MATH SOLVE

3 months ago

Q:
# Find 10^(5^101) (mod 21).note: 10^(5^101) is not 10^(501)

Accepted Solution

A:

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.