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Jun 23, 2019 at 3:43 comment added J. W. Tanner @ChrisCuster: sorry if wasn't clear; does this help? If $a\equiv1\pmod9$ then also $a\equiv1\pmod3$ so $3|a^2+a+1$ so $27|a^3-1=(a-1)(a^2+a+1)$ ; use that with $a=10^{\phi(2001)}$
Jun 23, 2019 at 3:06 comment added user403337 Could you explain the last step?
Jun 23, 2019 at 2:20 comment added J. W. Tanner so the number with $3696$ $1$s is divisible by $2001$
Jun 23, 2019 at 2:16 history undeleted J. W. Tanner
Jun 23, 2019 at 2:16 history deleted J. W. Tanner via Vote
Jun 23, 2019 at 2:16 history answered J. W. Tanner CC BY-SA 4.0