Are any set of 10 consecutive terms always divisible by 11 or does 11 increase as the numbers increase?
In: 3
As said by the other comment, it’s always 11, and I will try to give a little more information.
For this comment, I will assume that we take F(1) = 1, F(2) = 1, F(3) = 2, F(4) = 3, F(5) = 5, etc.
The Fibonacci sequence has a nice property which is that if you consider a set of N consecutive terms, then their sum can be decomposed as a multiple of F(N) plus a multiple of F(N+1)-1.
So with N=10, the sum is composed of a multiple of F(10)=55 and a multiple of F(11)-1=88. And since both are multiple of 11, then the sum is a multiple of 11. More generally:
* The sum of 3 consecutive terms is even, because F(3)=2 and F(4)-1=4
* The sum of 6 consecutive terms is a multiple of 4, because F(4)=8 and F(5)-1=12
* The sum of 8 consecutive terms is a multiple of 3, because F(8)=21 an F(9)-1=33
* The sum of 9 consecutive terms is even, because F(9)=34 and F(10)-1=54
* The sum of 10 consecutive terms is a multiple of 11, because F(10)=55 and F(11)-1=88
* The sum of 12 consecutive terms is a multiple of 8, because F(12)=144 and F(13)-1=232
* Etc
It’s a nice mathematical exercice to prove all of that, but the proof I know goes through the intermediary step of proving that F(x+y) = F(x)*F(y-1) + F(x+1)*F(y), and while it’s reasonably easy on a black board it’s a mess to write it in a reddit comment.
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