Compute minimum amount of 60s to obtain a 9-darts finish in 501 double darts out In 501 double out darts, it is possible to reach the final score of 501 with only 9 darts. The dartboard has fields 1 - 20, and also double and triple fields and the semi-bull (25) and bullseye (50).
The last dart needs to hit a double field, i.e. either of:
[2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 50]
I already know all possible 9-dart-finish combinations (you can look them up on the internet). I have noticed that you need a minimum of three 60s to obtain a 9-darts finish. Now I wonder: is there a way to prove this? Or is there at least a way to see that you need to hit a 60 at least once in order to obtain a 9-darts finish? Because in 301 darts you can obtain a 6-darts finish without ever hitting a 60.
Would anyone care to show an example calculation?
 A: Proof you need at least one 60:
The minimum scoring at each throw should be at least $501-8\times57=45$. Thus, the only double that suffices this condition is bullseye $50$. Now we have a problem getting $451$ points in 8 throws.
The score $451$ isn't a multiple of $3$, so you cannot cover it with triples. The largest points not divisible by 3 is again bullseye $50$. However, $451-50 = 401 > 57\times7$.
Edit: proof you need at least three:
By analogy, the minimum scoring at each throw is $501−6×57-2×60=39$. So you can finish the game either with $40$ or $50$.
With 40:
You need to cover $501-2×60-40=341$ points with 6 throws. It's not divisible by 3, so one throw have to be at least 50, $(341-50)/5=58.2 > 57$.
With 50:
You need to cover $501-2×60-50=331$ points with 6 throws. $331\equiv 1 \mod 3$, so we need at least one throw with a remainder 1 (the largest one is 40; $(331-40)/5 = 58.2>57$) or two throws with a remainder 2 (the largest one is 50; $(331-100)/4 = 57.75>57$),
A: The highest score with one dart is a triple $20$ $(60)$ and the second highest is triple $19$ $(57)$. The highest score with a final dart is a bullseye $(50)$ and the second highest is double $20$ $(40)$.
Suppose you want to finish with a non-bullseye double, and no more than two triple $20$s. Then your maximum score is $2 \times 60 + 6 \times 57 + 1 \times 40 = 502$ which is $1$ too much, and you cannot reduce this nine-dart total by exactly $1$ since any change reduces any of the triples by $3$ or the double by $2$, and a change from a triple to a double or single or bullseye is at least $7$.
Suppose you want to finish with a bullseye $(50)$, and no more than two triple $20$s. Then your maximum score is $2 \times 60 + 6 \times 57 + 1 \times 50 = 512$ which is $11$ too much, and you cannot reduce this nine-dart total by exactly $11$ since any change reduces any of the triples by a multiple of $3$, and a change from a triple to a double or single is at least $17$, while a change from a triple $20$ to a second bullseye leaves an impossible-to-close $1$ and a change from a triple $19$ to second bullseye leaves an impossible-to-close $4$.
So there is no  nine-dart finish without at least three triple $20$s.
