http://mathforum.org/kb List of forum topics en http://mathforum.org/kb/thread.jspa?messageID=7912035&tstart=0#7912035 Since 3^509=3(mod 509),  3^508=(3^509)/3=1 (mod 509) since inverses are unique]]> Oct 24, 2012 1:05:34 PM Oct 24, 2012 1:05:34 PM hp1702 0 http://mathforum.org/kb/thread.jspa?messageID=7911862&tstart=0#7911862 Edward,
In #1, your idea of adding/subtracting multiples of 9 (k +/ 9m for k = 0, 1, ..., 8 and m any integer) is solid.
Have forgotten]]> Oct 24, 2012 6:50:41 AM Oct 24, 2012 6:50:41 AM benb 0 http://mathforum.org/kb/thread.jspa?messageID=7911659&tstart=0#7911659 Oct 23, 2012 8:52:38 PM Oct 23, 2012 8:52:38 PM LaC0saNostra 2