Question

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

Answer 1

The original number is 58.

Step-by-step explanation:

We are given that a two-digit number and the number with digits interchanged add up 143. In the given number the digit in the unit's place is 3 more than the digit in the ten's place.

Let the digit at the unit's place be 'x' and the digit at the ten's place be 'y'.

So, the two-digit number so formed is 10y + x, then the number formed with digits interchanged will be 10x + y.

Now, according to the question;

• The first condition states that a two-digit number and the number with digits interchanged add up 143, that is;

                     $(10y + x) + (10x + y) = 143$

                         $11x + 11y = 143$

                          $11(x+y) = 143$

                             $x+y=\frac{143}{11}$

                             $x+y=13$

                             $x= 13-y$  ------------------ [Equation 1]

• The second condition states that the digit in the unit's place is 3 more than the digit in the ten's place, that is;

                               $x = y +3$

                              $13-y = y+3$     {using equation 1}

                              $2y = 13-3$

                               $y=\frac{10}{2} = 5$

Now, putting the value of y in equation 1 we get;

                              $x= 13-y$

                              $x= 13-5 = 8$

Hence, the original number is 58.