Question

1)In a three-digit number, the sum of the digits at hundred's place and
unit's place is 1 less than the digit at the ten's place. The number is
17 times the sum of the three digits. If 198 is added to the original
number, the order of the digits is reversed. Find the original
number.​

Answer 1

Answer:

153

Step-by-step explanation:

Let the no. be 100x+10y+z

Now, by the problem,

z+x = y-1 …………(1)

100x+10y+z = 17(x+y+z)

83x = 7y+16z …………(2)

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

99x+198 = 99z

x+2 = z …………(3)

Putting value of z from (3) into (1)

x+2+x = y-1

2x+3 = y …………(4)

Putting values of y and z from (4) and (3) respectively into (2)

83x = 7y+16z

83x = 7(2x+3)+16(x+2)

83x = 14x+21+16x+32

83x = 30x + 53

53x = 53

x=1

Putting value of x in (3)

x+2 = z = 1+2 = 3

Putting value of x in (4)

2x+3 = y = 2+3 = 5

x=1, y=5 and z=3

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