A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit's place is 3 more than the digit in the ten's place. Find the original number.

Answer: The original number is 58.

Let the unit's digit be \(x\) and the ten's digit be \(y\).

Number \(= 10y + x\); number with digits interchanged \(= 10x + y\).

First condition: \((10y + x) + (10x + y) = 143\), so \(11x + 11y = 143\) and \(x + y = 13\) ... (I)

Second condition: \(x = y + 3\), so \(x - y = 3\) ... (II)

Adding (I) and (II): \(2x = 16\), \(x = 8\). Then \(8 + y = 13\), \(y = 5\).

Number \(= 10y + x = 50 + 8 = 58\). Check: \(58 + 85 = 143\), which is true.

Boxes in order: 10, 10, 10x + y, 11, 11, 13, y, 16, 5, 50.