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

Options

A. 456
B. 454
C. 540
D. 544

Answer

To find the least number which, when divided by 12, 15, 20, and 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 \( N - 4 \) should be divisible by 12, 15, 20, and 54.

### Step 2: Calculate the LCM
We first need to find 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 \( 2 \): highest power is \( 2^2 \) (from 12 and 20)
- For \( 3 \): highest power is \( 3^3 \) (from 54)
- For \( 5 \): highest power is \( 5^1 \) (from 15 and 20)

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

Calculating this:
\[ = 4 \times 27 \times 5 \]
\[ = 108 \times 5 = 540 \]

### Step 3: Add the Remainder
Now that we have the LCM, we need to add the remainder of 4 to find \( N \):
\[ N = \text{LCM} + 4 = 540 + 4 = 544 \]

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

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