Chapter 9.7, Problem 26P

To determine

To solve: the problems about three-digit numbers.

Answer to Problem 26P

The 3-digit number is 342.

Explanation of Solution

Given information:

The sum of the three digits is 9. The tens digit is 1 more than the hundreds digit . When the digits are reversed, the new number is 99 less than the original number.

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:-

  $=h+t+u=9\text{ (i)}$

Since , tens digit is 1 more than the hundreds digit

  $=t=h+1\text{ (ii)}$

Therefore,

  $\begin{array}{l}=h+h+1+u=9\\ =2h+u=9-1\\ =2h+u=8\text{ (iii)}\end{array}$

	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,

  $=100u+10t+h=100h+10t+u-99$

  $\begin{array}{l}=100u+10t+h-100h-10t-u=-99\\ =99u-99h=-99\\ =\frac{99(u-h)}{99}=-\frac{99}{99}\\ =u-h=-1\\ =u=-1+h\text{ (iv)}\end{array}$

Substituting equation (vi) in equation (iii)

  $\begin{array}{l}=2h+u=8\\ =2h+(h-1)=8\\ =2h+h-1=8\\ =3h-1=8\\ =3h=8+1\\ =\frac{3h}{3}=\frac{9}{3}\\ =h=3\end{array}$

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

  $\begin{array}{l}=u=-1+3\\ =u=2\end{array}$

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

  $\begin{array}{l}=t=3+1\\ =t=4\end{array}$

	Hundreds	Tens	Units	Value
Original number	$3$	$$$4$	$2$	$\begin{array}{l}=3(100)+4(10)+2(1)\\ =300+40+2\\ =342\end{array}$

The 3-digit number is 342.