`(10^n - 1)` is always divisible by 11, if

Options

A. n is any number
B. n is an odd number.
C. n is an even no.
D. n is a multiple of 11

Answer

To determine when \(10^n - 1\) is always divisible by 11, we can analyze the expression based on the parity (even or odd) of \(n\).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \(10^n - 1\). We want to find conditions under which this expression is divisible by 11.

2. **Using the Difference of Powers**:
We can express \(10^n - 1\) as a difference of powers. The general formula for the difference of powers states:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + ab^{n-2} + b^{n-1}) \]
In our case, \(a = 10\) and \(b = 1\).

3. **Case 1: \(n\) is Even**:
If \(n\) is even, we can factor \(10^n - 1\) as follows:
\[ 10^n - 1 = (10 - 1)(10^{n-1} + 10^{n-2} + \ldots + 1) \]
Here, \(10 - 1 = 9\) and the second factor is a sum of terms. However, we are particularly interested in the factor \(10 - 1\).

4. **Checking Divisibility by 11**:
Now, we check the divisibility of \(10^n - 1\) by 11. We know that:
\[ 10 \equiv -1 \mod 11 \]
Therefore, if \(n\) is even:
\[ 10^n \equiv (-1)^n \equiv 1 \mod 11 \]
This implies:
\[ 10^n - 1 \equiv 0 \mod 11 \]
Hence, \(10^n - 1\) is divisible by 11 when \(n\) is even.

5. **Case 2: \(n\) is Odd**:
If \(n\) is odd, we have:
\[ 10^n \equiv -1 \mod 11 \]
Thus:
\[ 10^n - 1 \equiv -1 - 1 \equiv -2 \mod 11 \]
This shows that \(10^n - 1\) is not divisible by 11 when \(n\) is odd.

6. **Conclusion**:
From the analysis, we conclude that \(10^n - 1\) is always divisible by 11 if \(n\) is an even number.

### Final Answer:
Thus, the correct condition is that \(n\) should be an even number.