(In reply to
Trying to avoid 98 mods... by Jer)
You are correct on the number of zeroes, but I fail to see how this really helps?
As for a shorter way, I've been trying some things, but I can't seem to get very far.
I started with:
98! MOD 101
= (100!/(99*100)) MOD 101
and used the distributive property of MOD:
=((100! MOD 101)/((99*100) MOD 101)) MOD 101
and reduced:
=((100! MOD 101)/(9900 MOD 101)) MOD 101
=((100! MOD 101)/2) MOD 101
Where should I go from here, though?
Edited on September 3, 2011, 4:59 am

Posted by Joshua
on 20110902 18:00:41 