Vulture 10's Riddle

[jenae]

Hi, originally posted by Vulture 10:-

I have a problem that's not really aPC related ,but I hope some of the fine minds that use this site can help me .

Imagine a dart board divided into four segments with the numbers 1,3,5,9 in the different segments.You have six darts and all six must score.How can I make 21.

Thats the problem I was given and it has absolutely done my head in.I've been assured that there is a legitmate answer to this but I can't find it.CAN ANYONE HELP ME.I THINK I'M GOING MAD.......

Thanks

The answer is related to "how can I make 21" Darts has two or more players so you only need to throw 3 darts and your opponent 3 darts. You throw 2*9 and 1*3. This answer was not worked out be me (no way smart enough)

[greengoose1]

Hi jenae, If I understand you one player has 6 darts and they have to score as he/she throws in such a manner as to total 21 in the four segments labeled 1, 3, 5, and 9.

If this is correct then I submit the following:

3darts x 5=15

1dart x 4= 4

2darts x 1= 2

A totalof 21.

[jenae]

Hi, very creative ,however 1 dart by 4 is not on the board!!

[Rapmaster]

greengoose-- '4' is not one of the segments. 1, 3, 5 and 9

[imadreamer2]

How about

1 dart x 9 = 9
1 dart x 5 = 5
2 darts x 3 = 6
1 dart x 1 = 1

for a total of 21

Won't the problem really be actually getting the darts to go where you want them to go? It would be for me.

[Rapmaster]

Originally posted by imadreamer2
How about

1 dart x 9 = 9
1 dart x 5 = 5
2 darts x 3 = 6
1 dart x 1 = 1

for a total of 21

Won't the problem really be actually getting the darts to go where you want them to go? It would be for me.

Hmm but you need to use all 6 darts... that's only 5.

I think Jenae gave the answer above-- there are two players in darts. (???)

[104456]

The algebraic formula should look something like:

[a x 1] + [b x 3] + [c x 5] + [d x 9] = 21
a + b + c + d = 6


Where the top formula is the total score and the bottom is the number of darts.
So anybody remember how to do algebra? :D:D

[greengoose1]

Durn that 4 anyway. And I thought I had it. All I did was put a 3 and 1 together on my worksheet in the wrong place. Yuk is me.:mad: :rolleyes:

[104456]

I dont think its possible anyway the closest you can get with those numbers and six darts is 20 or 22

9+5+3+[3 x 1] = 20
[3 x 5] + 3 + [2 x 1] = 20

[K G G]

An option with one player, and 6 darts:

A dartboard usually has two rings, the outer counting a dart as double, the inner ring counting a dart as triple.

I.e.
5 darts into triple 1 = 15 pts
1 dart into double 3 = 6 pts
Total 21

[jslater25]

104456 --

If I remember high school algebra (its been a long time), you need at least one more equation before you can solve it using algebra. Since there are four unknowns (a,b,c,d), you must have at least three equations.