1 Answer

Canilho · Answer 1 · 2020-04-25T14:46:29+0000

We have
$\lambda(21)=6$ , where
$\lambda$ is the Carmichael function. So we have

$10^{5^(101)}\equiv10^{5^(101)\pmod6}\pmod{21}$

The powers of 5 modulo 6 follow a periodic pattern

$5^1\equiv5\pmod6$

$5^2\equiv25\equiv1\pmod6$

$5^3\equiv1\cdot5\equiv5\pmod6$

$5^4\equiv5^2\equiv1\pmod6$

and so on, with odd powers of 5 equivalent to 5 modulo 6. So

$10^{5^(101)}\equiv10^{5^(101)\pmod6}\equiv10^5\pmod{21}$

The rest is easy to deal with. We have

$10^2\equiv16\pmod{21}$

$10^3\equiv160\equiv13\pmod{21}$

$10^4\equiv130\equiv4\pmod{21}$

$10^5\equiv40\equiv19\pmod{21}$

and so the answer is 19.

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