Two times a two-digit number is 9 times the number obtained by reversing the…

Concept: For a two-digit number, if the tens digit is x and the units digit is y, the number equals 10x + y and its digit-reversal equals 10y + x. A relationship between twice the number and nine times its reverse translates directly into a linear equation in x and y, which combines with the digit-sum condition to solve for x and y uniquely.

1. Let the two-digit number be 10x + y, so its reverse is 10y + x.

2. The given condition (twice the number equals nine times the reverse) gives 2(10x + y) = 9(10y + x).

3. Expanding: 20x + 2y = 90y + 9x, which simplifies to 11x = 88y, that is x = 8y.

4. The digit-sum condition gives x + y = 9. Substituting x = 8y: 8y + y = 9, so y = 1 and x = 8.

5. Therefore, the number is 10(8) + 1 = 81.

Cross-check: For 81, the digit sum is 8 + 1 = 9 (matches), and its reverse is 18; twice 81 is 162 and nine times 18 is also 162, so both given conditions hold together.

Hence, the number is 81.