Find the smallest digit which gives remainder as 3 on being divided by 15, 20, 36 & 48 in each case ?

Options

A. 183
B. 243
C. 483
D. 723

Answer

To solve the problem of finding the smallest digit that gives a remainder of 3 when divided by 15, 20, 36, and 48, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- \( x \equiv 3 \mod 15 \)
- \( x \equiv 3 \mod 20 \)
- \( x \equiv 3 \mod 36 \)
- \( x \equiv 3 \mod 48 \)

This means that \( x - 3 \) should be divisible by each of these numbers.

### Step 2: Subtract the Remainder
Let’s define \( y = x - 3 \). Then we need to find \( y \) such that:
- \( y \equiv 0 \mod 15 \)
- \( y \equiv 0 \mod 20 \)
- \( y \equiv 0 \mod 36 \)
- \( y \equiv 0 \mod 48 \)

This means \( y \) should be a common multiple of 15, 20, 36, and 48.

### Step 3: Find the LCM
To find \( y \), we need to calculate the least common multiple (LCM) of the numbers 15, 20, 36, and 48.

1. **Prime Factorization**:
- \( 15 = 3 \times 5 \)
- \( 20 = 2^2 \times 5 \)
- \( 36 = 2^2 \times 3^2 \)
- \( 48 = 2^4 \times 3 \)

2. **Identify the highest powers of each prime**:
- For \( 2 \): highest power is \( 2^4 \) (from 48)
- For \( 3 \): highest power is \( 3^2 \) (from 36)
- For \( 5 \): highest power is \( 5^1 \) (from 15 and 20)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 \]
\[ = 144 \times 5 = 720 \]

### Step 4: Find \( x \)
Since \( y = 720 \), we can find \( x \):
\[ x = y + 3 = 720 + 3 = 723 \]

### Step 5: Conclusion
The smallest digit that gives a remainder of 3 when divided by 15, 20, 36, and 48 is \( 723 \).