Verified Answer :

  • The problem depends on the interpretation of \(\sigma_n\).
  • Using the definition \(\sigma_n = \sum_{k=1}^n k = \frac{n(n+1)}{2}\), the equation \(12 \sigma_n = 8 \sigma_{10}\) yields no integer solution for \(n\).
  • Using the sum of divisors definition \(\sigma(n) = \sum_{d|n} d\), the equation yields \(\sigma(n) = 12\), which is satisfied by \(n=6\).
  • The final answer is \(\boxed{n=6}\).

Explanation

  1. Analyzing the Definition of \(\sigma_n\)

To solve the equation \(12 \sigma_n = 8 \sigma_{10}\), we must first address the ambiguity of the notation \(\sigma_n\). Since \(\sigma_n\) is not explicitly defined in the problem, we consider two common mathematical interpretations:

  1. The sum of the first \(n\) positive integers, defined as \(\sigma_n = \sum_{k=1}^n k = \frac{n(n+1)}{2}\).
  2. The sum of divisors function, commonly denoted as \(\sigma_1(n)\) or simply \(\sigma(n)\), which calculates the sum of all positive divisors of \(n\).

We analyze the equation \(12 \sigma_n = 8 \sigma_{10}\) under both interpretations. Simplified, the equation is \(\sigma_n = \frac{8}{12} \sigma_{10} = \frac{2}{3} \sigma_{10}\).

  1. Case 1: Sum of the First \(n\) Integers

First, consider \(\sigma_n = \frac{n(n+1)}{2}\).
We calculate \(\sigma_{10} = \frac{10(10+1)}{2} = \frac{110}{2} = 55\).
Substituting this into our equation \(\sigma_n = \frac{2}{3} \sigma_{10}\):
\(\sigma_n = \frac{2}{3} \times 55 = \frac{110}{3} \approx 36.67\).
For \(\sigma_n = \frac{n(n+1)}{2} = \frac{110}{3}\), we solve for \(n\):
\(n^2 + n = \frac{220}{3}\)
\(3n^2 + 3n - 220 = 0\).
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(n = \frac{-3 \pm \sqrt{3^2 - 4(3)(-220)}}{2(3)} = \frac{-3 \pm \sqrt{9 + 2640}}{6} = \frac{-3 \pm \sqrt{2649}}{6}\).
Since \(\sqrt{2649} \approx 51.47\), the positive solution for \(n\) is approximately \(\frac{48.47}{6} \approx 8.08\). Since \(n\) must be an integer, there is no integer solution under this definition.

  1. Case 2: Sum of Divisors Function

Next, consider \(\sigma(n)\) as the sum of divisors function.
We calculate \(\sigma(10)\): the divisors of 10 are 1, 2, 5, and 10. Thus, \(\sigma(10) = 1 + 2 + 5 + 10 = 18\).
Substituting this into our equation \(\sigma_n = \frac{2}{3} \sigma_{10}\):
\(\sigma_n = \frac{2}{3} \times 18 = 12\).
Now, we find \(n\) such that the sum of its divisors is 12.
Testing small values of \(n\):

  • For \(n=6\), the divisors are 1, 2, 3, 6. Sum: \(1+2+3+6 = 12\).

Thus, \(n=6\) is a valid solution under this interpretation.

  1. Conclusion

Based on the analysis, the solution depends on the definition of \(\sigma_n\). If \(\sigma_n\) represents the sum of the first \(n\) integers, there is no integer solution for \(n\). If \(\sigma_n\) represents the sum of divisors of \(n\), then \(\boxed{n=6}\) is the solution. Given the standard mathematical context where \(\sigma(n)\) is a common notation for the sum of divisors, \(n=6\) is the most likely intended answer.

Examples

In number theory, the divisor function \(\sigma(n)\) is used to explore properties of integers, such as identifying perfect numbers. For instance, a perfect number is one where the sum of its proper divisors equals the number itself (e.g., for \(n=6\), divisors 1+2+3=6). Problems like this one, involving equations of divisor sums, are common in competitions to test one's ability to relate algebraic structures to number theoretic functions.

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