Chapter 9.7, Problem 23P

To determine

To solve: the problems about three-digit numbers.

Answer to Problem 23P

The 3-digit number is 360.

Explanation of Solution

Given information:

A number between 300 and 400 is forty times the sum of its digits. The tens digit is 6 more than the units digit.

Formula used:

Digit problems are based on our decimal system of numeration.

Step1. The problem asks for the original number.

Step2. Let h = the hundreds digit of the original number.

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 h = the hundreds digit of the original number.

  $\text{So,}h=3$

Let u = the ones digit of the original number.

Let t = $6(u)=6+u$ .

	Hundreds	Tens	Units	Value
Original number	$3$	$$$6+u$	$u$	$\begin{array}{l}=3(100)+(6+u)(10)+u(1)\\ =300+10(6+u)+u\\ =300+60+10u+u\\ =360+11u\end{array}$

According to the question:-

  $\begin{array}{l}=360+11u=40\left(3+u+6+u\right)\\ =360+11u=40(9+2u)\\ =360+11u=360+80u\\ =360-360=80u-11u\\ =0=69u\\ =u=0\end{array}$

The tens digit is

  $\begin{array}{l}=6+0\\ =6\end{array}$

	Hundreds	Tens	Units	Value
Original number	$3$	$$$6$	$0$	$\begin{array}{l}=3(100)+(6)(10)+(0)(1)\\ =300+60+0\\ =360\end{array}$

The 3-digit number is 360.