Chapter 9.7, Problem 25P

To determine

To solve: the problems about three-digit numbers.

Answer to Problem 25P

The 3-digit number is 345.

Explanation of Solution

Given information:

The sum of the three digits is 12 . The tens digit exceeds the hundreds digit by the same amount that the units digit exceeds the tens digit. I the digits are reversed, the new number exceeds the original number by 198.

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.

Let t = the tens digit of the original number.

Let u = the units digit of the original number.

	Hundreds	Tens	Units	Value
Original number	$h$	$$$t$	$u$	$\begin{array}{l}=h(100)+t(10)+u(1)\\ =100h+10t+u\end{array}$

According to the question:-

  $=h+t+u=12\text{ (i)}$

Since , tens digit exceeds hundreds digit by the same amount that the units digit exceeds the tens digit.

  $\begin{array}{l}=t-h=u-t\\ =t-h-u+t=0\\ =2t-h-u=0\text{ (ii)}\end{array}$

By solving two equations we get,

  $\begin{array}{l}=2t-h-u=0\\ =t+h+u=12\\ \text{Adding the equations we get},\\ =\frac{3t}{3}=\frac{12}{3}\\ =t=4\text{ (iii)}\end{array}$

	Hundreds	Tens	Units	Value
Original number	$h$	$$$t$	$u$	$\begin{array}{l}=h(100)+t(10)+u(1)\\ =100h+10t+u\end{array}$
Number reversed	u	t	h	$\begin{array}{l}=u(100)+t(10)+h(1)\\ =100u+10t+h\end{array}$

Now,

  $=100u+10t+h=100h+10t+u+198$

  $\begin{array}{l}=100u+10t+h-100h-10t-u=198\\ =99u-99h=198\\ =\frac{99(u-h)}{99}=\frac{198}{99}\\ =u-h=2\\ =u=2+h\text{ (iv)}\end{array}$

Substituting equation (iii) and (iv) in equation (i)

  $\begin{array}{l}=h+4+2+h=12\\ =2h+6=12\\ =2h=12-6\\ =\frac{2h}{2}=\frac{6}{2}\\ =h=3\end{array}$

Putting value of h in equation (iv) we have,

  $\begin{array}{l}=u=2+3\\ =u=5\end{array}$

	Hundreds	Tens	Units	Value
Original number	$3$	$$$4$	$5$	$\begin{array}{l}=3(100)+4(10)+5(1)\\ =300+40+5\\ =345\end{array}$

The 3-digit number is 345.