- status: open --> closed
Current CVS (2006/04/27) has 2*3*2^k becoming 6*2^k.
Should this be 3*2^(k+1)? It would be nice, but....
What about a*b^k where a is a large number? Do we want
to check if a contains any powers of b so we can return
a new result of the form (a/b^k)*b^(k+1)?
Logged In: YES
user_id=28849
This particular bug is fixed. We basically handle the case
of m/n*b^k and convert that to m1/n1*b^k1 by removing powers
of b from m and n.
Log in to post a comment.