Find the sum of ten's and hundred's place digit of smallest number which when divided by 5,6,7,8,9 Leaves remainder 2 in each case ?

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to find the smallest number that leaves a remainder of 2 when divided by 5, 6, 7, 8, and 9. Then, we will find the sum of the digits in the ten's and hundred's places of that number.

### Step-by-Step Solution:

1. **Find the LCM of the Divisors**:
We need to find the least common multiple (LCM) of the numbers 5, 6, 7, 8, and 9.
- The prime factorization of each number is:
- 5 = 5
- 6 = 2 × 3
- 7 = 7
- 8 = 2^3
- 9 = 3^2
- The LCM is found by taking the highest power of each prime:
- LCM = 2^3 × 3^2 × 5 × 7 = 8 × 9 × 5 × 7

2. **Calculate the LCM**:
- First, calculate 8 × 9 = 72
- Then, calculate 72 × 5 = 360
- Finally, calculate 360 × 7 = 2520
- So, LCM(5, 6, 7, 8, 9) = 2520

3. **Adjust for the Remainder**:
Since we need a number that leaves a remainder of 2, we add 2 to the LCM:
- Smallest number = 2520 + 2 = 2522

4. **Identify the Digits**:
Now, we need to find the digits in the ten's and hundred's places of the number 2522:
- The hundred's place digit is 5 (in 2522).
- The ten's place digit is 2 (in 2522).

5. **Calculate the Sum**:
- Sum of the ten's and hundred's place digits = 5 + 2 = 7

### Final Answer:
The sum of the ten's and hundred's place digits of the smallest number is **7**.

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