
\[ n-1 = \frac{895}{5} = 179 \]
\[ n = 179 + 1 = 180 \]
- Therefore, there are 180 terms in this sequence.

### Step 4: Calculate the sum of the arithmetic progression.
- The sum \( S_n \) of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (a + a_n) \]
- Substituting the values we found:
\[ S_{180} = \frac{180}{2} \cdot (102 + 997) \]
\[ S_{180} = 90 \cdot (1099) \]
\[ S_{180} = 98910 \]

### Conclusion:
The sum of all three-digit numbers that leave a remainder of 2 when divided by 5 is **98910**.

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