Consider a three-digit number where the middle digit

Solution

Let the digit at one's place and at hundredth place be 'y' and 'x' respectively. So, original number = (100x + 30 + y) [Since, 3 is at tens place] New number = (100y + 30 + x) ATQ, (100y + 30 + x) - (100x + 30 + y) = 198 Or, (99y - 99x) = 198 So, (y - x) = 2..........(I) Given, (y + x + 3) = 13 So, (y + x) = 10.............(II) On adding equation (I) and (II) , we get, y - x + y + x = 2 + 10 So, 'y' = (12/2) = 6 Using value of 'y' in equation (II) , we get, 'x' = 10 - 6 = 4 Therefore, original number = (100x + 30 + y) = 100 X 4 + 30 + 6 = 436