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We have to find the remainder of

$$32^{({32^{32}})}\mod 9$$

$\gcd(9,32)=1$ and by Euler's theorem we find $\phi(9)=9\cdot\frac{2}{3}=6$

Now

$$32\equiv5\mod 9 \ \\ 32 \equiv 2 \mod 6$$

$$\therefore32^{32}\equiv 2^{32}\mod 6$$

Now here comes a part where I am stuck. We know $2\equiv 2\mod 6$,

hence $2^{32}\equiv2^{32}\mod 6 \\ \Rightarrow 2^{31}\equiv (-1)^{31}\equiv 2\mod 3$

Now if I multiply both sides by $2$, then will it be $2^{32}\equiv4\mod 3$
or $2^{32}\equiv4\mod 6$?

I know that $a\cdot c\equiv b\cdot c \mod m$ but we don't multiply $m$ with $c$. Here I am little confused.

But I know the method to solve the whole but I am stuck in this little area.
Please help me. Any help is appreciated.

vbm
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    See this thread and other threads linked to it for a compendium of techniques for handling questions of this type. If I were a dictator I would close this as a dupe, but I have promised not to be the first close voter. Also, the iterated power makes it a bit trickier (though we have handled such problems also in the threads linked to that mother question). – Jyrki Lahtonen Mar 07 '18 at 19:43
  • Why would you thind $a \equiv b \mod 3\implies a \equiv b \mod 6$? – fleablood Mar 07 '18 at 19:47

4 Answers4

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$\gcd(32,9) = 1$ so $32^{\phi (9)} = 32^{6} \equiv 1 \mod 9$ via Eulers theorem.

So if $32^{32} \equiv k \mod 6$ then $32^{32^{32}} \equiv 32^k \equiv 5^k \mod 9$.

$32^{32} \equiv 0 \mod 2$ and $32^{32} \equiv (-1)^{32} \equiv 1 \mod 3$ so by Chinese remained theorem $32^{32} \equiv 4 \mod 6$.

So $32^{32^{32}} \equiv 32^4 \equiv 5^4 \equiv 25^2 \equiv (-2)^2 \equiv 4 \mod 9$.

fleablood
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  • Thank you for your solution. I know the answer and the method but I am not sure in the second hilighted part . Would you like to help me there? – vbm Mar 07 '18 at 19:38
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    Well, if you want to wrestle: $2^3 \equiv 2 \mod 6$ so $2^{32}\equiv (2^3)^{10}2^2 \equiv 2^{10}2^2 \equiv 2^{12} \equiv (2^3)^4 \equiv 2^4 \equiv 2^3 2 \equiv 22 \equiv 4$. But just use CRT. It's easier. – fleablood Mar 07 '18 at 19:42
  • That's incredible and helpful. Nice technique. – vbm Mar 07 '18 at 19:46
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To answer your particular question: you can use the Chinese Remainder Theorem at the end. You've already found that $2^{31}\equiv 2\pmod 3$; this implies that $2^{32}\equiv 1\pmod 3$. (Even more quickly, you could note that $2^2\equiv 1\pmod 3$ and therefore $2^{32}=\left(2^2\right)^{16}\equiv1^{16}\equiv 1\pmod 3$.) Also, clearly, $2^{32}\equiv 0\pmod 2$. So you just need to find the residue class $\pmod 6$ that's $\equiv 1\pmod 3$ and $\equiv 0\pmod 2$. In this case, the quickest is to work 'by sight'; the two $\bmod 6$ residue classes that are $\equiv 1\pmod 3$ are $\langle1\rangle$ and $\langle4\rangle$, and clearly only one of these is even.

Steven Stadnicki
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  • Got it and thank you. – vbm Mar 07 '18 at 19:52
  • "Even more quickly, you could note...." Just to be thorough: $2 \equiv -1\mod 3$ so $2^{32} \equiv (-1)^{32}$ and as $32$ is even $2^{32}\equiv 1 \mod 3$ is even quicker than that. – fleablood Mar 07 '18 at 21:21
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Notice that $32 \equiv 5 \pmod 9$

So now we have $5^{32^{32}}$

We also have $5^3 \equiv -1 \pmod 9$

We know that $32^{32} =3 \lfloor \frac {32^{32}}{3} \rfloor + r$ where r is the remainder of dividing $32^{32}$ by $3$

So now let's try to find r

$32^{32} \equiv 2^{32} \equiv (2^2)^{16} \equiv 4^{16} \equiv 1^{16} \equiv 1 \pmod 3$

So $r=1$. But we also know that $3 \lfloor \frac {32^{32}}{3} \rfloor + r$ must be even because it equals $32^{32}$

So now we can say that $3 \lfloor \frac {32^{32}}{3} \rfloor$ must be odd, since we get an even number when we add 1 to it.

A number of the type $3k$ can only be odd if k is odd, so $\lfloor \frac {32^{32}}{3} \rfloor$ must then also be odd. So we'll write $\lfloor \frac {32^{32}}{3} \rfloor = 2k+1$ for our final proof below

$5^{32^{32}} \equiv 5^{3 \lfloor \frac {32^{32}}{3} \rfloor + 1} \equiv 5^{3(2k+1) +1} \equiv (5^3)^{2k+1} \cdot 5 \equiv (-1)^{2k+1} \cdot 5 \equiv -5 \equiv 4 \pmod 9$

Francisco José Letterio
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  • That's a great unique approach. Beautifully done. – vbm Mar 07 '18 at 20:23
  • thanks! It's really odd because I feel like much of this solution would be useless anywhere else. I'm just lucky I could get a $-1$ in there and an easy parity test to check the result – Francisco José Letterio Mar 07 '18 at 20:24
  • Yes. Right now I was trying to solve in another method and I got a result $4^n\equiv 4\mod 6$ and using that I proved the whole. And appreciate your approach truely. – vbm Mar 07 '18 at 20:27
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    Also thanks for the problem. I just realized there's an easily provable identity I was never given in class. (The one you use to go from mod 6 to mod 3) – Francisco José Letterio Mar 07 '18 at 20:35
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Another way without theorems. $$32^N=(27+5)^N\equiv5^N\pmod9\\32^{32}\equiv5^{32}\pmod9$$ $$\begin{cases}5^1\equiv5\pmod9\\5^2\equiv7\pmod9\\5^3\equiv8\pmod9\\5^4\equiv4\pmod9\\5^5\equiv2\pmod9\\5^6\equiv1\pmod9\end{cases}\Rightarrow \begin{cases}5^6\equiv1\pmod9\\5^9\equiv8\pmod9\\5^{9(2n+1)}\equiv8\pmod9\\5^{9(2n)}\equiv1^\pmod9\end{cases}$$ It follows $$5^{32}=5^{30+2}\equiv5^2\equiv7^\pmod9\iff5^{32}=7+9M$$ So we have $$32^{32^{32}}=32^{7+9M}\Rightarrow\text{$M$ is odd }$$ It follows $$32^{32^{32}}=5^{7+9(2n+1)}=5^7\cdot8\equiv5\cdot8\equiv\color{red}{4}\pmod9$$

Ataulfo
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  • Marvellous. Another great approach. Thank you. – vbm Mar 08 '18 at 07:32
  • I agree with you for your sensitivity to the particular properties of the case. But it must be said, in honor of the truth, that there was some luck in the solution. – Ataulfo Mar 08 '18 at 12:40