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

Yiyuan Lee
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Satvik Mashkaria
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  • http://math.stackexchange.com/questions/545759/what-is-the-remainder-when-21990-is-divided-by-1990 – lab bhattacharjee Mar 30 '14 at 06:17
  • A quick and painless way: (2 ^ 1990) \mod` 1990` in Haskell says "1024". – Dan Shved Mar 30 '14 at 06:40
  • 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. – Greg Martin Mar 30 '14 at 06:42

1 Answers1

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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.

wendy.krieger
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