A three digit number is equal to 17 times the sum of its digits. If 198 is added to the number, the digits get reserved. The sum of the extreme digits is 1 less than the midle digit. Find the number.

Explanation

Let the digit at the hundreds place be `x` and the digit at the units place be `y`.
The sum of the extreme digits is 1 less than the middle digit.
`:.` middle digit (digit at tens place) is `(x+y+1)`
`:.` the number is `100x+10(x+y+1)+y`
`=100x+10x+10y+10+y`
`=110x+11y+10`
The number obtained by reversing the digits
`=100y+10(x+y+1)+x`
`=100y+10x+10y+10+x`
`=110y+11x+10`
From the first condition.
`110x+11y+10=17(x+x+y+1+y)`
`:.110x+11y+10=37x=34y+17`
`:.110x-34x+11y-34y=17-10`
`:. 76x-23y=7`.............1
From the second condition
`110x+11y+10+198=110y+11x+10`
`:.110x-11x+11y-110y=-198`
`:.99x-99y=-198`
`:. x-y=-2` .... (Dividing both the sides by 99) ..............2
Multiplying equation (2) by 23
`23x-23y=-46`...................3
Subtracting equation 3 from equation 1

Substituting `x=1` in equation 2
`1-y=-2 :. -y=-2-1 :.-y=-3 :. y=3`
`:.110x+11y+10`
`=110(1)+11(3)+10`
`=110+33+10`
`=153`
The required number is 153.