Are any set of 10 consecutive terms always divisible by 11 or does 11 increase as the numbers increase?
In: 3
It’s always 11.
The sum of 10 consecutive numbers in a Fibonacci Sequence where the first number of the sequence is a and the second number in the sequence is b is 55a + 88b = 11(5a + 8b).
It actually doesn’t matter what the first two numbers are! As long as you follow the Fibonacci Rule where the next number is equal to the sum of the previous two, you get a number that is divisible by 11.
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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