Given r and n, find first n numbers of x where moving first digit of x to last gives x/r = y
Haskell, 182 179

Second version, probably further golfable, but with "proper" algorithm this time.  In particular, it finishes within a few minutes with r=4 and n=65535, but then again my computer is neither reasonable nor a desktop, so chances are this stays within a minute on other machines.

n#r=take n$[s(10^k*a+d)++'/':s r++'=':s d++s a|k&lt;-[0..],a&lt;-[1..9],let(d,m)=divMod(a*(10^k-r))(10*r-1),m&lt;1]
s=show
main=interact$unlines.(\(r:n:_)-&gt;n#fromIntegral r).map read.words


It is based on the idea that x=10^k*a + m, where its first digit 0≤a≤9 is moved to the end to obtain y=10*m+a.  A little maths reveals that m can be obtained as a*(10^k-r)/(10*r-1), so we simply scan a over [1..9] for every k from 0 to infinity, and keep and print the first n results for which the above expression for m is integral.

The fromIntegral is required because reading a list with n as one of its elements in main, in combination with the use of n in take, would force r to Int throughout, which results in nasty overflows with the big numbers in question.  I could have used genericTake, but that requires an import.

This code also has the benefit of being almost trivial to expand to bases other than 10.

Input is read from stdin, the two values can be separated by any whitespace.