
Word problem
A dart board is designed to have two scoring areas, as shown. If an unlimited number of darts is allowed, what is the largest finite score that cannot be attained?
I couldn't draw the two circles so I'll explain. If the dart landed on the outer circle if would count for 7 points and the inner circle would count for 15.
I see that I will be able to get all the multiples of 7
7 14 21 28 35 42
and then if i get a 15 and then all 7 i can get all the
15 22 29 36 43
and I get 7 and all 15..
7 22 37 52...
and I can't see how to get the largest finite score that can't be attained.
I tried other combinations also. Maybe i am doing it too randomly.
Could you please explain the logic of this problem step by step.
Thanks.
Vicky.

If you put x darts into the 15 point area and y darts into the 7 point area, you get 15x+ 7y points. So the mathematics question you are asking is "what is the largest number that cannot[ be written 15x+ 7y for nonnegative integers x and y."
7 divides into 15 twice with remainder 1 so 15(1)+ 7(2)= 1. If A is any number then 15(A)+ 7(2A)= A. x= A, y= 2A will give any number A. Of course, 2A is not a nonnegative number. In fact, x= A 7k and y= 2A+ 15k is also a solution for any integer k: 15(A 7k)+ 7(2A+ 15k)= 15A+ 7(15)k 14A+ 7(15)k= A since the "7(15)k" terms cancel out.
So we must have $\displaystyle A 7k\ge 0$ and $\displaystyle 2A+ 15k\ge 0$. Those are the same as $\displaystyle \frac{A}{7}\ge k$ and $\displaystyle k\ge \frac{2A}{15}$: $\displaystyle \frac{2A}{15}\le k\le\frac{A}{7}$
Well, obviously, that is always possible if there is an integer between $\displaystyle \frac{2A}{15}$ and $\displaystyle \frac{A}{7}$. In particular it is certainly possible if $\displaystyle \frac{A}{7} \frac{2A}{15}= A\left(\frac{1}{7} \frac{2}{15}\right)$$\displaystyle = \frac{1}{105}A> 1$. That is, it is possible for any A> 105.
But that doesn't mean it isn't possible for any number less than or equal to 105. But at least we have only a finite number of cases to check:
If A= 105, $\displaystyle \frac{2A}{15}= \frac{2(105)}{15}= \frac{210}{15}= \frac{42}{3}= 14$ and $\displaystyle \frac{A}{7}= \frac{105}{7}= 15$ so k= 14 or 15 will work. If k=14, x= 105 7(14)= 105 98= 7 and y= 210+ 15(14)= 210+ 210= 0. You can get 105 points by putting 7 darts in the 15 point circle and none in the 7 point circle. If k= 15, x= 105 7(15)= 0 while y= 201+ 15(15)= 15. You can get 105 points by putting 0 darts in the 15 point circle and 15 points in the 7 point circle.
If A= 104, $\displaystyle \frac{2A}{15}= \frac{2(104)}{15}= \frac{208}{15}$[tex]= 13 and 13/15 while $\displaystyle \frac{A}{7}= \frac{104}{7}$= 14 and 6/7. k= 14 is between them. x= 104 7(14)= 6 and now y= 2(104)+ 14(15)= 208+ 210= 2. You can get 104 points by putting 6 darts in the 15 point circle and 2 darts in the 7 point circle.
If A= 103, $\displaystyle \frac{2A}{15}= \frac{2(103)}{15}= \frac{206}{15}$= 13 and 11/15 while $\displaystyle \frac{A}{7}= \frac{104}{7}$= 14 and 6/7. k= 14 is still between those two numbers. x= 103 7(14)= 5 and y= 2(103)+ 14(15)= 206+ 210= 4. You can get 103 points by putting 5 darts in the 15 point circle and 4 darts in the 7 point circle.
If A= 102, $\displaystyle \frac{2A}{15}= \frac{2(102)}{15}= \frac{204}{15}$= 13 and 3/5 while $\displaystyle \frac{A}{7}= \frac{102}{7}$= 14 and 4/7. k= 14 is still between those two numbers. x= 102 7(14)= 4 and y= 2(102)+ 14(15)= 204+ 210= 6. You can get 102 points by putting 4 darts in the 15 circle and 6 darts in the 7 point circle.
If A= 101, $\displaystyle \frac{2A}{15}= \frac{2(101)}{15}= \frac{202}{15}$= 13 and 7/15 while $\displaystyle \frac{101}{7}$= 14 and 3/7. k= 14 is still between those two integers. x= 101 7(14)= 3 and y= 2(101)+ 14(15)= 202+ 210= 8. You can get 101 points by putting 3 darts in the 15 circle and 6 darts in the 7 circle.
If A= 100, $\displaystyle \frac{2A}{15}= \frac{2(100)}{15}= \frac{200}{15}$= 13 and 1/3 while $\displaystyle \frac{100}{7}$= 14 and 2/7. k= 14 is still between those two integers. x= 100 7(14)= 2 and y= 2(100)+ 14(15)= 200+ 210= 10. You can get 100 points by putting 2 darts in the 15 point circle and 10 darts in the 7 point circle.
If A= 99, $\displaystyle \frac{2A}{15}= \frac{2(99)}{15}= \frac{198}{15}$= 13 and 1/5 while $\displaystyle \frac{99}{7}$= 14 and 1/7. k= 14 is still between those two integers. x= 99 7(14)= 1 and y= 2(99)+ 14(15)= 198+ 210= 12. You can get 99 points by putting 1 dart in the 15 point circle and 12 darts in the 7 point circle.
If A= 98, $\displaystyle \frac{2A}{15}= \frac{2(98)}{15}= \frac{196}{15}$= 13 and 1/15 while $\displaystyle \frac{98}{7}$= 14. k= 14 again. x= 98 7(14)= 0 and y= 2(98)+ 14(15)= 196+ 210= 14. You can get 98 points by putting 0 darts in the 15 point circle and 14 darts in the 7 point circle.
If A= 97, $\displaystyle \frac{2A}{15}= \frac{2(97)}{15}= \frac{194}{15}$= 12 and 14/15 while $\displaystyle \frac{97}{7}$= 13 and 6/7. Now k= 13 will work.
If A= 96, $\displaystyle \frac{2A}{15}= \frac{2(96)}{15}= \frac{192}{15}$= 12 and 4/5 while $\displaystyle \frac{96}{7}$= 13 and 5/7. Again, k= 13 will work.
If A= 95, $\displaystyle \frac{2A}{15}= \frac{2(95)}{15}= \frac{190}{15}$= 12 and 2/3 while $\displaystyle \frac{95}{7}$= 13 and 4/7. k= 13 will work.
If A= 94, $\displaystyle \frac{2A}{15}= \frac{2(94)}{15}= \frac{188}{15}$= 12 and 8/15 while $\displaystyle \frac{94}{7}$= 13 and 3/7. k= 13 will work.
If A= 93, $\displaystyle \frac{2A}{15}= \frac{2(93)}{15}= \frac{186}{15}$= 12 and 2/15 while $\displaystyle \frac{93}{7}$= 13 and 2/7. k= 13 will work.
If A= 92, $\displaystyle \frac{2A}{15}= \frac{2(92)}{15}= \frac{184}{15}$= 12 and 4/15 while $\displaystyle \frac{92}{7}$= 13 and 1/7. k= 13 will work.
If A= 91, $\displaystyle \frac{2A}{15}= \frac{2(91)}{15}= \frac{182}{15}$= 12 and 2/15 while $\displaystyle \frac{91}{7}$= 13. k= 13 will work.
If A= 90, $\displaystyle \frac{2A}{15}= \frac{2(90)}{15}= \frac{180}{15}$ while $\displaystyle \frac{90}{7}$= 12 and 6/7. k= 12 will work.
If A= 89, $\displaystyle \frac{2A}{15}= \frac{2(89)}{15}= \frac{178}{15}$= 11 and 13/15 while $\displaystyle \frac{89}{7}$= 12 and 5/7. k= 12 will work.
If A= 88, $\displaystyle \frac{2A}{15}= \frac{2(88)}{15}= \frac{175}{15}$= 11 and 2/3 while $\displaystyle \frac{88}{7}$= 12 and 4/7. k= 12 will work.
If A= 87, $\displaystyle \frac{2A}{15}= \frac{2(87)}{15}= \frac{174}{15}$= 11 and 3/5 while $\displaystyle \frac{87}{7}$= 12 and 3/7. k= 12 will work.
If A= 86, $\displaystyle \frac{2A}{15}= \frac{2(86)}{15}= \frac{172}{15}$= 11 and 7/15 while $\displaystyle \frac{86}{7}$= 12 and 2/7. k= 12 will work.
If A= 85, $\displaystyle \frac{2A}{15}= \frac{2(85)}{15}= \frac{170}{15}$= 11 and 1/3 while $\displaystyle \frac{85}{7}$= 12 and 1/7. k= 12 will work.
If A= 84, $\displaystyle \frac{2A}{15}= \frac{2(84)}{15}= \frac{168}{15}$= 11 and 1/5 while $\displaystyle \frac{84}{7}$= 12. k= 12 will work.
If A= 83, $\displaystyle \frac{2A}{15}= \frac{2(83)}{15}= \frac{166}{15}$= 11 and 1/15 while $\displaystyle \frac{83}{7}$= 11 and 6/7.
Finally! There is no integer between 11 and 1/15 and 11 and 6/7 so there are no x and y that will make 15x+ 7y= 83.
Since every number large than 83 can be written that way, 83 is the largest number that cannot be written 15x+ 7y and so is the largest number of points that cannot be achieved by throwing darts at 15 point and 7 point circles.

Thank you so much for your help. But I have to admit I don't fully understand the solution yet. I think I will be able to if I study it a little more with some guessing and checking.
You also saved me a lot of time cause I was starting to write down all the possible numbers and it was getting very messy.
Vicky.