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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...
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got the answer through guess and check, but there has to be a better way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ive got up until n^2+(2a1)n  200=0 but now what?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just as clarification, i started with a + (a + 1) + (a + 2) + ... + (a + n  1)= \frac{[a+(a+n1)]\cdot n}{2}=10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To me, a = 18 and n =5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.018 +(18+1)+(18+2)+(18+3)+(18+4) =100 5*18 +1+2+3+4 =100

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And it is not a guess!! I have my logic on it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Crazyandbeautiful do you think you could tell me how you went about it? I'm really curious...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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+(n1)/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
 one year ago
Best ResponseYou've already chosen the best response.0that'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
 one year ago
Best ResponseYou've already chosen the best response.0But i cant help but wonder... is there a more direct way to do it?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 what do you think? I have come to the point of finding a such that "g(a)=4a^24a+801=perfect square", which verifies a=9 and 18 correctly. Do you have a way to solve for a apart from brut force?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ive got it! We can go through a process of simple elimination...

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@emeyluv99 Do you mind sharing you elimination method with us?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yes sorry, of course

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take a step back from n^2+(2a1)n  200=0, we get n(n+2a1) = 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it does reduces the number of candidates for checking. Oh, thank you so much! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank YOU! you gave me idea :D

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0We both learned something! :)
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