A two-digit number is 21 times the difference of its digits. If each digit is…

Step-by-Step Solution

To solve this, let the two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit.

1. Formulate the equations based on the conditions:

   • Condition 1: The number is 21 times the difference of its digits. Equation: 10x + y = 21 * (x - y) 10x + y = 21x - 21y 22y = 11x x = 2y (Equation 1)

   • Condition 2: If each digit is increased by 2, the new number is 7 more than 6 times the sum of its digits. The new number is 10(x + 2) + (y + 2). The sum of the digits is (x + 2) + (y + 2) = x + y + 4. Equation: 10(x + 2) + (y + 2) = 6(x + y + 4) + 7 10x + 20 + y + 2 = 6x + 6y + 24 + 7 10x + y + 22 = 6x + 6y + 31 4x - 5y = 9 (Equation 2)

2. Solve the system of equations: Substitute x = 2y (from Equation 1) into Equation 2: 4(2y) - 5y = 9 8y - 5y = 9 3y = 9 y = 3

   Now find x: x = 2y = 2 * 3 = 6

   The number is 10x + y = 10(6) + 3 = 63.