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Ind the last 7 digits in the binary expansion of $27^{1986}$. (give your answer in binary.)

User Arrow Cen
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Since 1986 = 62*32 + 2, we have


27^(1986)\equiv\left(27^(62)\right)^(32)\cdot27^2\equiv27^2\pmod{2^6}

by Euler's theorem, since
\varphi(64)=32. Then


27^(1986)\equiv27^2\equiv729\equiv25\pmod{64}

which tells us the 7th digit from the right is 1.

Then


25\equiv25\pmod{2^5}

so the next digit must be 0;


25\equiv9\pmod{2^4}

so the next digit is 1;


9\equiv1\pmod{2^3}

so the next digit is 1;


1\equiv1\pmod{2^2}

so the next digit is 0;


1\equiv1\pmod2

so the next digit is 0; and


1\equiv0\pmod1

so the last digit is 1.

So the last 7 digits of this number in binary are 1011001.

User Amr Osama
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