Question

A three digit number is equal to 17 times the sum of the digits if the digits are reversed the new number is 198 more than the original number the sum of the extreme digits 1 less than the middle digit find the original number ​

Answer 1

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$100x+10y+z = 17(x+y+z) 100x+10y+z = 17x + 17y + 17z100x-17x+10y-17y+z-17z = 17x + 17y + 17z83x -7y -16z = 0-----------------------(1)$

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