What is the least natural number which when divided by 8, 12, 15 or 16 leaves a remainder of 7?

Options

A. 247
B. 548
C. 257
D. 614

Answer

To solve the problem of finding the least natural number that leaves a remainder of 7 when divided by 8, 12, 15, or 16, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find a number \( N \) such that:
- \( N \mod 8 = 7 \)
- \( N \mod 12 = 7 \)
- \( N \mod 15 = 7 \)
- \( N \mod 16 = 7 \)

This means that \( N - 7 \) must be divisible by each of these numbers.

2. **Setting Up the Equation**: Let \( M = N - 7 \). Then, we need to find \( M \) such that:
- \( M \mod 8 = 0 \)
- \( M \mod 12 = 0 \)
- \( M \mod 15 = 0 \)
- \( M \mod 16 = 0 \)

This means \( M \) must be a common multiple of 8, 12, 15, and 16.

3. **Finding the LCM**: We need to calculate the Least Common Multiple (LCM) of the numbers 8, 12, 15, and 16.

- **Prime Factorization**:
- \( 8 = 2^3 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 16 = 2^4 \)

- **Taking the highest power of each prime**:
- For \( 2 \): highest power is \( 2^4 \) (from 16)
- For \( 3 \): highest power is \( 3^1 \) (from 12 and 15)
- For \( 5 \): highest power is \( 5^1 \) (from 15)

- **Calculating the LCM**:
\[ \text{LCM} = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 \]
\[ = 48 \times 5 = 240 \]

4. **Finding \( N \)**: Since \( M = 240 \), we can find \( N \) by substituting back:
\[ N = M + 7 = 240 + 7 = 247 \]

5. **Conclusion**: The least natural number which when divided by 8, 12, 15, or 16 leaves a remainder of 7 is \( \boxed{247} \).