What 3-digit number when doubled is, by value, one more than the same 3-digit number reversed? There can be two answers.

Let ones digit = z

Let tens digit = y

Let hundreds digit = x

Then three digit number xyz will be equivalent to (100x+10y+z)

If we reverse the digits then new number zyx will be equivalent to (100z+10y+x)

2 (100x+10y+z) = (100z+10y+x) + 1

200x+20y+2z=100z+10y+x+1

199x+10y-98z-1=0

There are no more conditions given so that we can use them to minimize calculation in order to find the answer.

So now we have only choice to plug random numbers for x, y, z which can vary from 0 to 9.

Working as explained above gives x=3, y=9, z=7 is satisfying the above equation.

Hence required number is 397.