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Math Help - Word problem

  1. #1
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    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.

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  2. #2
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    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 non-negative 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 non-negative 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 A- 7k\ge 0 and -2A+ 15k\ge 0. Those are the same as \frac{A}{7}\ge k and k\ge \frac{2A}{15}: \frac{2A}{15}\le k\le\frac{A}{7}

    Well, obviously, that is always possible if there is an integer between \frac{2A}{15} and \frac{A}{7}. In particular it is certainly possible if \frac{A}{7}- \frac{2A}{15}= A\left(\frac{1}{7}- \frac{2}{15}\right) = \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, \frac{2A}{15}= \frac{2(105)}{15}= \frac{210}{15}= \frac{42}{3}= 14 and \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, \frac{2A}{15}= \frac{2(104)}{15}= \frac{208}{15}[tex]= 13 and 13/15 while \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, \frac{2A}{15}= \frac{2(103)}{15}= \frac{206}{15}= 13 and 11/15 while \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, \frac{2A}{15}= \frac{2(102)}{15}= \frac{204}{15}= 13 and 3/5 while \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, \frac{2A}{15}= \frac{2(101)}{15}= \frac{202}{15}= 13 and 7/15 while \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, \frac{2A}{15}= \frac{2(100)}{15}= \frac{200}{15}= 13 and 1/3 while \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, \frac{2A}{15}= \frac{2(99)}{15}= \frac{198}{15}= 13 and 1/5 while \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, \frac{2A}{15}= \frac{2(98)}{15}= \frac{196}{15}= 13 and 1/15 while \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, \frac{2A}{15}= \frac{2(97)}{15}= \frac{194}{15}= 12 and 14/15 while \frac{97}{7}= 13 and 6/7. Now k= 13 will work.

    If A= 96, \frac{2A}{15}= \frac{2(96)}{15}= \frac{192}{15}= 12 and 4/5 while \frac{96}{7}= 13 and 5/7. Again, k= 13 will work.

    If A= 95, \frac{2A}{15}= \frac{2(95)}{15}= \frac{190}{15}= 12 and 2/3 while \frac{95}{7}= 13 and 4/7. k= 13 will work.

    If A= 94, \frac{2A}{15}= \frac{2(94)}{15}= \frac{188}{15}= 12 and 8/15 while \frac{94}{7}= 13 and 3/7. k= 13 will work.

    If A= 93, \frac{2A}{15}= \frac{2(93)}{15}= \frac{186}{15}= 12 and 2/15 while \frac{93}{7}= 13 and 2/7. k= 13 will work.

    If A= 92, \frac{2A}{15}= \frac{2(92)}{15}= \frac{184}{15}= 12 and 4/15 while \frac{92}{7}= 13 and 1/7. k= 13 will work.

    If A= 91, \frac{2A}{15}= \frac{2(91)}{15}= \frac{182}{15}= 12 and 2/15 while \frac{91}{7}= 13. k= 13 will work.

    If A= 90, \frac{2A}{15}= \frac{2(90)}{15}= \frac{180}{15} while \frac{90}{7}= 12 and 6/7. k= 12 will work.

    If A= 89, \frac{2A}{15}= \frac{2(89)}{15}= \frac{178}{15}= 11 and 13/15 while \frac{89}{7}= 12 and 5/7. k= 12 will work.

    If A= 88, \frac{2A}{15}= \frac{2(88)}{15}= \frac{175}{15}= 11 and 2/3 while \frac{88}{7}= 12 and 4/7. k= 12 will work.

    If A= 87, \frac{2A}{15}= \frac{2(87)}{15}= \frac{174}{15}= 11 and 3/5 while \frac{87}{7}= 12 and 3/7. k= 12 will work.

    If A= 86, \frac{2A}{15}= \frac{2(86)}{15}= \frac{172}{15}= 11 and 7/15 while \frac{86}{7}= 12 and 2/7. k= 12 will work.

    If A= 85, \frac{2A}{15}= \frac{2(85)}{15}= \frac{170}{15}= 11 and 1/3 while \frac{85}{7}= 12 and 1/7. k= 12 will work.

    If A= 84, \frac{2A}{15}= \frac{2(84)}{15}= \frac{168}{15}= 11 and 1/5 while \frac{84}{7}= 12. k= 12 will work.

    If A= 83, \frac{2A}{15}= \frac{2(83)}{15}= \frac{166}{15}= 11 and 1/15 while \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.
    Last edited by HallsofIvy; July 28th 2009 at 02:23 AM.
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  3. #3
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    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.
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