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How to compute $75^{75^{75}}$ modulo $32$?

I tried:

It's $\mathrm{gcd}(75,32)=1$, so with Euler's phi function I get

$\varphi(32)=32(1-\frac{1}{2})=16$.

Then it's $75^{16}\equiv1 \ (\mathrm{mod} \ 32)$.

So I computed $75^{75}$ modulo $16$:

It's $\varphi(16)=16(1-\frac{1}{2})=8$

So $75^{75}\equiv 1 \ (\mathrm{mod} \ 8)$ and $75^{75^{75}} \equiv 1 \ (\mathrm{mod} \ 8)$

Then it's $75^{75^{75}}\equiv 75^1 \equiv 75 \ (\mathrm{mod} \ 32)$.

I'm not sure if this is correct. Is there something wrong in this calculation or can it be done like this?

Tartulop
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1 Answers1

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$$75^{75 } \equiv 75^{72} 75^3 \equiv (75^{8})^{9}11^3 \equiv 3 \mod 16$$

so we can say $$ {75^{75}}^{75} \equiv 75^3 \equiv 11^3 \equiv 19 \mod 32$$

  • How to see fast that $75^3 \equiv 11^3 \equiv 19 \ (\mathrm{mod} \ 32)$? – Tartulop May 16 '20 at 11:44
  • Well $75 = 2(32) + 11$ a simple application of the euclidean algorithm. As for $11^3$ I just computed it to be $1331$ and I noticed at first $1331 = 1024 + 307 = 2^{10} + 307 = 32^2 + 307$. Then just euclidean algorithm on $307$ o get $307 = 9(32) + 19 $ – Ziad Fakhoury May 17 '20 at 20:06