A 4-digit number is formed by repeating a 2 -digit number such as 1515, 3737 , etc . Any number of this form is exactly divisible by

Options

A. 7
B. 11
C. 13
D. 101

Answer

To solve the problem, we need to analyze the structure of the 4-digit number formed by repeating a 2-digit number. Let's denote the 2-digit number as \( AB \), where \( A \) and \( B \) are the digits.

### Step-by-Step Solution:

1. **Understanding the 4-Digit Number**:
The 4-digit number formed by repeating the 2-digit number \( AB \) can be represented as \( ABAB \).

2. **Expressing \( ABAB \) Mathematically**:
The number \( AB \) can be expressed as \( 10A + B \). Therefore, the 4-digit number \( ABAB \) can be expressed as:
\[ ABAB = AB \times 100 + AB = (10A + B) \times 100 + (10A + B) \]

3. **Simplifying the Expression**:
We can factor out \( AB \):
\[ ABAB = (10A + B) \times (100 + 1) = (10A + B) \times 101 \]

4. **Conclusion on Divisibility**:
Since \( ABAB \) can be expressed as \( (10A + B) \times 101 \), it shows that \( ABAB \) is divisible by 101 for any digits \( A \) and \( B \).

5. **Final Answer**:
Therefore, any 4-digit number formed by repeating a 2-digit number is exactly divisible by 101.