Question

A three digit number is equal to 17 times the sum of its digit; if the digits reversed, the new number is 198 more than the old number; also the sum of extreme digits is less than the middle digit by unity. find the original number by using the method of linear ewuation in two variables

Answer 1

number is 153 ,solution in pic

Answer 2

Answer:

153

Step-by-step explanation:

let the hundred place digit be x , tens digit be y and unit digit be z

So Number = 100x+10y+z

We are given that A three digit number is equal to 17 times the sum of its digit

 $100x+10y+z = 17(x+y+z)$

 $100x+10y+z = 17x + 17y + 17z$

$100x-17x+10y-17y+z-17z = 17x + 17y + 17z$

$83x -7y -16z = 0$-----------------------(1)

if the digits reversed, the new number is 198 more than the old number

$100x+10y+z + 198 = 100z + 10y + x$

$100x- x +10y-10y +z - 100z +198 = 0$

$99x - 99z+198 = 0$

 $z-x=2$

 $z = x + 2----------------------------(2)$

the sum of extreme digits is less than the middle digit by unity.

And,  

z + x = y-1

By putting (2)

$z+z-2=y-1$

 y = 2z-1                                                       ----------------------------(3)

Putting (3) and (2) in (1) we get

 $83(z-2) -7(2z-1) -16z = 0$

 $83(z-2) -7(2z-1) -16z = 0$

z = 3

put z = 3 in (3)

y = 2(3) -1

y = 5

Put z = 3 in (2) 

x = 3-2=1

So. number = 100(1) + 10(5)  + 3 = 153

So, number = 153

Hence the number is 153