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How would you approach the following problem?

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I've tried to focus on last digits for example but that didn't lead me to the answer (Like, 2021^(any number) will have last digit 1). Can you guys give me a clear hint on how to approach it?

Bill Dubuque
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Prim3numbah
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  • Focusing on the last digit would give you an answer if the question were asking about the remainder when you divide by $10$. Do you know about modular arithmetic? – Chris Lewis Jun 13 '24 at 14:10
  • @ChrisLewis I have in the past, but forgot most of it. It was closed I see, hopefully the suggested post will give me a clue – Prim3numbah Jun 13 '24 at 14:16
  • @BillDubuque I'll do so – Prim3numbah Jun 13 '24 at 14:18
  • @ChrisLewis Although I gotta say, this one feel a lot more complicated – Prim3numbah Jun 13 '24 at 14:34
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    There are three results of mod arithmetic that can be a mantra: Fermat's last theroem which tells us as $19$ is prime $2024^{M}+2025^{N}\equiv (2024%19)^{M%18}+2025^{N%18}\pmod{19}$. Eulers Theorem that tells us as $18=2\times 3^2$ so $\phi(18)=1\times 2\times 3=6$ so $2024^{2003^K}+2025^{2021^J}\equiv 10^{7^{K%6}}+11^{5^{J%6}}\pmod{19}$. The Chinese remainder th. won't be of much use here but it's useful with the base of exponents isn't relatively prime to what you are dividing by. Good luck finding your answer in those linked posts. – fleablood Jun 13 '24 at 14:37
  • It looks intimidating but it needn't be. $2024\equiv 10\pmod {19}$. So $2024^{2023^{2022^{...}}}\equiv 10^{2023^{2022^{..}}}\pmod{19}$. $\phi {19}=18$ so $10^{2023^{2022^{..}}}\equiv 10^{(2023\pmod {18})^{2022^{....}}}\pmod {19}$. $2023\equiv 7\pmod {18}$, so $10^{(2023\pmod {18})^{2022^{....}}}\equiv 10^{(7^{2022^{...}}\pmod {18}}\pmod{19}$. And $\phi(18)=6$ and $2022\equiv 0\pmod 6$ so $10^{(7^{(2022\pmod 6)^{...}}\pmod {18}}\pmod{19}\equiv 10^{7^{0^{...}}}\pmod {19}\equiv 10^{7^0}\equiv 10^1\equiv 10 \pmod{19}$. So we are half done. It's kind of like magic. – fleablood Jun 13 '24 at 14:53
  • It looks more complicated to write it out then it is to do it. Actually, yours is easier and more straightforward than the two linked problems. – fleablood Jun 13 '24 at 14:59
  • e.g. as in the linked dupes, using Euler totient $,\phi!:, 19\to \color{#0a0}{18}\to\color{#c00} 6$ $$2025^{\Large 2021^{\LARGE 2017^N}}!!!!!!!!\bmod 19 ,=, {2025^{\phantom{|^|}}}^{!!!\Large 2021^{\LARGE 2017^N_{\phantom|}!\bmod\ \color{#c00}6}!!!!!!!!!!!!!!!!!!!!!!\large \bmod \ \color{#0a0}{18}}!!!!!!!!!!!!!!!!!!!!\bmod 19 ,\equiv 11^{\large\color{#0a0}{5}^{\Large \color{#c00}1^N}}!!!!\equiv 11^{\large 5} \equiv 7!!\pmod{!19}\qquad\qquad$$ – Bill Dubuque Jun 13 '24 at 16:49
  • e.g. as in the linked dupes, using Euler totient $,\phi!:, 19\to \color{#0a0}{18}\to\color{#c00} 6$ $$2024^{\Large 2023^{\LARGE 2022^N}}!!!!!!!!\bmod 19 ,=, {2024^{\phantom{|^|}}}^{!!!\Large 2023^{\LARGE 2022^N_{\phantom|}!\bmod\ \color{#c00}6}!!!!!!!!!!!!!!!!!!!!!!\large \bmod \ \color{#0a0}{18}}!!!!!!!!!!!!!!!!!!!!\bmod 19 ,\equiv 10^{\large\color{#0a0}{7}^{\Large \color{#c00}0^N}}!!!!\equiv 10^{\large 1} \equiv 10!!\pmod{!19}\qquad\qquad$$ – Bill Dubuque Jun 13 '24 at 16:55
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    @BillDubuque I ... was being one hundred percent sincere. I truly hope the op can learn from the posts to understand and solve there question. I don't see how it come across as sarcastic. – fleablood Jun 14 '24 at 08:19
  • Thanks both for the methods you showed me. And @fleablood No worries, I actually took it sincere :) – Prim3numbah Jun 14 '24 at 13:41
  • If anything remains unclear please feel welcome to ask questions in comments (here or in the dupes - that's how we improve prior answers). – Bill Dubuque Jun 14 '24 at 20:10

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