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 original number .The sum of the middle digit.Find the original number (please give full sum explanation)

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