The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is:

Options

A. A) 450
B. B) 454
C. C) 540
D. D) 544

Answer

To find the least number that, when divided by 12, 15, 20, or 54, leaves a remainder of 4 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 12 = 4 \)
- \( N \mod 15 = 4 \)
- \( N \mod 20 = 4 \)
- \( N \mod 54 = 4 \)

This means that \( N - 4 \) must be divisible by each of these numbers.

### Step 2: Set Up the Equation
Let \( M = N - 4 \). Then, \( M \) must be divisible by 12, 15, 20, and 54. We can express this as:
- \( M \mod 12 = 0 \)
- \( M \mod 15 = 0 \)
- \( M \mod 20 = 0 \)
- \( M \mod 54 = 0 \)

### Step 3: Find the Least Common Multiple (LCM)
To find \( M \), we need to calculate the least common multiple (LCM) of the numbers 12, 15, 20, and 54.

#### Prime Factorization:
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 20 = 2^2 \times 5^1 \)
- \( 54 = 2^1 \times 3^3 \)

#### LCM Calculation:
To find the LCM, we take the highest power of each prime factor:
- For \( 2 \): the highest power is \( 2^2 \) (from 12 and 20)
- For \( 3 \): the highest power is \( 3^3 \) (from 54)
- For \( 5 \): the highest power is \( 5^1 \) (from 15 and 20)

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^3 \times 5^1 = 4 \times 27 \times 5 \]

Calculating this step-by-step:
1. \( 4 \times 27 = 108 \)
2. \( 108 \times 5 = 540 \)

So, \( \text{LCM}(12, 15, 20, 54) = 540 \).

### Step 4: Calculate \( N \)
Now that we have \( M = 540 \), we can find \( N \):
\[ N = M + 4 = 540 + 4 = 544 \]

### Conclusion
The least number which, when divided by 12, 15, 20, or 54, leaves a remainder of 4 in each case is:
\[ \boxed{544} \]