Let the tens digit of the two-digit number be \(x\). ∴ The units digit is \((x + 6)\). So, the number is \(10x + (x + 6)\). And the product of the digits is \(x(x + 6)\). ∴ According to the question: \[ x(x + 6) = 10x + (x + 6) - 12 \] \[ x^2 + 6x = 10x + x + 6 - 12 \] \[ x^2 + 6x = 11x - 6 \] \[ x^2 + 6x - 11x + 6 = 0 \] \[ x^2 - 5x + 6 = 0 \] \[ x^2 - 3x - 2x + 6 = 0 \] \[ x(x - 3) - 2(x - 3) = 0 \] \[ (x - 3)(x - 2) = 0 \] ∴ Either \((x - 3) = 0\) or \((x - 2) = 0\) When \((x - 3) = 0\), then \(x = 3\); the number is \(10 \times 3 + (3 + 6) = 39\) When \((x - 2) = 0\), then \(x = 2\); the number is \(10 \times 2 + (2 + 6) = 28\) ∴ The number could be 39 or 28.