Chapter 3.1, Problem 9P

To determine

To Find: All possibilities for N such that N is odd and exceeds K by more than 18.

Answer to Problem 9P

Final Answer: All the possible values of N satisfying the given condition are {41,51,61,71,81,91,63,73,83,93,85,95}

Explanation of Solution

Given Information: It is given that N is a two digit positive integer and on interchanging its digits we get an integer K such that N is odd and exceeds K by more than 18.

Formula Used: A two digit number is expressed as $10x+y$ if its unit digit is $y$ and tens digit is $x$ and on interchanging its digits it becomes $10y+x$

Calculation: According to the question, we have

N= $10x+y$ ........ (1)

K= $10y+x$ ........(2)

Now N exceeds K by more than 18 means,

  $N-K>18$

From (1) and (2), we have

  $(10x+y)-(10y+x)>18$

Or  $9x-9y>18$

Or  $x-y>2$

Now since N is odd soycannot be even

So, possible values of y is {1,3,5,7,9}

Now if y = 1

xcan have values such as {4,5,6,7,8,9}

So, N can have the values such as {41,51,61,71,81,91} ..........(3)

Now if y = 3

Ix can have values such as {6,7,8,9}

So, N can have the values such as {63,73,83,93} ..........(4)

Now if y = 5

xcan have values such as {8,9}

So, N can have the values such as {85,95}

Now if y = 7

There is no any possible value of x

So, all the possible values that N can have are 41,51,61,71,81,91,63,73,83,93,85,95