In a two-digit number, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is

### Concept & Logic Solving non-linear equations derived from digit place values and their products. Utilizing multiple constraints simultaneously. ### Step-by-Step Solution Let the ten's digit be $x$ and the unit's digit be $y$. The original number is $10x + y$. Condition 1: Unit's digit exceeds ten's digit by 2. $y = x + 2$ Condition 2: The product of the number and the sum of its digits is 144. $(10x + y) \times (x + y) = 144$ Substitute $y = x + 2$ into the product equation: $(10x + x + 2) \times (x + x + 2) = 144$ $(11x + 2) \times (2x + 2) = 144$ Divide the equation by 2 to simplify: $(11x + 2)(x + 1) = 72$ $11x^2 + 11x + 2x + 2 = 72$ $11x^2 + 13x - 70 = 0$ Solve the quadratic (or just test integers): $x = 2$ is a clear solution because $11(4) + 13(2) - 70 = 44 + 26 - 70 = 0$. If $x = 2$, then $y = 2 + 2 = 4$. The number is 24. ### Exam Strategy & Shortcut The quadratic equation takes too long. Immediately test the options using the given conditions! (a) 24: $4 - 2 = 2$ (Pass). Product: $24 \times (2+4) = 24 \times 6 = 144$. (Pass). We have our answer instantly. (b) 26: $6 - 2 \neq 2$. (Fail) (c) 42: $2 - 4 \neq 2$. (Fail) ### Common Pitfall Attempting to solve the full quadratic equation $11x^2 + 13x - 70 = 0$ under exam pressure instead of realizing that digit problems are almost always faster to solve by testing the four given options. ### Final Answer Therefore, the correct answer is **24**.