-1
$\begingroup$

Hearts is a game where the lowest score wins. We know this :

  • The fourth player scored a $105$
  • The first three players scored a combined value of $103$
  • No scores are zero
  • No score (except loser) can exceed 100

If the second score was greater than four times the first and the third score was greater than twice the score of the second, what was the scores of all four players ?

$\endgroup$
  • $\begingroup$ D = 105, A + B + C = 103, 4A < B, 2B < C. Stuck there $\endgroup$ – Mark Walsh Jul 7 '15 at 5:46
  • 1
    $\begingroup$ A=1, B=5, C=97, D=105 works, but there is not only one solution $\endgroup$ – BusyAnt Jul 7 '15 at 6:00
  • 1
    $\begingroup$ BusyAnt is correct, is there any more information? Are there any rules to Hearts that are relevant to solving this problem? $\endgroup$ – Mike Pierce Jul 7 '15 at 6:01
  • $\begingroup$ I guess there is more information I'm not seeing. $\endgroup$ – Mark Walsh Jul 7 '15 at 6:08
  • $\begingroup$ Are we assuming all scores are non-negative integers? This seems not to be stated in the body of the question. $\endgroup$ – Gerry Myerson Jul 7 '15 at 7:25
0
$\begingroup$

Assuming there are no other particular rules, we have several solutions. Let $A, B, C, D$ be the scores (in ascending order) of the four players.

We know that $D=105$, it is no more a problem for the rest of the reasoning. We also have $A+B+C=103$ and the two inequalities $4A<B$ and $2B<C$

First, since no scores are equal to $0$, $$ 3B=2B+B<C+B<103 ~~~~~(1) $$ Thus $$ B\le 34 $$ In the same way, $$ 5A=4A+A<B+A<103~~~~(2) $$ And then $$ A\le 20 $$ We can deduce that $B+A \le 54$, and repeat $(2)$ with $5A \le 54$. Doing this repeatidly, we should arrive (not proven) to $A\le 7$

Let $A=7$. Then we need $B\gt 28$
Let $B=29$. Then we need $C \gt 58$
And for $A+B+C =103$, we choose $C =67$

I think that with this way of thinking you can find many solutions, by testing different values of $A$ and $B$. This is probably not the most effective way but it gives you many solutions with not so much effort.

$\endgroup$
  • $\begingroup$ That's what I couldn't figure out. I got to A had to be 7 or less, but after that I think I need to use more rules of hearts (the total of each round can either be 26 (not shooting the moon) or 78 (shooting the moon, where 3 players receive 26 points each) $\endgroup$ – Mark Walsh Jul 7 '15 at 13:13
  • $\begingroup$ Maybe from here you can generate all possibilities and then eliminate from the rules you know, or you could translate these rules mathematically... $$^\text{If this answer has been helpful to you, please remember to accept and upvote}$$ $\endgroup$ – BusyAnt Jul 7 '15 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.