A six digit number of the form abcabc is written where `a, b, c in I^(+)` , then which statement is true about this number:

Options

A. It is always divisible by 7 and 11
B. it is divisible by 143
C. it is divisible by 1001
D. all of (a),(b) and (c ) are correct

Answer

To solve the problem, we need to analyze the six-digit number of the form \( abcabc \), where \( a, b, c \) are positive integers.

### Step-by-Step Solution:

1. **Understanding the Number Format**:
The six-digit number \( abcabc \) can be expressed mathematically. It represents the number formed by repeating the three digits \( abc \). This can be rewritten as:
\[ abcabc = 1000 \times abc + abc = 1001 \times abc \]
Here, \( abc \) is a three-digit number.

2. **Identifying the Divisibility**:
Since \( abcabc = 1001 \times abc \), we can analyze the properties of the number \( 1001 \). We need to check if \( 1001 \) is divisible by the numbers mentioned in the question: \( 7, 11, 143, \) and \( 1001 \).

3. **Checking Divisibility**:
- **Divisibility by 7**:
\[ 1001 \div 7 = 143 \quad (\text{exact division, hence divisible}) \]
- **Divisibility by 11**:
\[ 1001 \div 11 = 91 \quad (\text{exact division, hence divisible}) \]
- **Divisibility by 143**:
\[ 1001 \div 143 = 7 \quad (\text{exact division, hence divisible}) \]
- **Divisibility by 1001**:
\[ 1001 \div 1001 = 1 \quad (\text{exact division, hence divisible}) \]

4. **Conclusion**:
Since \( abcabc \) can be expressed as \( 1001 \times abc \), and since \( 1001 \) is divisible by \( 7, 11, 143, \) and \( 1001 \), it follows that \( abcabc \) will also be divisible by these numbers.

### Final Answer:
The correct statement is that the number \( abcabc \) is divisible by \( 7, 11, 143, \) and \( 1001 \).

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