Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete one round. After how much time do they meet at the starting point for the first time?

Options

A. 1800 seconds
B. 3600 seconds
C. 2400 seconds
D. 4800 seconds

Answer

To find out when the four runners meet at the starting point for the first time, we need to calculate the least common multiple (LCM) of the times taken by each runner to complete one round. The times are 200 seconds, 300 seconds, 360 seconds, and 450 seconds.

### Step-by-Step Solution:

1. **List the times taken by each runner**:
- Runner 1: 200 seconds
- Runner 2: 300 seconds
- Runner 3: 360 seconds
- Runner 4: 450 seconds

2. **Prime Factorization of each time**:
- **200**:
- 200 = 2 × 100
- 100 = 2 × 50
- 50 = 2 × 25
- 25 = 5 × 5
- Therefore, 200 = \(2^3 \times 5^2\)
- **300**:
- 300 = 3 × 100
- 100 = 2 × 50
- 50 = 2 × 25
- 25 = 5 × 5
- Therefore, 300 = \(2^2 \times 3^1 \times 5^2\)
- **360**:
- 360 = 36 × 10
- 36 = 6 × 6 = (2 × 3) × (2 × 3) = \(2^2 × 3^2\)
- 10 = 2 × 5
- Therefore, 360 = \(2^3 \times 3^2 \times 5^1\)
- **450**:
- 450 = 45 × 10
- 45 = 9 × 5 = \(3^2 × 5^1\)
- 10 = 2 × 5
- Therefore, 450 = \(2^1 \times 3^2 \times 5^2\)

3. **Determine the highest power of each prime factor**:
- For \(2\): The highest power is \(2^3\) (from 200 and 360).
- For \(3\): The highest power is \(3^2\) (from 360 and 450).
- For \(5\): The highest power is \(5^2\) (from 200, 300, and 450).

4. **Calculate the LCM**:
\[ \text{LCM} = 2^3 \times 3^2 \times 5^2 \]
- Calculate each part:
- \(2^3 = 8\)
- \(3^2 = 9\)
- \(5^2 = 25\)

Now multiply these together:
\[ LCM = 8 \times 9 \times 25 \]

- First, calculate \(8 \times 9 = 72\).
- Then, calculate \(72 \times 25\):
\[ 72 \times 25 = 1800 \]

5. **Conclusion**:
The runners will meet at the starting point for the first time after **1800 seconds**.