Find the least number which when divided by 24, 30 and 54 leaves 5 as remainder in each case.

Answer

To find the least number which, when divided by 24, 30, and 54, 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 \( x \) such that:
- \( x \mod 24 = 5 \)
- \( x \mod 30 = 5 \)
- \( x \mod 54 = 5 \)

This means that \( x - 5 \) should be divisible by 24, 30, and 54.

2. **Set Up the Equation**:
Let \( y = x - 5 \). Therefore, we need to find \( y \) such that:
- \( y \mod 24 = 0 \)
- \( y \mod 30 = 0 \)
- \( y \mod 54 = 0 \)

This means \( y \) is a common multiple of 24, 30, and 54.

3. **Find the LCM**:
We need to calculate the Least Common Multiple (LCM) of 24, 30, and 54.

- **Prime Factorization**:
- \( 24 = 2^3 \times 3^1 \)
- \( 30 = 2^1 \times 3^1 \times 5^1 \)
- \( 54 = 2^1 \times 3^3 \)

- **Take the highest power of each prime**:
- For \( 2 \): highest power is \( 2^3 \)
- For \( 3 \): highest power is \( 3^3 \)
- For \( 5 \): highest power is \( 5^1 \)

- **Calculate the LCM**:
\[ \text{LCM} = 2^3 \times 3^3 \times 5^1 = 8 \times 27 \times 5 \]
\[ 8 \times 27 = 216 \]
\[ 216 \times 5 = 1080 \]

4. **Finding \( x \)**:
Since \( y = 1080 \), we have:
\[ x = y + 5 = 1080 + 5 = 1085 \]

5. **Conclusion**:
The least number which when divided by 24, 30, and 54 leaves a remainder of 5 is \( \boxed{1085} \).