Some raffle tickets have been printed for a fund raising event. There are a
total of 100,000 tickets, each one with a distinct 5-digit number ranging from
00000 to 99999. But some tickets are ambiguous. For example, we can’t
determine if the number of the ticket shown below is supposed to be 09081 or
18060. We define an ambiguous ticket to be one that can show two different numbers
depending on the ticket’s orientation. How many of the 100,000 tickets
are ambiguous?
Digits used on the tickets appear as shown: 
Solution 1 by Anthony Ballor, De La Salle Collegiate
There are 3050 ambiguous tickets. This is because five numbers, 0, 1, 6, 8,
and 9, when rotated 180 degrees, form another number. Each number can be used
up to five times in one ticket and 55=3125. However, among the 3125
numbers, some numbers like 11111, 11011, and 61819, are not ambiguous. In order
for one of the 3125 numbers not to be ambiguous, its third digit must form itself
when rotated 180 degrees. Only 0, 1, and 8 do this. To be unambiguous, the 4th
and 5th digits must form the 1st and 2nd, respectively, when rotated 180 degrees.
For all of the 3125 tickets, there are 25 two-number combinations. So, after
the first two digits are chosen, there are only 3 possibilities for the 3rd
digit. Therefore, we must subtract 52*3=25*3=75 from 3125, leaving
3050 ambiguous tickets.
Solution 2 by Matthew Stephens, Plymouth Christian
Academy
We know that the ambiguous numbers are 0, 1, 6, 8, and 9. Let each of the 5 digits composing a ticket be designated by A, B, C, D, and E. We know that there are only 5 choices for A, and these are the same for B, C, D, and E. Therefore, there are 5*5*5*5*5=3125 ambiguous numbers. However, some of these create the same number when flipped. For example, 10101. Exactly how many of the 3125 are not truly ambiguous? There are 5 possible choices for A; and once A is picked, E is picked. There are 5 possible choices for B; and once B is picked, D is picked. There are 3 possible choices for C. Therefore, 5*5*3*1*1=75 of the 3125 are not ambiguous, which leaves us with 3125 – 75 = 3050 truly ambiguous numbers.
Partially correct solutions were submitted by: Alexander Tang
(Ann Arbor Huron H.S.); Kelvin Shuneson (Forest Hills Eastern
H.S.); Kevin Soubly (Detroit Catholic Central H.S.); Anupama
Prasad (Troy Athens H.S.); Matthew Brownell (Plymouth
Christian Academy); Brandon Turnbull (Birch Run H.S.); Mike
Meindertsma (Forrest Hills Northern H.S.); Jim Dunneback, Liz
Freund (Lansing Catholic Central H.S.); Rebe Henry
(C.M.U. Accelerated Math Program); and Pascal Carole (Saginaw
Arts & Sciences Academy).