Q.In a two-digit number, the unit digit is 6 more than the tens digit, and the product of the two digits is 12 less than the number itself. What could the number be?

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.
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