AIME I 2010 #7

[stupidityismygam]

Define an ordered triple [unparseable or potentially dangerous latex formula] of sets to be minimally intersecting if [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]. For example, [unparseable or potentially dangerous latex formula] is a minimally intersecting triple. Let [unparseable or potentially dangerous latex formula] be the number of minimally intersecting ordered triples of sets for which each set is a subset of [unparseable or potentially dangerous latex formula]. Find the remainder when [unparseable or potentially dangerous latex formula] is divided by [unparseable or potentially dangerous latex formula]

[klebian]

We see that triples of the form [unparseable or potentially dangerous latex formula] are minimally intersecting (Let's call this a "basic triple"). Additionally, the other 4 unused numbers can be added to any one of the three sets and retain the property of being minimally intersecting. For example [unparseable or potentially dangerous latex formula] is minimally intersecting, and from there you can go to [unparseable or potentially dangerous latex formula] or anything like that. So for any basic triple that is minimally intersecting, we have a number of ways to add the 4 other integers. Each integer has 4 options- to go in the first set [unparseable or potentially dangerous latex formula], second set, third set, or in none. Thus for each basic triple there are [unparseable or potentially dangerous latex formula] unordered minimally intersecting triples, and we multiply that by 6 because the triples are ordered (which is really pointless and adds an unnecessary step that seems to only serve a purpose of tricking kids who realize that there is no reason for the triples to be ordered), and we multiply that by 7C3=35 for the number of basic triples there are. So the answer is [unparseable or potentially dangerous latex formula]