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I'm trying to work out a question found in my textbook for college on modular exponentiation, {11^399 mod 19}.

These are the steps I took:

  1. 11^399 mod 19
  2. 8^339 mod 19 [19 - 11 = 8]
  3. (8^388 * 8) mod 19 [subtracted the exponent by 1 and multiplying the base by itself to make up for the subtracted exponent]
  4. [This part is where I'm stuck :] My next step was to multiply Step 3 () but when I put 8^338 into the calculator it either showed a very very long number or mostly infinity on the online calculators.

Also the answer is supposed to be 7 from what I'm seeing, using the online calculators but I don't even know how to arrive at that. I have tried this same process for other questions and they have worked without any obnoxious remainders (at most 2 digits). If anyone could help me out with this one that'd be much appreciated!

miracle173
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TheQuietlyDepressedCS_Student
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  • Use FLT(fermat's little theorem), then you would easily get it. – jaemmin Dec 13 '22 at 03:19
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    Is the point to work it out by hand? If so, what theorems do you have at your disposal? Hard to see how to do this without Fermat's Little (or Euler's) theorem. – Randall Dec 13 '22 at 03:22
  • @jaemmin HI, I search up this theorem and I got the answer solved. The textbook I have does not show anything at all about this theorem which is surprising considering how fast I got it out. Thanks for the tip anyways! – TheQuietlyDepressedCS_Student Dec 13 '22 at 03:28
  • @TheQuietlyDepressedCS_Student Hi, happy to hear that. If you were to do it by hand, then there's a way, too. Is it fine with using FLT or need hand-work solution? – jaemmin Dec 13 '22 at 03:30
  • @Randall i used the fermats theorem suggested to me by one of the commenters and it worked. Thank you too for the help :) – TheQuietlyDepressedCS_Student Dec 13 '22 at 03:30
  • @jaemmin I did similar questions using hand work but this question had me stuck so sure, that might help :) – TheQuietlyDepressedCS_Student Dec 13 '22 at 03:32
  • We can reduce the exponent to $,339 \bmod 18 = \color{#c00}{-3},$ by mod order reduction - see the Theorem in the linked dupe, yielding $,11^{339}\equiv 1/(11^{\color{#c00}{3}})\equiv 1/1.,$ Follow the links there for many other methods. – Bill Dubuque Dec 13 '22 at 08:32
  • Or use $,11\equiv (\color{#c00}{-2})^{\large 3}\equiv (\color{#c00}{6^{\large 2}})^{\large 3},$ so $,11^{\large 3K}\equiv (\color{#c00}6^{\large 18})^{\large K}\equiv 1^{\large K}\equiv 1,,$ by little Fermat. Of course we could also notice by luck that $,11^{\large 3}\equiv 1,$ by enumerating small powers. – Bill Dubuque Dec 13 '22 at 08:51

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I recommend you to use FLT, but if you need hand-work solution, it would help.

\begin{align} 11^{339} &\equiv 11^{338}* 11&\mod{19} \\ &\equiv (11^2)^{169} * 11&\mod{19} \\ &\equiv 7^{169} * 11&\mod{19} \\ &\equiv 7^{168} * 7 * 11&\mod{19} \\ &\equiv (7^2)^{84} * (7 * 11)&\mod{19} \\ &\equiv 11^{84} * 1&\mod{19} \\ &\equiv (11^2)^{42} &\mod{19} \\ &\equiv 7^{21} &\mod{19} \\ &\equiv (7^3)^{7} &\mod{19} \\ &\equiv 1^{7} &\mod{19} \\ &\equiv 1 &\mod{19} \\ \end{align}

jaemmin
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Dec 13 '22 at 08:00
  • @Bill Dubuque Sorry, as the questioner wanted other hand-work solution, I thought it is not the very same question cuz it needs different answer. – jaemmin Dec 13 '22 at 23:42
  • @BillDubuque my apologies, I saw questions close to the pertaining to the topic but not the same format of question. I will try to not do this as much as possible if I can. I'm a bit new to stack so I'm working on getting used to the rules and stuff, hope you understand. – TheQuietlyDepressedCS_Student Dec 14 '22 at 06:09