Vulture 10's Riddle

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)

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.

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

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

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.

Quote:

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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.

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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. (???)

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

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:

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

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

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.