Chapter 9.7, Problem 9P

To determine

To solve: the problems about two-digit numbers.

Answer to Problem 9P

The number is 48.

Explanation of Solution

Given information:

The sum of the digits of a two-digit is one third of the number. The units digit is 5 more than the tens digit.

Formula used:

Digit problems are based on our decimal system of numeration.

Step1. The problem asks for the original number.

Step2. Let t = the tens digit of the original number.

Let u = the units digit of the original number.

Step3. Use the facts of the problem to write two equations.

Step4. Solve the equations.

Calculation:

Consider ,

Let t = the tens digit of the original number.

Let u = the units digit of the original number.

	Tens	Units	Value
Original number	t	u	$10t+u$
Number reversed	u	t	$10u+t$

According to the question,

  $\begin{array}{l}=10u+t=10t+u+36\\ =10u+t-10t-u=36\\ =9u-9t=36\\ =9(u-t)=36\\ =\frac{9(u-t)}{9}=\frac{36}{9}\\ \text{=}(u-t)=4\text{ (i)}\end{array}$

Also,

Let t = the tens digit of the original number.

Therefore,

u = the ones digit will be

  $=u=2t\text{ (ii)}$

Substituting equation (ii) in (i) we get,

  $\begin{array}{l}=u-t=4\\ =2t-t=4\\ =t=4\end{array}$

Putting value of u in equation (ii) we get,

  $\begin{array}{l}=u=2(4)\\ =u=8\end{array}$

Thus , the 2-digit number is 48.