(5) A three digit number is equal to 17 times the sum of its digits; If the digits are
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.
Answers
Step-by-step explanation:
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=17(x+y+z)
100x+10y+z = 17x + 17y + 17z100x+10y+z=17x+17y+17z
100x-17x+10y-17y+z-17z = 17x + 17y + 17z100x−17x+10y−17y+z−17z=17x+17y+17z
83x -7y -16z = 083x−7y−16z=0 -----------------------(1)
if the digits reversed, the new number is 198 more than the old number
100x+10y+z + 198 = 100z + 10y + x100x+10y+z+198=100z+10y+x
100x- x +10y-10y +z - 100z +198 = 0100x−x+10y−10y+z−100z+198=0
99x - 99z+198 = 099x−99z+198=0
z-x=2z−x=2
z = x + 2----------------------------(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-1z+z−2=y−1
y = 2z-1 ----------------------------(3)
Putting (3) and (2) in (1) we get
83(z-2) -7(2z-1) -16z = 083(z−2)−7(2z−1)−16z=0
83(z-2) -7(2z-1) -16z = 083(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