The least number which when divided by 12, 16 and 18 leaves 5 as remainder in each case. Find the number.

Options

A. 139
B. 144
C. 149
D. 154

Answer

To find the least number which, when divided by 12, 16, and 18, leaves a remainder of 5 in each case, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find a number \( N \) such that:
- \( N \mod 12 = 5 \)
- \( N \mod 16 = 5 \)
- \( N \mod 18 = 5 \)

This means that \( N - 5 \) should be divisible by 12, 16, and 18.

2. **Set Up the Equation**: Let \( M = N - 5 \). Then, we need to find \( M \) such that:
- \( M \mod 12 = 0 \)
- \( M \mod 16 = 0 \)
- \( M \mod 18 = 0 \)

This means \( M \) must be a common multiple of 12, 16, and 18.

3. **Find the LCM**: We need to calculate the Least Common Multiple (LCM) of 12, 16, and 18.

- **Prime Factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)
- \( 18 = 2^1 \times 3^2 \)

- **Take the highest powers of all prime factors**:
- For \( 2 \): highest power is \( 2^4 \) (from 16)
- For \( 3 \): highest power is \( 3^2 \) (from 18)

- **Calculate LCM**:
\[ \text{LCM} = 2^4 \times 3^2 = 16 \times 9 = 144 \]

4. **Calculate \( N \)**: Since \( M = N - 5 \), we have:
\[ M = 144 \]
Therefore,
\[ N = M + 5 = 144 + 5 = 149 \]

5. **Final Answer**: The least number which when divided by 12, 16, and 18 leaves a remainder of 5 is:
\[ \boxed{149} \]