A 6 digit number is formed by repeating 2 digits for 3 times (ex. 282828, 131313). This number will always be a multiple of?

Options

A. 10101
B. 11001
C. 101010
D. 11100

Answer

To solve the problem of determining what a 6-digit number formed by repeating 2 digits three times is always a multiple of, we can follow these steps:

### Step 1: Understand the Structure of the Number
A 6-digit number formed by repeating 2 digits three times can be expressed in the form of \( \overline{ab} \overline{ab} \overline{ab} \), where \( a \) and \( b \) are the digits. This can also be represented mathematically as:
\[ N = 100000a + 10000b + 1000a + 100b + 10a + b \]
This simplifies to:
\[ N = 101010a + 10101b \]

### Step 2: Factor the Expression
Now, we can factor out the common terms:
\[ N = 10101(10a + b) \]
This shows that \( N \) is a product of \( 10101 \) and \( (10a + b) \).

### Step 3: Determine the Multiplicity
Since \( (10a + b) \) is a 2-digit number (ranging from 10 to 99), \( N \) is guaranteed to be a multiple of \( 10101 \) regardless of the values of \( a \) and \( b \).

### Conclusion
Therefore, the 6-digit number formed by repeating 2 digits three times will always be a multiple of \( 10101 \).

### Final Answer
The answer is that the number will always be a multiple of **10101**.

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