anonymous
  • anonymous
a + (a + 1) + (a + 2) + ... + (a + n - 1) = 100 such that a and n are positive integers and n >= 2...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i got the answer through guess and check, but there has to be a better way
anonymous
  • anonymous
ive got up until n^2+(2a-1)n - 200=0 but now what?
anonymous
  • anonymous
just as clarification, i started with a + (a + 1) + (a + 2) + ... + (a + n - 1)= \frac{[a+(a+n-1)]\cdot n}{2}=10

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anonymous
  • anonymous
To me, a = 18 and n =5
anonymous
  • anonymous
18 +(18+1)+(18+2)+(18+3)+(18+4) =100 5*18 +1+2+3+4 =100
anonymous
  • anonymous
And it is not a guess!! I have my logic on it.
anonymous
  • anonymous
@Crazyandbeautiful do you think you could tell me how you went about it? I'm really curious...
anonymous
  • anonymous
(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
anonymous
  • anonymous
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
anonymous
  • anonymous
But i cant help but wonder... is there a more direct way to do it?
mathmate
  • mathmate
@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?
anonymous
  • anonymous
Ive got it! We can go through a process of simple elimination...
mathmate
  • mathmate
@emeyluv99 Do you mind sharing you elimination method with us?
anonymous
  • anonymous
oh yes sorry, of course
anonymous
  • anonymous
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,
anonymous
  • anonymous
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.
mathmate
  • mathmate
Yes, it does reduces the number of candidates for checking. Oh, thank you so much! :)
anonymous
  • anonymous
Thank YOU! you gave me idea :D
mathmate
  • mathmate
We both learned something! :)

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