Modular arithmetic modulo $10^{2011}.$

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Is there a natural number $n$ such that $3^{n} \equiv 7\ (\text {mod}\ 10^{2011})$

I can't come up with an idea how to solve it...

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Simply to get this out of the unanswered queue. No. the original congruence can be rewritten as $$10^{2011}y+7=3^n$$ which works for any divisor of $10^{2011}$ 8 is such a divisor. The problem is no power of 3, is 7 mod 8. It therefore fails to be true.