# I try to find the last two digits of $2019^{102^{830}}$ in basis 7 [duplicate]

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 :)

• This is hard to read. here is a tutorial on formatting for this site.
– lulu
Commented Jan 6, 2023 at 15:20
• 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 Commented Jan 6, 2023 at 15:49