Hearts Game Word Problem

Question

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 ?

Answer 1

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.