If the square of a two-digit number is reduced by the square of the number formed by reversing the digits of the number, the final result is

### Concept & Formula This problem leverages the algebraic identity for the difference of two squares applied to the standard properties of reversing two-digit numbers. $$ a^2 - b^2 = (a - b)(a + b) $$ ### Step-by-Step Solution **Given:** * An expression for the difference between the square of a two-digit number and the square of its reverse. **Calculation / Deduction:** * Let the original number be $N = 10x + y$ and the reversed number be $M = 10y + x$. * The question asks for the properties of $N^2 - M^2$. * Expand using the difference of squares: $N^2 - M^2 = (N - M)(N + M)$. * We know that the difference between a number and its reverse is always a multiple of 9: $N - M = 9(x - y)$. * We know that the sum of a number and its reverse is always a multiple of 11: $N + M = 11(x + y)$. * Substitute these back: $N^2 - M^2 = 9(x - y) \times 11(x + y) = 99(x - y)(x + y)$. * Since the final expression contains the factor 99, the result must be divisible by 99, meaning it is simultaneously divisible by both 9 and 11. ### Exam Strategy & Shortcut **Test with Simple Numbers:** Pick a simple two-digit number like 21. Its reverse is 12. Calculate $21^2 - 12^2 = 441 - 144 = 297$. Check divisibility by 11: $297 / 11 = 27$ (Divisible). Check divisibility by 9: $297 / 9 = 33$ (Divisible). Since it is divisible by both, the answer is immediately "Both (a) and (b)". ### Common Pitfall Trying to brute-force square the algebraic term $(10x + y)^2 = 100x^2 + 20xy + y^2$ and subtract the reversed squared term. This leads to a massive, easily-bungled polynomial instead of the clean factored form. ### Final Answer **Therefore, the correct answer is Both (a) and (b).**