Let the tens digit of the two-digit number be \(x\) ∴ The units digit will be \(x + 6\) So, the number is \(10x + (x + 6)\) And the product of the digits is \(x(x + 6)\) ∴ According to the question: \(x(x + 6) = 10x + (x + 6) − 12\) i.e., \(x^2 + 6x = 10x + x + 6 − 12\) i.e., \(x^2 + 6x = 11x − 6\) i.e., \(x^2 + 6x − 11x + 6 = 0\) i.e., \(x^2 − 5x + 6 = 0\) i.e., \(x^2 − 3x − 2x + 6 = 0\) i.e., \(x(x − 3) − 2(x − 3) = 0\) i.e., \((x − 3)(x − 2) = 0\) ∴ Either \(x − 3 = 0\) ⇒ \(x = 3\) Or \(x − 2 = 0\) ⇒ \(x = 2\) When \(x = 3\), the units digit = \(3 + 6 = 9\) When \(x = 2\), the units digit = \(2 + 6 = 8\) ∴ The possible values of the units digit are 8 or 9.