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Find the remainder when $2^{1990}$ is divided by $1990$.

I have obtained that $2^{1990}\equiv{29}\pmod{199}$ and $2^{1990}\equiv{4}\pmod{10}$ but I am unable to proceed after this. So how can we proceed the given problem?

Bill Dubuque
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Sillyasker
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  • By the Chinese Remainder Theorem, there is a unique remainder modulo $1990$ which is $29 \bmod{199}$ and $4 \bmod{10}$. You can find this remainder using the Euclidean algorithm. – Haran Oct 29 '23 at 05:27
  • I am sorry I would like to get an answer without the use of Chinese remainder theorem @Haran. But thank you for the idea. – Sillyasker Oct 29 '23 at 05:31

1 Answers1

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You know that there exist $m,n$ such that $$2^{1990}=199m+29=10n+4.$$ This gives you the equation $$\tag1199m-10n=25.$$ As $199$ and $10$ are coprime, we can see that $199\times(-1)+10\times20=1$. Hence $$199\times(-25)+10\times500=25.$$ So the solutions to $(1)$ are of the form $$ m=-25+10k,\qquad\qquad n=500+199k,\qquad\qquad k\in\mathbb Z. $$ Then $$ 2^{1990}=10n+4=10(500+199k)+4=5004+1990k. $$ This tells us that $$ 2^{1990}\equiv 5004\pmod{1990}=1024\pmod{1990}. $$

Martin Argerami
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  • Note This answer is a special case of a well-known Bezout-based CRT solution method, which is described in many prior answers, e.g. see here. – Bill Dubuque Oct 29 '23 at 05:43
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing). – Bill Dubuque Oct 29 '23 at 05:47