Find the least number of five digits which when divided by 16,24,30 and 32 leaves a remainder 2 in each case .

Options

A. 10084
B. 10071
C. 10082
D. 10002

Answer

To find the least number of five digits which, when divided by 16, 24, 30, and 32, leaves a remainder of 2 in each case, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find a five-digit number \( N \) such that:
\[ N \mod 16 = 2, \quad N \mod 24 = 2, \quad N \mod 30 = 2, \quad N \mod 32 = 2 \]
This implies that \( N - 2 \) is divisible by 16, 24, 30, and 32.

2. **Calculate LCM**: We need to find the least common multiple (LCM) of the numbers 16, 24, 30, and 32.
- Prime factorization:
- \( 16 = 2^4 \)
- \( 24 = 2^3 \times 3^1 \)
- \( 30 = 2^1 \times 3^1 \times 5^1 \)
- \( 32 = 2^5 \)

- Taking the highest power of each prime:
- \( 2^5 \) from 32
- \( 3^1 \) from 24 or 30
- \( 5^1 \) from 30

- Therefore, the LCM is:
\[ \text{LCM} = 2^5 \times 3^1 \times 5^1 = 32 \times 3 \times 5 = 480 \]

3. **Finding the Least Five-Digit Number**: The least five-digit number is 10,000. We need to find the smallest multiple of 480 that is greater than or equal to 10,000.
- Divide 10,000 by 480:
\[ 10,000 \div 480 \approx 20.8333 \]
- The smallest integer greater than 20.8333 is 21. Therefore, we multiply:
\[ 21 \times 480 = 10,080 \]

4. **Adjust for Remainder**: Since we need \( N \) such that \( N \mod 16 = 2 \), we add 2 to our result:
\[ N = 10,080 + 2 = 10,082 \]

5. **Conclusion**: The least five-digit number that meets the criteria is:
\[ \boxed{10,082} \]