[creative555]

A dart board contains a region worth 9 points and a region worth 4 points as shown in the
Accompanying figure. If you are allowed to throw as many darts as you wish, then what is the largest possible total score that you cannot get?
The dart board figure has two circles. The larger circle has the number 9 and the smaller circle is inside the circle with the number nine and it carries the #4 in it.
I tried using all the common factors for numbers 4, 9, and 13 but I cannot come up with a formula which does not include any of them combined.

[stevetcg]

(this is not a correct answer I don't think)

[Unknown008]

The problem is.. I can't see any picture here... Have you attached your picture well?

Click on "Go advanced" under the post box, and there, click on manage attachments. Browse your computer for the picture and click on upload. Then, post the picture.

The description you gave is quite confusing, especially that part:

The larger circle has the number 9 and the smaller circle is inside the circle with the number nine and it carries the #4 in it.

If it were a real dart board, the smaller circle would be the most scoring... :confused:

[HelpinHere]

I'm not sure of the formula. But, here goes.

You already know you can't get 1-3, and 5-7.
You get the smallest number you have available, 4, and start multipling it by an increasing interger.
When you get to the first value where that will equal more than the greater number, you place the greater number there in the sequence, one at a time.
You proceed until the numbers you get are sequential, enough to equal the value of the smaller number.

Sorry if that doesn't make sense...

I'll show you an example, using 3 and 8:
1x3 = 3
2x3 = 6
1x8 = 8
3x3 = 9
8+(1x3) = 11
4x3 = 12
8+(2x3) = 14
5x3 = 15
2x8 = 16
8+(3x3) = 17
...
Because, starting at 14, all the numbers became sequential, and for the amount of the lowest number, 3 (14, 15, 16), then you can conclude that by combining different ammounts of eight and three, you can get every number from 14 and above. Thus, the highest number you can't get is 13.

Equation: X[lower interger]+Y[higher interger]=Z
You then find the highest total Z cannot equal.

Now, you do the same thing with 4 and 9.


(Note: This isn't my particular area of expertise, so if it is wrong, feel free to correct me.)

Unknown, as the OP never asked for probability, the size or placement of the various regions doesn't matter. Just that those two scoring regions, and only those two, are on the board.

[Unknown008]

Ah! Now I understand! That's the second question of this kind I'm 'meeting', so, you don't expect me to recognise such a type of question, and I now understand the problem better. The thing was that the title included 'probability' and I thought that was about probability.

And I've got the answer for that number too! ;)

Ans: 23

[HelpinHere]

Yep, my guess is that the word "probability" in the title is either a misused term by the OP, or this setup has 2 questions, and the "probability" is related to the second one.

Yep, I got the answer too. I'm not going to post though, waiting for the OP to show some work.

[hollywooddoc]

The equation for this wher 2X+1=Y for the first consecutive number is always (X-1)(Y-1). Therefore the greatest number not attainable is simply (X-1)(Y-1)-1!