Question

SOLUTION: A 3 digit number is equal to 17 times the sum of its digits. If 198 is added to the number the digits are interchanged.The addition of 1st and 3rd digit is 1 less than middle digit

Answer 1

Solution
Let ones digit = x  , tens digit = y , hundreds digit = zSo, number becomes = 100z+10y+xAccording 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 + z100z- z +10y-10y +x - 100x +198 = 099z - 99x +198 = 099x - 99z = 198   x - z = 2  x = z + 2                                                       ----------------------------(2)
And,  z + x = y-1By 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  = 053z = 53z = 1put z = 1 in (3)y = 2(1) +3y = 5
Put z = 1 in (2) x = 1+3= 3So. number = 100(1) + 10(5)  + 3 = 153So, number = 153

Answer 2

There may be easier ways to solve this problem, but here's one solution.
Clearly there are three variables (but I am not keen on solving a 3x3 system of linear equations), and we can reason as follows.

Let the number be "xyz", i.e. the numeric value is 100x+10y+z where x,y,z are the hundreds, tens and unit digits respectively.

The sum of first and last digits is 1 less than the middle digit means
x+z=y+1...............(1) 
If 198 is added to the number, the digits are "interchanged" or reversed, so
100x+10y+z+198=100z+10y+x
which simplifies to
99(z-x)=198, or
z-x=2..............(2)

Add (1) and (2) to get a simpler equation:
2z=y+1................(3)

From (3) we can make a possible list of values relating y and z, for example, when y=1, 2z=1+1, so z=1, etc.
(y,z)={(1,1),(3,2),(5,3),(7,4),(9,5)}
noting the fact that y cannot take on even values or else x will not be an integer.

Now that we have limited our solutions to five, it's time to substitute values of y and z into equation (2) to find x. For example, z=1 => x=-1 so reject negative values. Again, z=2, x=0, that makes a two digit number, so reject again. Try z=3, then x=1...
x,y,z)={(1,5,3),(2,7,4),(3,9,5)}

So the candidates are {153, 274, 395}
From this list, the number(s) that is(are) divisible by 17 will fit the bill!