Tricky question related to sequences and series I have the following question with me:

"In the sequence $$1,9,7,7,4,7,5,3,9,4,1,....$$ every term from the fifth one is the sum of the previous 4 modulo 10. Do the numbers $1977$(apart from beginning) and $0197$ ever occur in the sequence? If yes, does it occur a finite number of times or not?"

How do I solve this question?  The only advancement I have made is that when the sequence is written modulo 2, I get $$1,1,1,1,0,1,1,1,1,0....$$

How do I proceed from here?
Hint: you can prove results about sequences like this without explicitly computing anything using two facts:


The number of possible sequences of four digits is finite, while the sequence you are given is infinite.
Any four consecutive digits in the sequence (uniquely) determine the whole sequence both forwards and backwards


Obviously you have to show that these facts are true for the sequence in hand, and then work out how to use them to get the result you need.



The things about explicitly computing residues can help in determining the period of a recurring sequence.
First, you can consider the sequence with index in whole integers. For example, we can define $a_{0} = 0$ since $0 + 1 + 9 + 7\equiv 7$ mod 10 and $a_{-1} = 7$ since $7 + 0 + 1 + 9 \equiv 7$ mod 10. Now try to explain that there exists $m\neq n$ such that $(a_{m}, a_{m+1}, a_{m+2}, a_{m+3}) = (a_{n}, a_{n+1}, a_{n+2}, a_{n+3})$ by using the pigeonhole principle.