Question

The sum of the digits of a number is 10. The difference between this number and a number formed with the same digits, but in the reverse order, is 18. What is the square of the number formed by reversing its digits?

• A. 2116
• B. 4096
• C. 1764
• D. None of these

Hint:

When solving such problems, first set up equations based on the conditions given, and then solve systematically.

Answer 1

The Correct Option is A

Let the number be \( N = 10a + b \), where \( a \) and \( b \) are the digits of the number. The sum of the digits is given as: \[ a + b = 10. \] The number formed by reversing the digits is \( N' = 10b + a \). The difference between \( N \) and \( N' \) is given as: \[ |N - N'| = 18. \] This leads to the equation: \[ | (10a + b) - (10b + a) | = 18. \] Simplifying: \[ |9a - 9b| = 18 \quad \Rightarrow \quad |a - b| = 2. \] Now, solving the system of equations \( a + b = 10 \) and \( |a - b| = 2 \), we get two possible cases: 1. \( a - b = 2 \) leads to \( a = 6 \), \( b = 4 \). 2. \( b - a = 2 \) leads to \( a = 4 \), \( b = 6 \). So, the number \( N \) can be either 64 or 46. Reversing the digits, we get 46 and 64 respectively. The square of 46 is: \[ 46^2 = 2116. \] Thus, the correct answer is \( \boxed{2116} \).