The sum of all the 3-digit numbers, each of which on division by 5 leaves remainder 3

Options

A. 180
B. 1550
C. 6995
D. 99090

Answer

To find the sum of all the 3-digit numbers that leave a remainder of 3 when divided by 5, we can follow these steps:

### Step 1: Identify the first and last 3-digit numbers that leave a remainder of 3 when divided by 5.
- The smallest 3-digit number is 100. To find the first 3-digit number that leaves a remainder of 3 when divided by 5, we can check:
- 100 ÷ 5 = 20 remainder 0
- 101 ÷ 5 = 20 remainder 1
- 102 ÷ 5 = 20 remainder 2
- **103 ÷ 5 = 20 remainder 3** (This is our first number)

- The largest 3-digit number is 999. To find the largest 3-digit number that leaves a remainder of 3 when divided by 5, we can check:
- 999 ÷ 5 = 199 remainder 4
- 998 ÷ 5 = 199 remainder 3 (This is our last number)

### Step 2: List the sequence of numbers.
- The sequence of numbers that leave a remainder of 3 when divided by 5 is:
- 103, 108, 113, ..., 998
- This forms an arithmetic progression (AP) where:
- First term (a) = 103
- Common difference (d) = 5
- Last term (l) = 998

### Step 3: Find the number of terms (n) in the sequence.
- The nth term of an AP can be found using the formula:
\[ l = a + (n-1) \cdot d \]
Plugging in the values:
\[ 998 = 103 + (n-1) \cdot 5 \]
\[ 998 - 103 = (n-1) \cdot 5 \]
\[ 895 = (n-1) \cdot 5 \]
\[ n-1 = \frac{895}{5} = 179 \]
\[ n = 179 + 1 = 180 \]

### Step 4: Calculate the sum of the AP.
- The sum (S) of the first n terms of an AP is given by the formula:
\[ S_n = \frac{n}{2} \cdot (a + l) \]
Substituting the values we found:
\[ S_{180} = \frac{180}{2} \cdot (103 + 998) \]
\[ S_{180} = 90 \cdot 1101 \]
\[ S_{180} = 99090 \]

### Final Answer:
The sum of all the 3-digit numbers that leave a remainder of 3 when divided by 5 is **99090**.

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