Question

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.​

Answer 1

Answer:  153

Step-by-step explanation:

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

Answer 2

Step-by-step explanation:

Let the two-digit number be 10x + y

The number is five times the sum of its digits

10x+y=5(x+y)10x+y=5x+5y5x=4y ...(1)

When 9 is added to the number, its digits are reversed.

10x+y+9=10y+x9x+9=9yx+1=y ...(2)

Putting (2) in (1):

5x=4(x+1)5x=4x+4x=4y=x+1=4+1=5

Thus, the sum of the digits is 4 +5 = 9