Question

The sum of a two-digit number and the number obtained by reversing the order of its digits is 66. If the digits differ by 2, find the number.

Hint:

Always represent the digits of a two-digit number as \( 10x + y \). Reverse it as \( 10y + x \). Set up two equations based on given conditions and solve systematically.

Answer 1

Step 1: Assume the digits.
Let the tens digit be \( x \) and the ones (unit) digit be \( y \). Then, the number is \( 10x + y \). The reversed number is \( 10y + x \).
Step 2: Write the given conditions.
Sum of the two numbers is 66: \[ (10x + y) + (10y + x) = 66 \] \[ 11x + 11y = 66 \] \[ x + y = 6 \quad \text{(i)} \] The digits differ by 2: \[ x - y = 2 \quad \text{(ii)} \]
Step 3: Solve the two equations.
Add (i) and (ii): \[ 2x = 8 \Rightarrow x = 4 \] Substitute in (i): \[ 4 + y = 6 \Rightarrow y = 2 \]
Step 4: Form the number.
The number is \( 10x + y = 10(4) + 2 = 42 \). Step 5: Verification.
Reversed number = 24. Sum \( 42 + 24 = 66 \). ✓ Verified. Step 6: Final Answer.
\[ \boxed{\text{The required number is } 42.} \]