Question 1092151
<br>Interesting problem.  I thought it would be easy to solve algebraically, but I don't see a purely algebraic solution.  I will be interested to watch and see if someone has one for you.<br>
It seems simple enough.  If n and d are the numerator and denominator of the original fraction, then you want<br>
{{{(n+8)/(d-1) = d/n}}}<br>
But when you try to solve that using standard algebraic processes, what you get is
{{{n(n+8) = d(d-1)}}}<br>
you could rewrite this as
{{{n^2+8n = d^2-d}}}<br>
But that doesn't help any.  We have one equation with two unknowns.  That suggests Diophantine equations; but I don't know procedures for solving quadratic Diophantine equations.<br>
So I can find AN answer (I haven't been able to see why there couldn't be others) by plugging in values for n and seeing if the product n(n+8) is a number I recognize as being the product of two consecutive integers.<br>
It happens rather quickly.  Trying n=1 gives n(n+8)=9; that doesn't work.  But trying n=2 gives n(n+8) = 20, which I immediately recognize as 5*4.<br>
So "A" solution to your problem is that the numerator is 2 and the denominator is 5.<br>
The solution 2/5 can be seen to work:
{{{(n+8)/(d-1) = 10/4 = 5/2}}}
which is the reciprocal of 2/5.