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How to calculate: $$5^{5^{33}}\bmod{100}$$

I am aware of this quesiton, but still did not understand how to solve for my example.

I know how to use Euler's theorem, but I do not know how to go around $5^{33}$ and $100$ not being co-prime.

user412453
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    All powers of 5 from the square onwards are $\equiv 25$, maybe? – Oscar Lanzi Jun 14 '18 at 11:19
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    I think in this case, first work out $5^2$, then $5^3$, $5^4$, $5^5$, and so on until you reach $5^{5^{33}}$. Trust me when I say you'll get there faster than you think. – user361424 Jun 14 '18 at 11:28
  • I am aware of this, but I should solve this question using Euler's theorem or the Chinese remainder theorem. I am thinking about the answer below using the Chinese remainder theorem. – user412453 Jun 14 '18 at 11:31

2 Answers2

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We know that $25$ divides the given number. Hence, the only possible residues are $0,25,50,75$. Now, 2 definitely does not divide the given number, and so the possible residues are $25,75$. Now, we can just find out the residue mod $4$. The given number is congruent to $1^{5^{33}}=1\text{mod}~ 4$. Out of $25,75$ only $25$ has residue $1$ mod 4. Hence, the residue of the original number mod $100$ is 25.

The point here is that $100=25 \times 4$, and $(25,4)=1$. Hence, to find the residue mod $100$, just find the residue mod $25$ and mod $4$ and use the Chinese remainder theorem to get a unique solution mod $100$.

BDN
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Manan
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As $(5^{n+2},100)=25$ for integer $n\ge0$

let us find $5^n\left(\bmod\dfrac{100}{25}\right)$

As $5\equiv1\pmod4,5^n\equiv1^n\equiv?$

$\implies5^{n+2}\equiv5^2\pmod{4\cdot25}$

Here $n+2=5^{33}$

Reference :

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lab bhattacharjee
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