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

Home > Logic
Bullseye (Posted on 2013-06-25) Difficulty: 3 of 5
Three darts players playing 501 up have had three turns (A, B and C), each with three darts each turn.

From the information given below, which player can finish with a bull (50) on their 10th dart?
Player 1 
Turn A: Scored an odd score more than 70 points. 
Turn B: Scored 180. 
Turn C: Scored an even number that is more than 
player 3's score in turn B + 1 more point.

Player 2
Turn A: Scored one score less than the points 
scored in player 1's turn B.
Turn B: Scored two-thirds of the points scored 
in player 1's Turn B.
Turn C: Scored an even number of points that 
is more than 160 points.

Player 3
Turn A: Scored 17 more points than he did in Turn C.
Turn B: Scored the lowest even score of all nine 
turns taken by all 3 players.
Turn C: Scored an odd score that is more than 100 
point.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
From the fact that Player 1 score 180 points in turn B, we can determine player 2's eligibility to win with a 10th dart bull's eye. Player 2 scored one score less than 180 on turn A, which is 177 points. They also scored two-thirds of player 1's turn B on their own turn B; 2/3 * 180 = 120 points on turn B.

120 + 177 = 297 points

In order to be eligible to win with a bull's eye on the 10th dart, 451 points must be accounted for in the first 9 darts (3 turns). 451 - 297 = 154 points left in turn C. As the information given states that player 2 gets an even score over 160 on turn C, a bull's eye would put them over the 501 mark on their 10th dart. Player 2 can't win with bull's eye.

Moving on to player 3, we have less information to go off of. We know turn B is the lowest even number of all turns, so it must be less than 120 points (assuming it was not a tie for the lowest even). From the information for turns A and C, the highest pair that we can have is (174, 157) as 163 is not possible for a round score. The sum of those two values is 331. In order to be eligible, as above, we need the first 3 rounds to sum to 451. 451 - 331 = 120, and as we scored the lowest even score of all 9 turns in turn B, and that must be less than 120, we can't get to 451 before the end of the 3rd turn. Player 3 can't win with bull's eye.

Now, let's check to make sure player 1 can win, in case my understanding of no ties for lowest even is incorrect. Player 1 scored 180 points on turn B, leaving 271 points to be distributed between the other two turns. Turn C is simply an even number greater than whatever player 3 scored in their turn B. As round scores are not required to be distinct, player 1 could score 91 points on turn A, and 180 points, again, on turn C, to be able to win on the 10th dart via bull's eye. As player 3's turn B score is unknown, there is no limit to how low player 1 can score on turn C. There are, in fact, 36 total combinations possible that would result in them being the winner with a bull's eye on the 10th dart.

Player 1 can win with a bull's eye on their 10th dart

  Posted by Justin on 2013-06-29 10:22:15
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 (3)
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