Question

A three-digit number is equal to 17 times the sum of the digits. If the digits as
reversed the new number is 198 more than the original number. The sum of th
extreme digits is 1 less than the middle digit. Find the original number.
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Answer 1

Answer:

Let ones digit = x  , tens digit = y , hundreds digit = z

So, number becomes = 100z+10y+x

According to question-

  100z+10y+x   =  17(x+y+z)

  100z+10y+x   =   17x + 17y + 17z

  100z-17z +10y-17y + x-17x = 0

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

Also,

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

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

99z - 99x +198 = 0

99x - 99z = 198  

 x - z = 2 

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

And,  

z + x = y-1

By putting (2)

y = z+z+2+1      

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

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

 

83z -7(2z+3) -16(z+2) = 0

83z -14z -21 -16z -32  = 0

53z = 53

z = 1

put z = 1 in (3)

y = 2(1) +3

y = 5

Put z = 1 in (2) 

x = 1+3= 3

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

So, number = 153