Question

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

Answer 1

i got the answer through guess and check, but there has to be a better way

Answer 2

ive got up until n^2+(2a-1)n - 200=0 but now what?

Answer 3

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

Answer 4

To me, a = 18 and n =5

Answer 5

18 +(18+1)+(18+2)+(18+3)+(18+4) =100
5*18 +1+2+3+4 =100

Answer 6

And it is not a guess!! I have my logic on it.

Answer 7

@Crazyandbeautiful do you think you could tell me how you went about it? I'm really curious...

Answer 8

(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

Answer 9

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

Answer 10

But i cant help but wonder... is there a more direct way to do it?

Answer 11

@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?

Answer 12

Ive got it! We can go through a process of simple elimination...

Answer 13

@emeyluv99
Do you mind sharing you elimination method with us?

Answer 14

oh yes sorry, of course

Answer 15

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,

Answer 16

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.

Answer 17

Yes, it does reduces the number of candidates for checking.
Oh, thank you so much! :)

Answer 18

Thank YOU! you gave me idea :D

Answer 19

We both learned something! :)