The smallest number , which when increased by 5 is divisible by each of 24, 32, 36 and 54, is

Options

A. 69
B. 859
C. 4320
D. 427

Answer

To solve the problem of finding the smallest number which, when increased by 5, is divisible by each of 24, 32, 36, and 54, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the numbers**: We need to find a number that, when increased by 5, is divisible by 24, 32, 36, and 54.

2. **Calculate the LCM**: To find the smallest number that is divisible by all these numbers, we first need to calculate the Least Common Multiple (LCM) of 24, 32, 36, and 54.

- **Prime Factorization**:
- 24 = 2^3 × 3^1
- 32 = 2^5
- 36 = 2^2 × 3^2
- 54 = 2^1 × 3^3

- **Finding LCM**:
- For LCM, we take the highest power of each prime factor:
- For 2: max(3, 5, 2, 1) = 5 → 2^5
- For 3: max(1, 0, 2, 3) = 3 → 3^3
- Thus, LCM = 2^5 × 3^3

- **Calculating LCM**:
- 2^5 = 32
- 3^3 = 27
- LCM = 32 × 27 = 864

3. **Set up the equation**: Let the smallest number be \( x \). According to the problem, \( x + 5 \) should be divisible by 864. This can be expressed as:
\[ x + 5 = 864k \quad \text{(for some integer } k\text{)} \]
Rearranging gives:
\[ x = 864k - 5 \]

4. **Find the smallest value of \( x \)**: To find the smallest positive \( x \), we can set \( k = 1 \):
\[ x = 864 \times 1 - 5 = 864 - 5 = 859 \]

5. **Conclusion**: The smallest number which, when increased by 5, is divisible by each of 24, 32, 36, and 54 is **859**.

### Final Answer:
The required number is **859**.

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