All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Eight cards (Posted on 2005-06-14) Difficulty: 2 of 5
Eight cards are placed in the diagram shown below, one in each numbered cell. These are the facts:

a) There are 2 Aces, 2 Kings, 2 Queens and 2 Jacks.
b) Every Ace borders a King.
c) Every King borders a Queen.
d) Every Queen borders a Jack.
e) No Queen borders an Ace.
f) No two cards of the same kind border each other.

        +---+
        | 1 |
+---+---+---+
| 2 | 3 | 4 |
+---+---+---+---+
    | 5 | 6 | 7 |
    +---+---+---+
        | 8 |
        +---+
In what cells could you identify the card in it ?

  Submitted by pcbouhid    
Rating: 3.8889 (9 votes)
Solution: (Hide)

Though this is not a competition site, congratulations to Lisa (the first) and to the others who achieve the correct solution, even without showing their reasoning.

Let's start trying to deduce what card is in the "house" 6.

1) Suppose card number 6 were an Ace.

Then, neither cards number 7 nor 8 could be an Ace (from (f)), could be a Queen (from (e)), and could be a King (from (c)). Then, at most one card of cards number 7 and 8 must be a Jack, wich contradicts (d). So, card number 6 is not an Ace.


2) Suppose card number 6 were a Queen.

Then, none of cards number 4, 5, 7 and 8 could be a Queen (from (f)), and could be an Ace (from (e)). Then, two Aces and a Queen would be cards number 1, 2, and 3, wich, from (e) and (f), is impossible. So, card number 6 is not a Queen.


3) Suppose card number 6 were a Jack.

Then, neither cards number 7 nor 8 could be an Ace (from (b)), could be a Jack (from (f)), and could be a King (from (c)). Then, at most one card of cards number 7 and 8 must be a Queen, wich contradicts (c). So, card number 6 is not a Jack.


Then, card number 6 is a King.


Since card number 6 is a King, card number 5 or 4 is a Queen, from (c) and (d).

If card number 5 is a Queen, then, from (d), card number 3 is a Jack. Then, from (c), card number 2 is not a Queen, and card numbers 1 and 4 are a King and a Queen, respectively. Then, card number 2 must be a Jack, wich contradicts (f). So, card number 5 is not a Queen, and therefore card number 4 is a Queen.

Then : from (f), neither of cards number 1 and 3 is a Queen; from (d), neither of the cards number 7 and 8 is a Queen; and card number 5 is not a Queen, as showed above. So, card number 2 is a Queen, and : from (d), card number 3 is a Jack; from (c), card number 1 is a King; and from (f), card number 5 is an Ace.

A Jack and an Ace are left for cards number 7 and 8, and you are not able to deduce what card is in these cells, because both options fits.
                    
                       +---+
                       | K |
               +---+---+---+
               | Q | J | Q |  
               +---+---+---+---+
                   | A | K | ? |  
                   +---+---+---+
                       | ? | 
                       +---+                      
The two "?", one is an Ace and the other is a Jack (both option fits).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
OOOOOOOOOOpsvincent2005-09-09 10:48:26
awaiting a solutionvincent2005-09-09 10:44:20
re: What I think it is....David2005-06-19 22:56:30
re: Its time to declare a solutionpcbouhid2005-06-19 20:17:25
What I think it is....David2005-06-19 19:23:20
Its time to declare a solutionFrank2005-06-19 16:15:37
re(3): here's your answerpcbouhid2005-06-19 02:23:38
re: very difficultpcbouhid2005-06-19 02:21:45
re(3): To Lesley and NetGoofypcbouhid2005-06-19 02:17:56
re(2): here's your answerkc woolbright2005-06-19 00:09:11
very difficultkc woolbright2005-06-19 00:00:49
SolutionBob2005-06-18 19:54:31
re(2): To Lesley and NetGoofybrianjn2005-06-18 05:23:05
re: To Lesley and NetGoofypcbouhid2005-06-17 19:33:07
A hopefull Solution Here!NetGoofy2005-06-17 19:21:16
re: solution?pcbouhid2005-06-17 13:25:27
solution?Lesley2005-06-17 13:21:39
re(2): Full Solution - addendumLisa2005-06-17 09:49:17
re: Full Solution - addendumCharley2005-06-17 04:45:36
No Subjectchris2005-06-16 20:09:56
My GuessMason2005-06-16 18:30:34
re: Full Solutionbrianjn2005-06-15 00:54:58
Answer?Jon2005-06-14 21:44:16
LISA I THINK YOU ARE RIGHTsonya2005-06-14 19:18:53
re: To avoid misunderstandingLisa2005-06-14 15:57:37
To avoid misunderstandingpcbouhid2005-06-14 15:43:09
re: here's your answerLisa2005-06-14 15:36:50
re: here's your answerjohn2005-06-14 15:36:11
re(2): Does CO border AZ?john2005-06-14 15:30:36
re: Does CO border AZ?john2005-06-14 15:14:12
here's your answersonya2005-06-14 14:26:12
QuestionDoes CO border AZ?john2005-06-14 14:25:42
re: Impossiblepcbouhid2005-06-14 13:46:40
SolutionFull SolutionLisa2005-06-14 13:29:45
Some ThoughtsOverlooking Last conditionSyzygy2005-06-14 13:06:26
Some ThoughtsImpossibleCharley2005-06-14 11:39:13
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (13)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information