What is the sum of all two digit numbers which when divided by 3 leaves 2 as the remainder ?
(a)1565
(b)1585
(c)1635
(d)1655

Options

A. 1565
B. 1585
C. 1635
D. 1655

Answer

To find the sum of all two-digit numbers that leave a remainder of 2 when divided by 3, we can follow these steps:

### Step 1: Identify the first two-digit number
The smallest two-digit number is 10. We need to find the smallest two-digit number that leaves a remainder of 2 when divided by 3.

- When we divide 10 by 3, we get a remainder of 1.
- When we divide 11 by 3, we get a remainder of 2.

Thus, the first two-digit number that leaves a remainder of 2 when divided by 3 is **11**.

### Step 2: Identify the last two-digit number
The largest two-digit number is 99. We need to find the largest two-digit number that leaves a remainder of 2 when divided by 3.

- When we divide 99 by 3, we get a remainder of 0.
- When we divide 98 by 3, we get a remainder of 2.

Thus, the last two-digit number that leaves a remainder of 2 when divided by 3 is **98**.

### Step 3: Identify the sequence of numbers
The two-digit numbers that leave a remainder of 2 when divided by 3 form an arithmetic sequence:
- First term (A) = 11
- Last term (L) = 98
- Common difference (d) = 3

The sequence is: 11, 14, 17, ..., 98.

### Step 4: Find the number of terms (n)
To find the number of terms in the sequence, we use the formula for the nth term of an arithmetic sequence:
\[ n = \frac{L - A}{d} + 1 \]

Substituting the values:
\[ n = \frac{98 - 11}{3} + 1 \]
\[ n = \frac{87}{3} + 1 \]
\[ n = 29 + 1 = 30 \]

So, there are **30 terms** in this sequence.

### Step 5: Calculate the sum of the sequence
The sum \( S_n \) of the first n terms of an arithmetic sequence can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (A + L) \]

Substituting the values:
\[ S_{30} = \frac{30}{2} \times (11 + 98) \]
\[ S_{30} = 15 \times 109 \]
\[ S_{30} = 1635 \]

Thus, the sum of all two-digit numbers which when divided by 3 leaves 2 as the remainder is **1635**.

### Final Answer
The answer is **(c) 1635**.

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