The sequence 19 199 1999 starts off with three primes howev

The sequence 19, 199, 1999, ... starts off with three primes, however, most of the numbers in the sequence are composites. Prove that 19 divides infinitely many of the numbers in the sequence.

Solution

The numbers in the sequence are of the form 2 * 10ⁿ - 1 where the exponent n creates the value having n+1 digits.

All the numbers in the sequence having 18m + 2 digits will be divisible by 19. To see note that,

10^18m+1 = 10 (10¹⁸)^m = 10. 1^m (mod 19)

by fermet\'s little theorem, Thus 10^(18m + 1) = 10 (mod 19) for all postive integers m. This means that 2 * 10^(18m +1) -1 = 0(mod 19) or 19 divides all values in sequence having 18m+2 digits. Note further that 1(mod 18) is the only exponent on 10 in the last congruence that gives 0.