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I need to find that $10^{100}\equiv9 \pmod {19}.$

I started with $10\equiv-9 \pmod {19}.$

So $10^3\equiv-729 \pmod{19}.$

But $729\equiv7\pmod{19}$ so $10^3\equiv-7\pmod{19}.$

Which gives me $10^{100}\equiv (-7)^{33} \cdot( -9) \pmod {19},$ using $100=3\cdot 33+1.$

If this is correct we should normally have $(-7)^{33} \equiv -1 \pmod {19}$ to match with the final solution.

But I do not see why?

Thank you.

Bill Dubuque
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Sara
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  • I think the question you are asking is about cancellation. Specifically, why a is congruent to c mod 19 if ab is congruent to cb mod 19. This is because 19 and b have no factors in common. That said, you should use Fermat's Little Theorem to solve such questions. – Dhawal Patil Jan 31 '24 at 06:35
  • Check: $7^{3} \equiv (49)(7) \equiv (-8)(7)\equiv -56 \equiv 1 \pmod{19}$. Thus $7^{33} \equiv 1 \pmod{19}$. – Anurag A Jan 31 '24 at 06:37
  • You actually know $10^9$ is either $+1$ or $-1$ modulo $19.$ In general, $a^{100}\equiv \pm a\pmod {19}.$ If you know quadratic reciprocity, you can compute $10^{9}\bmod{19}$ directly. – Thomas Andrews Jan 31 '24 at 06:40
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    Also, have a look at Euler's theorem - you can use it to simplify the problem quite a bit. Another suggestion is to start from $10^2=20 \cdot 5\equiv 5 \pmod{19}$, which results in a sequence of "more friendly" numbers, at least in my opinion. – dvdgrgrtt Jan 31 '24 at 07:10
  • $!\bmod 19!:\ 10^{100}!\equiv (-3^2)^{100}\overset{\small\rm\color{#c00}{M}} = 3^{200 \bmod 18}!\equiv 3^{2(100\bmod 9)}!\equiv 3^2,$ by $\rm\color{#c00}{M}=$ mod order reduction $\ \ $ – Bill Dubuque Feb 02 '24 at 00:34
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    Duplicate of Mod of numbers with large exponents [modular order reduction], cf. prior comment. – Bill Dubuque Feb 02 '24 at 00:34
  • If you don't know any number theory (such as Fermat's little theorem) you can note $20\equiv 1 \pmod {19}$ and $10^{100} = 2^{100}5^{100} = 4^{50}5^{100} = 20^{50}\times 5^{50}$. So now you just have to fine $5^{50}$ and as $5^2\equiv 6$ we need $6^{25}$ and as $6\times 3 \equiv -1$ we can kind of noodle this to ridiculous extremes. – fleablood Feb 02 '24 at 03:59

1 Answers1

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By Fermat's little theorem $10^{19}\equiv 10$. Multiplying both sides by $10$ gives us $10^{20}\equiv 5$ since $100\equiv 5$. Now $(10^{20})^{5}\equiv 5^{5}$, so $10^{100}\equiv 66\equiv 9$.

For last statement note that $5^{5}\equiv 5*6*6\equiv 66$ because $25\equiv 6$ and $30\equiv 11$.

$\color{red}{\star}$$\pmod {19}$ for all congruence relations.

schneiderlog
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