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Helloo... Im kinda dumb in Discrete Mathematics since I just started studying it currently and I'm trying to figure out studying material to get remainders from operations like $11^{14^{4578369}}\pmod {41}$ or $11^{3928454} \pmod {3293}$. Can someone gimme hints or even the path to follow for their solutions?

I've tried using fermat's little theorem but I couldn't solve them anyway , cuz in the first case it's kinda power from power situation and the other one the mod isn't prime.
Note: Here's a transcription of the exercises that uses these operations that im talking about. Thanks, though.

Considering $a=11^{3928454}$, $b=3293$ and $x$ as the remainder from division of $a$ by $b$. Which is the addition of all $x$ digits?
a. 1
b. 2
c. 4
d. 3

Considering $a=11^{14^{4578369}}$ and $b=41$. Which is the remainder from division of $a$ by $b$.
a.40
b.38
c.37
d.39

EXTRA NOTE: I don't want the correct alternatives, I only want to figure out how to use Fermat's Little theorem in these situations !!!!

UmbQbify
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Rubenszinho
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    You could use the upgraded version of Fermat's Little Theorem known as Euler's Theorem. There are other approaches, though. – diracdeltafunk Jul 21 '20 at 23:54
  • Do you know the Chinese Remainder Theorem? – diracdeltafunk Jul 22 '20 at 00:00
  • Hey, u mean to use Euler's theorem in the (mod 3293) case which has 37 and 89 as its coprimes?? I thought phi only gives divisors, how do I get the remainders then? What about power of power case? – Rubenszinho Jul 22 '20 at 00:02
  • About chinese remainder I just have gave a peek in a pdf but it was kinda bad explained, u know any good article with good examples? – Rubenszinho Jul 22 '20 at 00:05
  • I'm not sure you've totally understood the statement of Euler's Theorem. Here, since $11$ and $3293$ are coprime, Euler's Theorem applies to tell us that $11^{\varphi(b)} \equiv 1 \pmod{b}$. Letting $a = \varphi(b) q + r$ with $0 \leq r < \varphi(b)$, we therefore get $11^a = 11^{\varphi(b) q + r} = (11^{\varphi(b)})^q 11^r \equiv 1^q 11^r = 11^r \pmod{b}$, so if you can find the remainder of $a$ when divided by $\varphi(b)$, you'll have a simpler problem.

    Notice that $\varphi(3293) = \varphi(37 \cdot 89) = \varphi(37) \cdot \varphi(89) = 36 \cdot 88 = 3168$.

     – diracdeltafunk Jul 22 '20 at 00:10
  • I know very much of the theory but I can't abstract without seeing exercises examples. That's where my difficult is you know, even with my researches I didn't find exercises alike them. – Rubenszinho Jul 22 '20 at 00:11
  • Hey, where did $a=φ(b)q+r$ come from? – Rubenszinho Jul 22 '20 at 00:16
  • $q$ and $r$ come from dividing $a$ by $\varphi(b)$ with remainder. – enedil Jul 22 '20 at 00:22
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    I used brilliant.org's quizzes and wikis to learn the basics of it. You can try it as well. – UmbQbify Jul 22 '20 at 00:27
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    @Rubenszinho Welcome to Math SE. I believe you'll find at least a few of the answers in How do I compute $a^b,\bmod c$ by hand?, as well as in some of its many linked posts, to be helpful. – John Omielan Jul 22 '20 at 01:14

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