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

Question 1023860: 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. find the number
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
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. find the number

Solution:

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!

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