Respuesta :

We have [tex]\lambda(21)=6[/tex], where [tex]\lambda[/tex] is the Carmichael function. So we have

[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\pmod{21}[/tex]

The powers of 5 modulo 6 follow a periodic pattern

[tex]5^1\equiv5\pmod6[/tex]

[tex]5^2\equiv25\equiv1\pmod6[/tex]

[tex]5^3\equiv1\cdot5\equiv5\pmod6[/tex]

[tex]5^4\equiv5^2\equiv1\pmod6[/tex]

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

[tex]10^{5^{101}}\equiv10^{5^{101}\pmod6}\equiv10^5\pmod{21}[/tex]

The rest is easy to deal with. We have

[tex]10^2\equiv16\pmod{21}[/tex]

[tex]10^3\equiv160\equiv13\pmod{21}[/tex]

[tex]10^4\equiv130\equiv4\pmod{21}[/tex]

[tex]10^5\equiv40\equiv19\pmod{21}[/tex]

and so the answer is 19.

Answer:

19

Step-by-step explanation: