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

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

Mathematics
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i got the answer through guess and check, but there has to be a better way
ive got up until n^2+(2a-1)n - 200=0 but now what?
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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To me, a = 18 and n =5
18 +(18+1)+(18+2)+(18+3)+(18+4) =100 5*18 +1+2+3+4 =100
And it is not a guess!! I have my logic on it.
@Crazyandbeautiful do you think you could tell me how you went about it? I'm really curious...
(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
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
But i cant help but wonder... is there a more direct way to do it?
@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?
Ive got it! We can go through a process of simple elimination...
@emeyluv99 Do you mind sharing you elimination method with us?
oh yes sorry, of course
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,
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.
Yes, it does reduces the number of candidates for checking. Oh, thank you so much! :)
Thank YOU! you gave me idea :D
We both learned something! :)

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