If a six digit number is formed by repeating a two digit number three times (e.g. 141414), then that number will be divisible by :

Options

A. 1001
B. 101
C. 10001
D. 10101

Answer

To solve the problem of finding out what a six-digit number formed by repeating a two-digit number three times is divisible by, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of the Number:**
- A six-digit number formed by repeating a two-digit number (let's denote the two-digit number as AB) three times would look like: ABABAB.
- For example, if AB = 14, the six-digit number would be 141414.

2. **Expressing the Number Mathematically:**
- We can express the number ABABAB in terms of A and B, where A is the first digit and B is the second digit of the two-digit number.
- The number can be represented as:
\[ 100000A + 10000B + 1000A + 100B + 10A + B \]

3. **Combining Like Terms:**
- Now, let's combine the terms:
\[ = (100000A + 1000A + 10A) + (10000B + 100B + B) \]
\[ = 100010A + 10101B \]

4. **Factoring the Expression:**
- We can factor out common terms from the expression:
\[ = 10101(10A + B) \]
- Here, \(10101\) is a factor of the six-digit number.

5. **Checking Divisibility:**
- Since \(10101\) is a product of prime factors, we can check its divisibility:
- \(10101 = 3 \times 7 \times 13 \times 37\)
- Therefore, any number formed by repeating a two-digit number three times will be divisible by \(10101\).

6. **Conclusion:**
- Thus, the six-digit number formed by repeating a two-digit number three times is divisible by \(10101\).