If $(12^n + 1)$ is divisible by $13$, then $n$ is

### Concept & Formula This problem tests the algebraic identity for the sum of $n$-th powers. The expression $x^n + y^n$ is divisible by $(x + y)$ **if and only if** $n$ is an odd integer. ### Step-by-Step Solution * **Given:** The expression $(12^n + 1)$ is perfectly divisible by $13$. * **Format to match identity:** We can rewrite $13$ as $(12 + 1)$ and the expression as $(12^n + 1^n)$. * **Apply the rule:** We know that $x^n + a^n$ is divisible by $(x + a)$ only when $n$ is odd. * Here, $x = 12$ and $a = 1$. * So, $(12^n + 1^n)$ is divisible by $(12 + 1) = 13$ only when $n$ is an odd integer. ### Exam Strategy & Shortcut Use modular arithmetic to prove it instantly. We want $12^n + 1 \equiv 0 \pmod{13}$. Since $12 \equiv -1 \pmod{13}$, substitute this into the equation: $(-1)^n + 1 \equiv 0 \pmod{13}$ $(-1)^n \equiv -1 \pmod{13}$ For $(-1)^n$ to equal $-1$, $n$ must absolutely be an odd number (like 1, 3, 5). ### Common Pitfall A student might test $n = 1$, see that $12^1 + 1 = 13$ works, and hastily select "1 only" (option a). Always remember that algebraic divisibility rules apply universally to sets of numbers (like all odds or all evens), not just a single instance. ### Final Answer Therefore, the correct answer is **any odd integer**.