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anonymous

  • one year ago

a + (a + 1) + (a + 2) + ... + (a + n - 1) = 100 such that a and n are positive integers and n >= 2...

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  1. anonymous
    • one year ago
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    i got the answer through guess and check, but there has to be a better way

  2. anonymous
    • one year ago
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    ive got up until n^2+(2a-1)n - 200=0 but now what?

  3. anonymous
    • one year ago
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    just as clarification, i started with a + (a + 1) + (a + 2) + ... + (a + n - 1)= \frac{[a+(a+n-1)]\cdot n}{2}=10

  4. anonymous
    • one year ago
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    To me, a = 18 and n =5

  5. anonymous
    • one year ago
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    18 +(18+1)+(18+2)+(18+3)+(18+4) =100 5*18 +1+2+3+4 =100

  6. anonymous
    • one year ago
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    And it is not a guess!! I have my logic on it.

  7. anonymous
    • one year ago
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    @Crazyandbeautiful do you think you could tell me how you went about it? I'm really curious...

  8. anonymous
    • one year ago
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    (a+1)+(a+n -1) = 2a + n (a +2) +(a + n -2) = 2a + n ... let say we have m of them, then m (2a + n) =100 that is m is a factor of 100. Now test m = 1 start at a +(a+(n-1)/2) =100 and that forces n is an odd. hence, if n =3, we have a +(a+1)+(a+2) =3a +3 =100--> 3(a+1) =100 --> a+1 is a factor of 100 or a+1 = 1, 2, 4, 5, 10 that is a = 0 , 1, 3, 4, 9. Plug back we can see none of them satisfy 3(a +1 ) =100 Same argument until I got m = 5, that gives me a +(a+1) +(a+2) +(a+3)+(a+4) =100, solve for a , I have a =18 and hence n = 5

  9. anonymous
    • one year ago
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    that's a cool way to do it! I tried something similar and got (18,5) as well as (9,8)... which you would have gotten with a few more trials

  10. anonymous
    • one year ago
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    But i cant help but wonder... is there a more direct way to do it?

  11. mathmate
    • one year ago
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    @dan815 what do you think? I have come to the point of finding a such that "g(a)=4a^2-4a+801=perfect square", which verifies a=9 and 18 correctly. Do you have a way to solve for a apart from brut force?

  12. anonymous
    • one year ago
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    Ive got it! We can go through a process of simple elimination...

  13. mathmate
    • one year ago
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    @emeyluv99 Do you mind sharing you elimination method with us?

  14. anonymous
    • one year ago
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    oh yes sorry, of course

  15. anonymous
    • one year ago
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    Take a step back from n^2+(2a-1)n - 200=0, we get n(n+2a-1) = 200 Meaning that n is a factor of 200. the factors of 200 are 1,2,4,5,8,10,20,25,40,50,100,200,

  16. anonymous
    • one year ago
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    so we cant have an n less than 2, nor can we have an n that has a value larger than 200 when squared. so that leaves us with 2,4,5,8,10. We can just test these to find which give the positive value of a.

  17. mathmate
    • one year ago
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    Yes, it does reduces the number of candidates for checking. Oh, thank you so much! :)

  18. anonymous
    • one year ago
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    Thank YOU! you gave me idea :D

  19. mathmate
    • one year ago
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    We both learned something! :)

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