We use the sum formula for an AP given the first and last terms: S_n = (n)/(2)(a + l). Substituting n = 180, a = 102, and l = 997, we get S_180 = 90 × 1099 = 98910.

find the sum of all three digit numbers which leaves remainder 2 when divided by 5

Answer: $98910$

Step-by-step solution

Step 1: Identify the first and last three-digit terms

Three-digit numbers range from $100$ to $999$ . The smallest three-digit number leaving remainder $2$ upon division by $5$ is $100 + 2 = 102$ . The largest three-digit number divisible by $5$ is $995$ , so adding $2$ gives the largest three-digit number leaving remainder $2$ , which is $995 + 2 = 997$ .

Step 2: Formulate the Arithmetic Progression

Since each subsequent number that leaves a remainder of $2$ when divided by $5$ increases by $5$ , the sequence forms an Arithmetic Progression (AP) with first term $a = 102$ and common difference $d = 5$ .

Step 3: Find the number of terms

Using the $n^{\text{th}}$ term formula of an AP, $a_n = a + (n - 1)d$ , we substitute $a = 102$ , $d = 5$ , and $a_n = 997$ . Subtracting $102$ from both sides gives $895 = (n - 1)5$ , which leads to $n - 1 = 179$ , so $n = 180$ .

Step 4: Calculate the sum of the AP

We use the sum formula for an AP given the first and last terms: $S_n = \frac{n}{2}(a + l)$ . Substituting $n = 180$ , $a = 102$ , and $l = 997$ , we get $S_{180} = 90 \times 1099 = 98910$ .

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