Question 570006
<pre>
Let the test begin at A:BC and end at D:EF

1. The 45 minutes tell us that C and F differ by 5. 
2. D is 3 more than A, and EF is 45 more than B, or else D is 4 more than A,
   and EF is 15 less than BC.

If D is 3 more than A, then A=2, and D=5, for they are the only ones of the
given numbers which differ by 3.  So F=1 and C=6, or F=6 and C=1.  So the first
case that might seem possible is for the test to begin at 2:01, but that would
make the test end at 5:46, but we can't use 4.  The next seemingly possible
starting time would be 2:06, but that would make the test end at 5:51, which is
out because that uses 5 twice.  Any later starting time would likewise cause 5
to be used twice.  So D is not 3 more than A.

So that means D is 4 more than A, and EF is 15 less than BC. 

So either A=1 and D=5 or A=2 and D=6

If A=1 and D=5 then none of the other given numbers differ by 5, but C&F must
differ by 5.

Therefore A=2 and D=6, and the only 2 left which differ by 5 are 0 and 5, so
either C=5&F=0 or C=0&F=5

Since EF is 15 less than BC, the starting time must be 2:15 or later.  It can't
start at 2:15, for it would end at 6:00 which uses two 0's. It can't start at
2:20 or 2:25, for they use two 2's. If it started at 2:30 it would end at 6:15.

Hey, that is the answer!

But could there be more than one possibility?  Let's see:

It can't start at 2:35 for it would end at 6:20 and that uses 2 twice. There is
no 4, so it can't start at 2:40 or 2:45.  It can't start at 2:50 or 2:55
because C or F must be 5, so B can't be 5.

So we have ruled out every possibility except one:

The test starts at 2:30 and ends at 6:15.

Edwin</pre>