Consider a three-digit number where the middle digit (tens place) is 3. If the digits in the hundreds and ones places are swapped, the new number becomes 198 more than the original. Additionally, the sum of all the digits in the number is 13. Determine the original three-digit number.
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
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