Chapter 9, Problem 29CLR

To determine

To Find: The original number.

Answer to Problem 29CLR

  $15$

Explanation of Solution

Given: The sum of the digits of a two digit number is $6$ when the digits are reversed, the resulting number is $6$ greater than $3$ times the original number.

Let $t$ be the tens digit and $u$ be the units digit.

Then

The original number is $10t+u$

The resulting number is $6$ greater than $3$ times the original number.

  $10u+t=6+3(10t+u)$

The sum of the digits is $6$

i.e.

  $\begin{array}{l}t+u=6\\ u=6-t\end{array}$

Substitute $u=6-t$ in $10u+t=6+3(10t+u)$

Thus

  $\begin{array}{l}10u+t=6+3(10t+u)\\ 10(6-t)+t=6+3(10t+u)\\ 60-10t+t=6+3(9t+6)\\ 60-9t=27t+24\\ 36=36t\\ t=1\\ then\\ u=6-t\\ u=6-1\\ u=5\\ i.e.\\ t=1\\ u=5\\ number=15\end{array}$

Hence, the original number is $15$