Bad Fraction Reduction That Actually Works

$$\frac{16}{64}=\frac{1\rlap{/}6}{\rlap{/}64}=\frac{1}{4}$$

This is certainly not a correct technique for reducing fractions to lowest terms, but it happens to work in this case, and I believe there are other such examples. Is there a systematic way to generate examples of this kind of bad fraction reduction?

Actually this is Project Euler's Problem 33 (in a more general form). In this blog entry it seems a (rather brute force) method is presented to generate your fractions. More examples are $16/64$, $49/98$, $19/95$, and $26/65$. Maybe there is a better way to generate them all, I saw a few promising ones when looking at the solutions on project euler maybe you want to check yourself.

Edit: For bigger numbers you have to say how you allow to cancel terms. For example do you allow $199/995=1/5$ where you cancel two digits or just one digit, do the digits have to be next to each other and so on, also do numerator and denominator need to have the same size or is $39/975=3/75$ allowed too?

It's easy to find them all. Suppose $\rm\: (10\ a + n)/(10\ n + b) = a/b\:.\:$ Thus $\rm\ (10\ a-b)\ n = 9\:a\:b\:.$

Case 1: $\rm\:(9,n) = 1\::\ \: 9\ |\ 10\:a-b\ \Rightarrow\ 9\ |\ a-b\ \Rightarrow a=b\ \Rightarrow\ 9\:a\:n = 9\:a^2\ \Rightarrow\ n=a=b\: $ (trivial)

Case 2: $\rm\:(9,n) = 9\::\ 10\:a-b = a\:b\ \Rightarrow\ a|b,\ 10 = (b/a)\:(a+1)\:$ so $\rm\ a,b = 1,5\:$ or $\: 4,8\:$

which yields the solutions: $\:\ 19/95 = 1/5\:,\:$ and $ 49/98 = 1/2\:.\ $ Similar analysis of the remaining

Case 3: $\rm\: (9,n) = 3\::\: $ yields $\:16/64 = 1/4\:,\:$ and $\: 26/65 = 2/5\:.$

Prior work on this question can be found in:

Ruekberg, B. Simplified mathematics. Journal of Irreproducible Results 35 (1).

If your institution (scandalously) doesn't carry this journal, a copy can be found here.