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 are the possible values of the unit digit?

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.
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