Let $(a_n )_{n=1} ^{\infty} $ be a sequence of elements in $\{0, 1\}$such that for all positive integers $n,\Sigma_{i=n}^{ n+9}a_i$ is divisible by 3 . Then there exist a positive integer $k$ such that $a_{n+k}= a_n $ for all positive integers $n$ (True/ False)
I think it is true .
My Idea :- Basically the question is asking if a binary sequence with sum of every 10 consecutive terms divisible by $3$ is necessarily $k-$ periodic or not for some positive $k$ .
Clearly if the sequence is zero sequence then the result is satisfied.
Claim : If it is not a zero sequence then it cant be eventually constant sequence .
Proof :
Suppose the last repeating digits is 0Then there must exist a 10 length string of the form ..1000000000 whose sum is not divisible by 3.
Again if the last repeating digit is 1 then there exist a 10 length string of all 1's , again a contradiction.
So the sequence can't be eventually constant
Claim is proved .
Sum of every 10 consecutive terms must be either 0, 3 ,6 ,9 .So there are exactly either 0 or 3 or 6 or 9 ones and rest zeros in every string of bits of length 10 .
There are exactly 1 , 10C3 , 10C6 and 10C9 strings of bits of length 10 and whose digits sums are 0 , 3 , 6 and 9 respectively .
Partitioning the sequence into consecutive strings of length 10 we can saysince each of the set of strings with digits sum as a given multiple of 3 is finite in number and the sequence is infinite (not eventually constant) by Pigeonhole principle , at least one of the combination of 10 length string must repeat ..
How may I proceed further ?