TITLE: Scores achievable in a game of darts
QUESTION [10 upvotes]: This is from an italian math competition. In the game of darts below a player is able to reach points only of 105, 30, 14 (either exact center or on the border). The final score is the sum or all points obtained.
There is a maximum score that cannot be reached in a game, above which all scores are feasible. What is this number ?

I think I have a solution, but it is not really elegant. Looking on the internet for inspiration I found this.
Chicken-Nugget lemma: Given a,b coprimes, we consider the set $S=\{am+bn,m\ge0,n\ge0\}$.  Than the maximum number not belonging to $S$ is $ab-(a+b)$.
Using this we can proceed like that. A generic score is of the form $s=105a+14b+30c$, where $a,b,c$ are positive. We can write is as $105a+2(7b+15c)$. According to the lemma all numbers higher than $105-15-7=83$ are of the form $7b+15c$, therefore we subsitute $7b+15c \rightarrow 84+t$, where $t\ge 0$ (note that also smaller numbers are feasible, but we do not consider this). Substituting $s=168+(105a+2t)$. Now again according to lemma all numbers higher than $210-105-2=103$ can be written as $105a+2t$. Therefore we substitute  $105a+2t \rightarrow 104 +q$, with q positive and obtain $s=272+q$. This shows that all scores larger than $272$ are feasible.
It remains to show that $271$ is not feasible. At this point I think it is mere luck if it is not, but we can give it a try. We suppose:
$271=105a+14b+30c$
We have $a \le 2$ and by parity we must have $a=1$. Therefore we have $166=14b+30c$, or $83=7b+15c$. This shows that $c \le 5$, but for neither $c \in \{0,1,2,3,4,5\}$ is $b$ integer. Therefore $271$ is the answer. c.v.d.
What I do not like in this solution is the second part. Is there some "theory", like in the chicken nugget theorem that assures that 271 should not be achievable ? I read some proofs of the chicken nugget theorem for two variables, but I am not sure that they apply to the 3-variables case. As it is, it looks to me by chance that the score 271 is not achievable. It could have been, and than I would have needed to check by hand also 270, 269,... which is not really elegant nor general ...
Any ideas/comments/proposals for other solutions ?

REPLY [10 votes]: $105$ is the only odd point value, so it is a necessary inclusion to reach any odd total. As $2\cdot 105 = 15\cdot 14$ is reachable without $105$, any reachable odd number can be reached with exactly one $105$, and any reachable even number can be reached with no $105$.
As a consequence, if $e$ is an unreachable even number, then $e+105$ is an unreachable odd number, and vice versa. So clearly, the highest unreachable number $x$ is odd. And $x-105$ is the highest unreachable even number.
This means that $\frac{x-105}2$ is the highest unreachable number using $\frac{14}2, \frac{30}2, \frac{42}2$ and $\frac{70}2$ (the last two are redundant, so I will ignore them). And the largest unreachable number using $7$ and $15$ is $7\cdot 15 - 7 - 15 = 83$. So we have $\frac{x-105}2 = 83$, which gives us $x = 271$.