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I have tried questions like this by expanding them in a binomial state but this question appears to be different

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

It is hard because 1990 is not prime and I have solved prolems only when divisors were primes.

Please help me.

Atul Mishra
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4 Answers4

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Using Fermat's Little Theorem,

$2^4\equiv1\pmod 5\implies 2^{1990}=2^2\cdot(2^4)^{997}\equiv4\cdot1^{997}\pmod 5\equiv4\ \ \ \ (1)$

$2^{198}\equiv1\pmod {199}\implies 2^{1990}=2^{10}\cdot(2^{198})^{10}\equiv1024\cdot1^{10}\pmod{199}\equiv 29\ \ \ \ (2)$

Clearly, $2^{1990}\equiv0\pmod2\ \ \ \ $

Continuing further, we have $2^{1990}\equiv4\pmod5$. This gives us that $2^{1990}\equiv4\pmod{10}$. Now using the fact that $a\equiv4\pmod{10}$ and $a\equiv{29}\pmod{199}$. Notice that $20*10+(-1)*199=1$ thus ${20*10}\equiv1\pmod{199}$ and $-199\equiv1\pmod{100}$. As we have $a\equiv4\pmod{10}$ and $a\equiv{29}\pmod{199}$, let us assume $a=29*(20\times 10) +4\times (-199)$. We can see that is satisfies these two conditions. Hence, we can proceed.

Thus $a= 5024 \equiv{1024}\pmod{1990}$.

2

$2^{10}\equiv1024\pmod{1990}\implies$

$2^{20}\equiv1024^2\equiv1048576\equiv1836\pmod{1990}\implies$

$2^{40}\equiv1836^2\equiv3370896\equiv1826\pmod{1990}\implies$

$2^{80}\equiv1826^2\equiv3334276\equiv1026\pmod{1990}\implies$

$2^{160}\equiv1026^2\equiv1052676\equiv1956\pmod{1990}\implies$

$2^{320}\equiv1956^2\equiv3825936\equiv1156\pmod{1990}\implies$

$2^{640}\equiv1156^2\equiv1336336\equiv1046\pmod{1990}\implies$

$2^{1280}\equiv1046^2\equiv1094116\equiv1606\pmod{1990}$

$2^{1990}\equiv$

$2^{1280+640+40+20+10}\equiv$

$2^{1280}\cdot2^{640}\cdot2^{40}\cdot2^{20}\cdot2^{10}\equiv$

$1606\cdot1046\cdot1826\cdot1836\cdot1024\equiv$

$5767009039908864\equiv1024\pmod{1990}$

barak manos
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  • It didn't striked my mind before seeing your solution.. Can you help me in knowing this How did you think this way??? – Atul Mishra Dec 08 '16 at 08:03
  • @AtulMishra: i Can't really explain my thinking process (I don't think anybody can, to be honest). I just tackled the problem straightforward (instead of via Fermat, Euler, etc). BTW, the other answers are simpler and shorter. If you're expected to make use of FLT or CRT, then I guess that one of them is the most suitable one for your question. My answer can be obtained without any pre-knowledge of Fermat's Little Theorem, Euler's Theorem and so forth. I initially meant it just to be used as a confirmation of the other answers, but if you find it best then you may as well use it... – barak manos Dec 08 '16 at 08:22
  • @AtulMishra The brute-force powering above is essentially a special case of Exponentiation by repeated squaring. I agree with barak that if this is meant to be an exercise in number theory then it is almost surely meant to exercise your knowledge on Fermat or Euler's Theorem (and possibly also CRT), so you should proceed using these approaches (used in some of the other answers). – Bill Dubuque Dec 16 '16 at 17:13
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By Fermat $\, 2^{\large 2} \equiv 1\pmod{\! 5},\,\ 2^{\large 198}\!\equiv 1\pmod{\!199}\ $ so $\ \color{#c00}{2^{\large 396}\equiv \bf 1}\pmod {5\cdot 199}$

So we have $\ {\rm mod}\,\ \color{#0a0}{5\cdot 199\!:\,\ 2^{\large 1998}}\! \equiv 2^{\large 9+5(396)}\! \equiv 2^{\large 9} (\color{#c00}{2^{\large 396}})^{\large 5}\!\equiv 2^{\large 9}{\color{#c00}{\bf 1}}^{\large 5}\!\equiv \color{#0a0}{2^{\large 9}} $

So $\,\ 2^{\large 1990}\!\bmod 1990\, =\, 2\,(\color{#0a0}{2^{\large 1989}\!\bmod\ 5\cdot 199})\, =\, 2(\color{#0a0}{2^{\large 9}}) = 2^{\large 10} =1024$

We used $\ ca\bmod cn =\, c\,(a\bmod n)\ $ in the prior line. See here for more on that.

Community
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Bill Dubuque
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The prime factorization of $1990$ is $1990 = 2 \cdot 5 \cdot 199$. So calculate $2^{1990}$ modulo $2,5$, and $199$ and then solve the resulting system of congruences to find $2^{1990}$ modulo $1990$.

Ethan Alwaise
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