A number of the form `10^(n)` - 1` is always divisible by 11 for every n is a natural number, when :

Options

A. n is odd
B. n is prime
C. n is even
D. can't say

Answer

To determine when a number of the form \(10^n - 1\) is divisible by 11 for every natural number \(n\), we can analyze the expression step by step.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression \(10^n - 1\). We need to find out when this expression is divisible by 11.

2. **Testing Small Values of \(n\)**:
Let's test small values of \(n\) to see if we can find a pattern.

- For \(n = 1\):
\[ 10^1 - 1 = 10 - 1 = 9 \]
\(9\) is not divisible by \(11\).

- For \(n = 2\):
\[ 10^2 - 1 = 100 - 1 = 99 \]
\(99\) is divisible by \(11\) (since \(99 \div 11 = 9\)).

- For \(n = 3\):
\[ 10^3 - 1 = 1000 - 1 = 999 \]
\(999\) is not divisible by \(11\) (since \(999 \div 11 \approx 90.82\)).

- For \(n = 4\):
\[ 10^4 - 1 = 10000 - 1 = 9999 \]
\(9999\) is divisible by \(11\) (since \(9999 \div 11 = 909\)).

- For \(n = 5\):
\[ 10^5 - 1 = 100000 - 1 = 99999 \]
\(99999\) is not divisible by \(11\) (since \(99999 \div 11 \approx 9090.82\)).

3. **Identifying the Pattern**:
From our tests, we see that \(10^n - 1\) is divisible by \(11\) when \(n\) is even (specifically \(n = 2, 4\)).

4. **Conclusion**:
The number \(10^n - 1\) is divisible by \(11\) when \(n\) is an even natural number.

### Final Answer:
The number of the form \(10^n - 1\) is always divisible by \(11\) when \(n\) is an even natural number.

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