Question

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

Hint:

When solving problems involving digits of a number, express the number in terms of its digits and use the given conditions to form a system of equations.

Answer 1

Let the two-digit number be \( 10a + b \), where \( a \) is the tens digit and \( b \) is the ones digit. The number obtained by reversing the digits is \( 10b + a \). We are given two conditions: 1. The sum of the number and the number obtained by reversing the digits is 66: \[ (10a + b) + (10b + a) = 66. \] Simplifying: \[ 11a + 11b = 66 \quad \implies \quad a + b = 6. \] 2. The difference of the digits is 2: \[ a - b = 2. \] Step 1: Solve the system of equations. We now solve the system of equations: 1. \( a + b = 6 \) 2. \( a - b = 2 \) Add the two equations: \[ (a + b) + (a - b) = 6 + 2 \quad \implies \quad 2a = 8 \quad \implies \quad a = 4. \] Substitute \( a = 4 \) into \( a + b = 6 \): \[ 4 + b = 6 \quad \implies \quad b = 2. \] Thus, the number is: \[ 10a + b = 10(4) + 2 = 42. \]
Conclusion:
The number is 42.