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I've solved many of these examples, but I couldn't solve these problems:

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I would be very thankful if you would solve this problems. Thank you very much for your help.

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    $\begingroup$ You can't do $2^3$ mod $11$? $\endgroup$
    – anon
    Nov 5, 2014 at 18:43

1 Answer 1

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First problem is by Fermat: $123^{123}\equiv 2^{123}=(2^{10})^{12}2^3\equiv2^3\equiv 8 \bmod 11$


Second problem by Euler, using $\varphi(27)=18$:

$10^{188}= ((10)^{18})^{10}10^8\equiv 10^8 \bmod 27$

we now evaluate this as $((10^2)^2)^2\equiv((19)^2)^2\equiv(10)^2\equiv19 \bmod 27$

Note the second problem can also be solved by noting $10^3\equiv 1 \bmod 27$ but this cannot be obtained by Euler.


Third problem is by Euler using $\varphi(44)=20$:

$(4447)^{2018}\equiv 3^{2018}=(3^{20})^{100}3^{18}\equiv 3^{18} \bmod 44$

We now evaluate this as $(3^6)^3\equiv(729)^3\equiv(25)^3\equiv 5 \bmod 44$

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