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.
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Submitted by pcbouhid
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Rating: 2.7500 (8 votes)
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Solution:
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(Hide)
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+---+---+---+
| 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 |
+---+---+---+
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