Is there any way I can compute $2^{2014} \mod 11$?
I tried squaring the terms every time, but can't seem to get to $2014$.
Would appreciate your help. Thanks!
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1Have you heard about Fermat's little theorem? – Arthur Nov 18 '16 at 18:26
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Unfortunately no, my professor hasn't taught that yet :( – NAA Nov 18 '16 at 18:27
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Well, that's just evil, telling you to calculate this without giving you the theorem that helps you. At any rate, the answers below tell you how to use the theorem in this specific case. – Arthur Nov 18 '16 at 18:28
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See http://math.stackexchange.com/questions/491576/simplifying-large-exponents-in-modular-arithmetic-like-1007-in-41007-pmod – Hans Hüttel Nov 18 '16 at 18:29
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1@Arthur I included my answer without Fermat's Little Theorem, making use of binary and consecutive squaring. – RGS Nov 18 '16 at 18:31
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$$2^{10}\equiv 1 \mod 11$$
$$2^{2014} \equiv 2^{2010+4} \equiv 2^4 \equiv 16 \equiv 5 \mod 11$$
Siong Thye Goh
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Note that $$2^5 \equiv -1 \pmod{11} $$ $$\implies (2^5)^{402} \equiv (-1)^{402} \pmod{11} $$ $$\implies 2^{2010}\cdot 2^4 \equiv 1\cdot 2^4 \pmod{11} $$ $$\implies 2^{2014} \equiv 16 \pmod{11}$$ $$\implies \boxed{2^{2014} \equiv 5 \pmod{11} }$$
SchrodingersCat
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If you are not allowed to use Fermat's little theorem, then just note that $2014 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2$ so
$2^{2014}\ =\ 2^{1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4 + 2}\ =\ 2^{1024}\cdot2^{512}\cdot2^{256}\cdot2^{128}\cdot2^{64}\cdot2^{16}\cdot2^{8}\cdot2^{4}\cdot2^{2}$
And each one of those factors can be computed by consecutive squaring.
RGS
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