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Prove that 111...121...111(2n+1 digits) is a composite number with n is natural number, n isn't equal to 0?
1 Answer
- Anonymous1 decade agoFavorite Answer
To be perfectly clear this is the sequence of numbers
121, 11211, 1112111, 111121111, 11111211111, etc.
I will work with the 4th number in the list but any can be done in the same way.
111121111 = 111110000 + 11111 = 11111*(10000 + 1) = 11111*10001
In other words they all have 1111... (n + 1) digits as a factor and so are composite.
EDIT. By the way, they all have 1000...0001 with (n - 1) zeros as another factor.