Solve for $n$:
$126_n = 86_{10}$
The base \(n\) representation \(126_n\) represents the number \(1 \cdot n^2 + 2 \cdot n^1 + 6 \cdot n^0\). We are given that this value is equal to \(86\) in base \(10\). Thus, we can write the equation:
\[n^2 + 2n + 6 = 86\]
Since \(n\) is a base, it must be an integer greater than the largest digit present in the number, which is \(6\). Therefore, we must have \(n > 6\).
To solve for \(n\), we first subtract \(86\) from both sides of the equation to bring it into the standard quadratic form \(an^2 + bn + c = 0\):
\[n^2 + 2n + 6 - 86 = 0\]
\[n^2 + 2n - 80 = 0\]
We can factor this quadratic expression by looking for two numbers that multiply to \(-80\) and add up to \(2\). These numbers are \(10\) and \(-8\). Thus, the equation factors as:
\[(n + 10)(n - 8) = 0\]
The solutions to the equation \((n + 10)(n - 8) = 0\) are \(n = -10\) and \(n = 8\). Since a base \(n\) must be a positive integer and, specifically, must satisfy the condition \(n > 6\), we reject \(n = -10\). Therefore, the only valid solution is \(n = 8\). We can verify this:
\[1 \cdot 8^2 + 2 \cdot 8 + 6 = 64 + 16 + 6 = 86\]
This confirms that \(126_8 = 86_{10}\).
The value of the base \(n\) that satisfies the equation \(126_n = 86\) is \(\boxed{8}\).
Understanding base systems is foundational in computer science, as data is stored and processed in binary (base 2), octal (base 8), and hexadecimal (base 16). For example, if a legacy system outputs data in a specific base code, you might need to identify that base to correctly interpret the numerical information. Solving equations like \(126_n = 86\) helps build the mathematical intuition required to translate between these different numerical environments, ensuring data is converted and stored accurately in modern applications.