A six digit number is formed by repeating a three digit number, for example, 625625 or 867867 etc. Any number of this form is always exactly divisible by :

Options

A. 7 only
B. 11 only
C. 13 only
D. 1001

Answer

To solve the problem, we need to understand how a six-digit number formed by repeating a three-digit number can be expressed mathematically. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Structure of the Number:**
A six-digit number formed by repeating a three-digit number can be represented as:
\[ \text{Let the three-digit number be } xyz. \]
Thus, the six-digit number can be expressed as:
\[ \text{Number} = xyzxyz. \]

2. **Mathematical Representation:**
We can express the six-digit number \(xyzxyz\) in terms of \(xyz\):
\[ xyzxyz = xyz \times 1000 + xyz. \]
This simplifies to:
\[ xyzxyz = xyz \times (1000 + 1) = xyz \times 1001. \]

3. **Divisibility:**
Since the six-digit number can be expressed as \(xyz \times 1001\), it follows that:
\[ xyzxyz \text{ is divisible by } 1001. \]

4. **Factorization of 1001:**
We can further analyze \(1001\) to see if it has any prime factors:
\[ 1001 = 7 \times 11 \times 13. \]
This means that \(xyzxyz\) is also divisible by \(7\), \(11\), and \(13\).

5. **Conclusion:**
Therefore, any six-digit number formed by repeating a three-digit number is always exactly divisible by \(1001\) and its factors \(7\), \(11\), and \(13\).

### Final Answer:
The six-digit number formed by repeating a three-digit number is always exactly divisible by **1001**.

---