The solution involves setting up and solving a system of equations to find the values of the digits x and y. The final number is 28.
Let's break down the solution with detailed calculations:
Setting up the Equations:
Let the unit digit be represented by x and the tens digit by y.
The original number is 10y + x, and the new number is 10x + y.
The sum of the squares of the digits is given by x^2 + y^2 = 100.
From this equation, we derive the relationship x + y = 10 (Equation 1).
Equation for Three Times the Number:
The problem states that if 2 is subtracted from three times the original number, the digits are reversed:
3(10y + x) - 2 = 10x + y
Simplifying, we get 30y + 3x - 2 = 10x + y
Rearranging, we obtain 29y - 7x = 2 (Equation 2).
Solving the System of Equations:
Now, we have a system of equations:
x + y = 10 (Equation 1)
29y - 7x = 2 (Equation 2)
Multiplying Equation 1 by 7 to make the coefficients of x match, we get:
7x + 7y = 70 (Equation 3)
Adding Equation 3 to Equation 2 to eliminate x, we get:
36y = 72
Solving for y, we find y = 2.
Substituting Back:
Substitute the value of y = 2 into Equation 1:
x + 2 = 10
Solving for x, we find x = 8.
Final Number:
The original number is 10y + x = 10 * 2 + 8 = 28.