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I've just started a cryptography course, so i am not experienced at all how to calculate such big numbers. Clearly, i can't use a calculator, because the number is too big, so i have to calculate it by hand.

Since $101$ is a prime number, i think i should use here Fermat's little theorem. Found some example and tried to solve it this way, but i am totally not sure, if it is correct and if my approach must be this one.

Calculate $5^{3^{1000}}\bmod 101$.

First of all i think i should calculate $3^{1000}\bmod 101$. From Fermat's little theorem i get to $3^{100}\equiv 1\bmod 101$.

Thus $1000=x100+0$ and $x=10$.

$3^{1000}\equiv 3^{999^{10}} = 1 ^{10} \equiv 102\bmod 101 $

Later i have to calculate $5^{102}\bmod 101$. Again by Fermat $5^{100}\equiv 1\bmod 101$.

$$102=100\cdot 1 +2$$

Here i am not sure how to move on... I think that my solution is wrong, but i'd love to see your suggeststions and remarks how to solve the problem. Thank you very much in advance!

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By Fermat's little theorem we know that $5^{100} \equiv 1 \bmod 101$.

What exactly does this tell us? It tells us that the powers of $5$ when reduced mod $101$ repeat every $100$ times.

So to find out what $5^{3^{1000}}$ is mod $101$ we really need to find out what $3^{1000}$ is mod $100$.

You can use the generalisation of FlT mentioned in another answer to see that $3^{40} \equiv 1 \bmod 100$, so that $3^{1000} = (3^{40})^{25} \equiv 1^{25} \equiv 1 \bmod 100$.

Alternatively you can do it by little step by step calculations.

Either way we find that $5^{3^{1000}} \equiv 5 \bmod 101$.

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  • $\begingroup$ Or he can type 3^1000 into an appropriate software an look at the last two digits ;) $\endgroup$ – Carsten S Sep 29 '13 at 21:56
  • $\begingroup$ ...but surely with that attitude he can type $5^{3^{1000}} mod 101$ into appropriate software? $\endgroup$ – fretty Sep 29 '13 at 22:07
  • $\begingroup$ Fair enough. I just wanted to point out that while it is agains the spirit of the problem, it is quite feasible to compute 3^10 exactly. $\endgroup$ – Carsten S Sep 30 '13 at 10:52
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You should calculate $3^{1000}\mod \varphi(101)=100$. Here Fermat's little theorem doesn't apply, since $100$ is not prime, thus you must resort to Euler's theorem

If $(a,n)=1$, then $a^{\varphi(n)}=1\mod n$.

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You have seen that $5^{100} \equiv 1 \pmod{101}$. It follows that $5^{100a+b} \equiv 5^b \pmod{101}$. Now you should see to which modulus you need to calculate $3^{1000}$.

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