If $n$ is even, $(6^n - 1)$ is divisible by

### Concept & Formula This relies on standard algebraic identities regarding differences of powers. Specifically, for any integer $x$ and $y$, the expression $x^n - y^n$ is perfectly divisible by both $(x + y)$ and $(x - y)$ strictly when $n$ is an **even** integer. ### Step-by-Step Solution * **Given:** The expression $6^n - 1$ where $n$ is an even number. This can be written as $6^n - 1^n$. * **Apply Algebraic Rule:** Because $n$ is even, $x^n - y^n$ is divisible by $(x + y)$ and $(x - y)$. * Here, $x = 6$ and $y = 1$. * **Find the factors:** * Factor 1: $(6 + 1) = 7$ * Factor 2: $(6 - 1) = 5$ * **Determine compound divisibility:** Since the expression is divisible by both 7 and 5, and because 7 and 5 are co-prime (they share no common factors), the expression must be divisible by their product. $$7 \times 5 = 35$$ ### Exam Strategy & Shortcut The fastest way to solve "If $n$ is..." questions is to substitute the smallest valid value for $n$. The smallest even positive integer is $n = 2$. Plug it in: $6^2 - 1 = 36 - 1 = 35$. Look at the options: 35 is clearly divisible by 35. ### Common Pitfall Students might accidentally substitute an odd number like $n=1$, yielding $6^1 - 1 = 5$. While 5 is a factor, it doesn't map to the provided options and fails to capture the full divisibility condition granted by an *even* exponent. ### Final Answer Therefore, the correct answer is **35**.