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The base case in mathematical induction is the starting point of our proof. It helps us establish that the statement is true for the initial value of the variable, which in this case is the smallest value for which the statement is intended to hold true. For the problem at hand, we need to check whether the expression \(11^n - 6\) is divisible by 5 when \(n = 1\). This step acts as an initial anchor for our induction process.

When we plug \(n = 1\) into the expression, we get \(11^1 - 6 = 5\). Clearly, 5 is divisible by 5, thus verifying that our base case is indeed true. By confirming this, we've successfully established that the statement holds for the base case, which is fundamental in advancing to the next steps of the induction process.

The inductive hypothesis involves assuming that the statement is true for a particular integer \(n = k\), which serves as a foundation to predict the correctness of the subsequent step. This assumption is crucial as it provides the stepping stone from which we demonstrate the general truth of the statement.

In our scenario, we assume that \(11^k - 6\) is divisible by 5, implying that this expression can be represented as \(5m\) for some integer \(m\). This assumption, though seemingly unproven for this step, sets the stage for proving the statement for the next integer \(n = k + 1\). It's important to note that the validity of the proof hinges on this assumption, which is why it must be correctly established in every induction problem.

The inductive step is where the magic of mathematical induction occurs. Here, we must show that if the statement is true for \(n = k\), then it must also be true for \(n = k + 1\). This step is crucial as it creates a domino effect, pushing the truth from one case to the next.

For our statement, we need to prove that \(11^{k+1} - 6\) also follows the divisibility rule by 5. Beginning with the expression \(11^{k+1} - 6\), we rewrite it as \(11 \times 11^k - 6\). We then utilize our inductive hypothesis: replace \(11^k - 6\) with \(5m\) to get \(11 \times (5m + 6) - 6\). Simplifying, this turns into \(55m + 66 - 6\), which simplifies further to \(55m + 60\), ultimately equaling \(5(11m+12)\). This transformation confirms that \(11^{k+1} - 6\) is indeed divisible by 5, fulfilling the requirements for the inductive step.

Divisibility is at the heart of this proof, particularly focusing on whether the expression \(11^n - 6\) is divisible by 5. Throughout the solution, divisibility demonstrates that a particular statement is true across all natural numbers starting from our base case.

The concept of divisibility means that when one number can be divided by another without leaving a remainder. For our exercise, it indicates that \(11^n - 6\) can be consistently divided by 5, producing integers which validate our inductive reasoning.

• **Base Case**: Check divisibility for the smallest \(n\).
• **Hypothesis**: Assume divisibility is true for an arbitrary \(n = k\).
• **Inductive Step**: Show the next number, \(n = k + 1\), upholds divisibility.

Successfully proving divisibility through induction reflects a powerful method in mathematics to verify statements that hold true for infinite numbers.