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Find the remainder when $2^{1990}$ is divided by $1990$. I didn't get answer by Euler's generalization.

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  • $\begingroup$ math.stackexchange.com/questions/545759/… $\endgroup$ Mar 30, 2014 at 6:17
  • $\begingroup$ A quick and painless way: (2 ^ 1990) `mod` 1990 in Haskell says "1024". $\endgroup$
    – Dan Shved
    Mar 30, 2014 at 6:40
  • $\begingroup$ Always use the Chinese remainder theorem to consider only congruences modulo primes (or prime powers). Then use Fermat's little theorem (or Euler's theorem) to simplify the exponent. $\endgroup$ Mar 30, 2014 at 6:42

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Let's see. 199 is prime, so $b^{198x} = 1 \pmod{199}$. Since the period of $5$ is 4, we see that $b^{1981} = b^1 \pmod{1990} $ for all b. So it's a matter of finding the remainder of $2^{10} \pmod{1990}$

But since this is $1024$, which is less than $1990$, that is the sought answer.

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  • $\begingroup$ Thanks: it should read $b^{1981}$. Corrected in post. $\endgroup$ Mar 30, 2014 at 9:43

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