1

I me preparing an exam with an old exam question we received, and Im trying to find the last two digits of $2019^{102^{830}}$ in basis 7.

So here is what I ve got so far:

The last two digits are of the form $$a_{1}7+a_{0}$$ with $a\in \{1,…,6\}$ which can be maximally $6\bullet7+6= 48$ $$\implies x\equiv2019^{102^{830}} mod 49$$ Using now Euler makes it: $$x\equiv2019^{102^{830}mod\varphi(49)}mod49\equiv2019^{102^{830}mod42}mod49$$

Now since $gcd(830,42)\neq 1$ I split it up into $$y\equiv 102^{830}mod7 $$and$$y\equiv 102^{830}mod6$$ (here I m not sure anymore if I may do this) Then I apply Euler again: $$y\equiv 102^{830 mod\varphi(7)}\equiv 102^{830 mod6}\equiv 102^2$$ Then is $$x\equiv 2019^{102^2mod42}mod49\equiv 2019^{30}mod49$$ And this is the maximum I find it to be reducible, but it is still too big. Any help or correction would be very much appreciated . Thank you in advance :)

Bill Dubuque
  • 272,048
T_B
  • 155
  • 1
    This is hard to read. here is a tutorial on formatting for this site. – lulu Jan 06 '23 at 15:20
  • @lulu thank your for your link, I ll do this now – T_B Jan 06 '23 at 15:22
  • 2
    Related: https://math.stackexchange.com/questions/81228/modular-exponentiation-by-hand-ab-bmod-c – Arthur Jan 06 '23 at 15:26
  • 1
    Use Carmichael's function, the problem will left very easy using this.

    https://en.m.wikipedia.org/wiki/Carmichael_function

     –  Jan 06 '23 at 15:42
  • 1
    First replace 830 by 830 mod φ(42), i.e. 2. Then $102^2$ by $102^2\bmod{42},$ i.e. 30. Then $2019^{30}$ by $(2019\bmod{49})^{30}\bmod{49}=32^3\bmod49=36.$ But I think this post might be considered as a duplicate of the related one pointed above by @Arthur – Anne Bauval Jan 06 '23 at 15:49
  • Thanks very much for these great answers and the link – T_B Jan 06 '23 at 18:35

0 Answers0