The Hatter, the Hare, the Rabbit, the Mouse, Alice and I met for a Tea Party. The table had two sides, with three chairs on each side, numbered like so:

Side A: 1 2 3
Side B: 6 5 4

Chair one was across from chair 6, chair 2 across from chair 5, etc. We would sit down one at a time, in any chair we choose. The first guest to sit down was called the first guest. After the second guest sat down, however, everyone moved clockwise to the next chair. This behavior was continued after the fourth and sixth guest sat down.

My question is, after everyone had been seated, where was I seated?

(1) The first guest sat in chair 1.
(2) The mouse sat down last.
(3) After the first time everyone moved, no one had been on side B yet.
(4) The Hatter and Alice choose the same starting chair.
(5) When I was in chair 1, the Rabbit was in chair 6.
(6) I started on side B, but didn't end there.
(7) The hare moved all three times.
(8) Alice ended up between the mouse and the hare.
(9) I sat down after the rabbit but before the hatter.
(10) The mouse sat down in an odd numbered chair.

Also, a bonus question. Who am I?

Hi Zipp,

My deductions lead to "I" being in chair 2, which agrees with your posted solution. But I turned out not to need either clue (4) or clue (7) to get there (see solution below) – were those just freebies? I notice josh didn't use them either.

Also, when you post your official solution could you explain how the clues point to "I" being the Cheshire Cat? I thought since you'd already added the Rabbit to the party that the "I" addition could also be any other character (eg Mock Turtle, or Duchess, or Queen of Hearts, etc.)? Or am I misremembering my Alice?

HM

seven-step-solution if you're interested:

i - mouse is sixth guest (2) and rabbit, I, hatter is a sequence (9), therefore the rabbit can only be the first, the second, or the third guest (and I can only be the second, the third or the fourth)

ii - The first and second guests start in chairs 1 and 2 respectively (1) and (3) and therefore after the first move are in chairs 2 and 3.

iii - I cannot be the second guest, because the second guest does not start on side B (6). Therefore I am the third or fourth guest (two moves) and my chair sequence is either 561 or 612. By clue (5), the rabbit's sequence can then only be 3456 or 561. Neither sequence is compatible with the rabbit being the second guest, because neither sequence starts in chair 2. Therefore the rabbit is the third guest. From (i) it follows I am the fourth and the hatter is the fifth.

iv – Alice ends up next to the mouse (6). Because the mouse ends up in an even numbered chair (10), Alice must end up in an odd one. Between the first and the second guest, therefore, Alice can only be the second guest, in chairs 2345. By elimination, the Hare is the first.

We now have the order of the guests:
first – Hare 1234
second – Alice 2345
third – Rabbit
fourth – I
fifth – Hatter
sixth – Dormouse

v – From clue (6) the mouse's chair sequence must be 56 in order to end up on the other side of Alice (8).

vi – Of the rabbit's two possible sequences from (iii), only 561 fits – the rabbit cannot sit down in chair 3 because Alice will already be in it. Again by clue (5), my sequence is thenn 612.

vii – By simple elimination then, the hatter can only sit down in chair 2, leaving chair 5 free for the mouse.

first – Hare 1234
second – Alice 2345
(after first move, Hare is in 2, Alice is in 3)
third – Rabbit 561
fourth – I 612
(after second move, Hare is in 3, Alice is in 4, Rabbit is in 6, I am in 1)
fifth – Hatter 23
sixth – Dormouse 56
(after third move, Hare is 4, Alice is 5, Rabbit is 1, I am 2, Hatter is 3, Mouse is 6)

Posted by hm on 2005-07-20 14:35:47