\( n^{2}-10 n+a=(n-1)(n-a) \)

To solve this equation, we can start by expanding the right-hand side: \[ (n-1)(n-a) = n^2 - an - n + a = n^2 - (a+1)n + a. \] Now, we can equate both sides of the equation: \[ n^{2} - 10n + a = n^{2} - (a+1)n + a. \] Next, we can simplify this by cancelling out \(n^2\) on both sides: \[ -10n + a = -(a+1)n + a. \] Now, let's eliminate \(a\) from both sides: \[ -10n = -(a+1)n. \] Factoring out \(n\) from the right-hand side gives us: \[ -10 = -(a+1). \] Now, to solve for \(a\), we multiply both sides by -1: \[ 10 = a + 1. \] Now, subtracting 1 from both sides yields: \[ a = 9. \] So, the value of \(a\) is \(9\).