Solution to Problem 7: Order in the Court

I'm going to put the notes first on this problem, since it's mostly an apology for the way in which I'm solving this one.

Notes: I went into this problem with a triage mentality.  My assumption for all three of these problems was that the lowest three numbers were going to start with 1 and the highest three numbers would start with 9.  If that assumption led to a solution, it would take less than a minute and be worth five points.  If not, I would likely not stick around to work out what went wrong.  As it turned out, my assumptions were correct and I got 15 points in less than five minutes.

In that spirit, I'm not going to rigorously demonstrate that these answers are unique.  I think that speed was more important than logic for these small puzzles.

	#2 must be 1119 and #1 is 1111.  To be between #3 and #5, #4 must be 1911.  To have #7 be more than #6, #9 is 9911.
	Nothing but net.

	If #2 were 11xx, then it would have to be 1199 to be more than #1, but then #6 would be more than #7.  So #2 is 19xx.  Since #8 is more than #7, #7 is 9991 and #8 is 9999. #6 must be 919x, #5 must be 91xx to be "two" more than #3.  To have #4 be less than #5 but more than #3, it must be 1999, and #6 is 9191 to make it different from #5.
	And that was the hard one.

	#1 must be #11x9 to be less than #2.  #3 and #4 must be 1991 and 1999 respectively.  #5 must be 919x, so #6 is 9911 to make it more than #5 but less than #7.  #8 must be 9991 to make it more than #7.
	Three up, three down.