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

Step-by-Step Solution

Let the two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit.

1. Translate the first condition: "A two-digit number is 21 times the difference of its digits."

   • 10x + y = 21 * (x - y) (assuming x > y)

   • 10x + y = 21x - 21y

   • 22y = 11x

   • x = 2y ---(Equation 1)

2. Translate the second condition: "If each digit is increased by 2, the number thus obtained is seven more than six times the sum of its digits."

   • The new digits are (x + 2) and (y + 2).

   • The new number is 10(x + 2) + (y + 2) = 10x + 20 + y + 2 = 10x + y + 22.

   • The sum of new digits is (x + 2) + (y + 2) = x + y + 4.

   • Equation: (10x + y + 22) = 6 * (x + y + 4) + 7

   • 10x + y + 22 = 6x + 6y + 24 + 7

   • 4x - 5y = 9 ---(Equation 2)

3. Solve the system:

   • Substitute Equation 1 into Equation 2: 4(2y) - 5y = 9

   • 8y - 5y = 9

   • 3y = 9

   • y = 3

   • Since x = 2y, x = 6.

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