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 Nine cards (Posted on 2005-10-14)
Nine cards are disposed as shown:
```           +---+---+---+
| a | b | c |
+---+---+---+
| d | e | f |
+---+---+---+
| g | h | i |
+---+---+---+
```

1) There are, at least, two Aces, two Kings, two Queens and two Jacks.
2) Every Ace borders a King and a Queen.
3) Every King borders a Queen and a Jack.
4) Every Queen borders a Jack.
Note: "border" means "touch" horizontally or vertically, not diagonally.

Identify the nine cards.

 Submitted by pcbouhid Rating: 2.7500 (8 votes) Solution: (Hide) ``` +---+---+---+ | a | b | c | +---+---+---+ | d | e | f | +---+---+---+ | g | h | i | +---+---+---+```Any corner card (a, c, g, i) borders on just two cards, any mid-edge card (b, d, f, h) borders on just three cards, and the center card (e) borders on just four cards.From (1) and (2), each of the two Aces borders on a Queen; from (1) and (3), each of the two Kings borders on a Queen; and from (1) and (4), each of the two Queens borders on the same Jack or two different Jacks. So, the Queens border on at least five different cards.There are three basic ways for two cards to be bordered on by a total of five cards (the other ways are the same, just rotating the grid):``` Case I Case II Case III +---+---+---+ +---+---+---+ +---+---+---+ | | o | X | | o | X | o | | | o | | +---+---+---+ +---+---+---+ +---+---+---+ | | o | o | | | o | | | o | X | o | +---+---+---+ +---+---+---+ +---+---+---+ | o | X | o | | o | X | o | | o | X | o | +---+---+---+ +---+---+---+ +---+---+---+```The cards are identified by "X" and the bordering cards by dots.If there are only two Queens ("X´s"), then: Case I does not apply because two Jacks (dots) as well as two Aces (dots) and two Kings (dots) would be necessary; Case II does not apply because no Ace (dot) could border on a King (dot) as required by (2); and Case III does not apply because the two Queens could not border on the same Jack (dot) as well as two Aces (dots) and two Kings (dots).So, there are three Queens.Then five cards borders on two Jacks (X´s): three Queens (dots), from (4), and two Kings (dots), from (1) and (3).Then, in Cases I, II and III the unmarked cards are Aces. Then, Case I does not apply, from (2); Case II does not apply, from (3). So, Case III applies and a Jack is in the center.From previous reasoning, the "X´s" are Jacks and the unmarked cards are Aces:``` +---+---+---+ | A | o | A | +---+---+---+ | o | J | o | +---+---+---+ | o | J | o | +---+---+---+```Then, because every King borders on a Queen (from (3)), the top dot is not a King, so is a Queen:``` +---+---+---+ | A | Q | A | +---+---+---+ | o | J | o | +---+---+---+ | o | J | o | +---+---+---+```Then, because every Ace borders on a King (from (2)), the dots in the second row are Kings, and because there are 3 Queens, the remaining dots are Queens:``` +---+---+---+ | A | Q | A | +---+---+---+ | K | J | K | +---+---+---+ | Q | J | Q | +---+---+---+```

 Subject Author Date re(2): Wayyyyyyyyyyyyyyyyyyyyy Easy thegnome54 2006-02-03 06:01:17 re: No Subject - to saprom pcbouhid 2005-12-22 08:11:02 re: Wayyyyyyyyyyyyyyyyyyyyy Easy saprom 2005-12-21 12:27:54 No Subject saprom 2005-12-21 12:14:44 Wayyyyyyyyyyyyyyyyyyyyy Easy William Bettley 2005-12-09 20:14:53 Too easy.... Jas 2005-10-18 13:00:26 re(2): Pooter yet!! pcbouhid 2005-10-18 06:28:42 re: Pooter yet!! Ken Haley 2005-10-18 01:32:55 re: is this right? - not yet pcbouhid 2005-10-17 19:46:08 is this right? mayra villalobos 2005-10-17 17:29:11 Pooter yet!! pcbouhid 2005-10-17 06:33:27 re: pooter! Ken Haley 2005-10-17 00:24:54 re(2): my solution (step-by-step) Bender 2005-10-17 00:20:21 re: three queens?? pcbouhid 2005-10-16 13:19:44 three queens?? fay 2005-10-16 12:00:56 I THINK I GOT IT Corey Jackson 2005-10-15 13:18:29 re: No Subject Charlie 2005-10-15 03:27:46 No Subject Mary 2005-10-14 23:17:30 re: my solution (step-by-step) pcbouhid 2005-10-14 20:53:35 my solution (step-by-step) tanx 2005-10-14 20:43:23 re: No Subject pcbouhid 2005-10-14 18:13:51 No Subject Ady TZIDON 2005-10-14 15:55:31 computer logic Charlie 2005-10-14 15:25:20 pooter! pcbouhid 2005-10-14 15:25:18 re(3): solution pcbouhid 2005-10-14 15:12:21 re(2): solution derek 2005-10-14 15:06:19 re: solution pcbouhid 2005-10-14 15:01:49 re: solution pcbouhid 2005-10-14 14:59:48 re: solution derek 2005-10-14 14:58:50 solution derek 2005-10-14 14:57:32

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