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

Home > Logic
Poker Precept (Posted on 2013-04-19) Difficulty: 3 of 5
In a game of poker, one of the hands of five cards had the following features:
  1. There was no card above a 10 (an ace is above a 10 in poker).
  2. No two cards were of the same value.
  3. All four suits were represented.
  4. The total values of the odd and even cards were equal.
  5. No three cards were in sequence.
  6. The black cards totaled 10 in value.
  7. The hearts totaled 14.
  8. The lowest card was a spade.
What was the hand?

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Answer | Comment 1 of 4

Since the sum of the odd cards equaled the sum of the even cards, the sum of the odd cards was even. Therefore, there were an even number of odd cards. There could not be 0 odd cards because there would have to be 0 even cards, too. Then, there would be no hand. If there were 4 odd cards, then the remaining even card would be the sum of the odd cards. However, the lowest possible sum of 4 odd cards is 3+3+3+3=12>10. Therefore, there were 2 odd cards and 3 even cards. Since the hearts totaled 14, there were 2 hearts and 1 of the other suits. If one of the black cards was odd, then the other was odd since the sum of the black cards was 10. Since there were 2 odd cards, the sum of the odd cards would be 10. Then, the sum of the even cards would be 10, which is impossible since the hearts would both be even and add to 14. Therefore, the black cards were both even. The hearts could not both be even since there were only 3 even cards. Therefore, at least one heart was odd. Since the sum of the hearts was 14, they were both odd. Then, the total of the odd cards was 14, so the total of the even cards was 14, too. Since the total of the black cards was 10, the diamond was a 4. The hearts were different odd numbers under 10 that added to 14, so they were 5 and 9. The black cards added to 10, so they were either 2 and 8, or 4 and 6. They could not be 4 and 6 since there was a 4 of diamonds. Therefore, they were 2 and 8. Since the lowest card was a spade, the 2 was a spade and the 8 was a club.

Hand:8 of clubs, 4 of diamonds, 2 of spades, 5 of hearts, 9 of hearts

 


  Posted by Math Man on 2013-04-19 11:02:59
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 (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information