$$2851^{604^{20}}\equiv 9^{604^{20}}\equiv(-5)^{604^{20}}\equiv 5^{604^{20}}\pmod{14}$$ $$5^{\phi(14)}\equiv 1\pmod{14}$$ (Thanks for the remark in comments) $$\gcd(k,14)=1,k\in \mathbb N, k<14 ,k\in\{1,3,5,9,11,13\}\implies\phi(14)=6$$ $$604^{20}\equiv 4^{20}\equiv 4\pmod{6}$$ $$5^x\equiv 1\pmod{14}\implies 5^{x+4}\equiv5^4\pmod {14}\implies5^{604^{20}}\equiv5^4\equiv 9\equiv\pmod{14}$$ Therefore: $$2851^{604^{20}}\equiv 5\pmod{14}$$ Is this correct?
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Matcha Latte
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$\phi(14) = \phi (2)\phi(7) = 6$ and you forget $1$ as one of the integers relative to $14$. Then numbers are $1,3,5,9,11,13$ and there are six of them, not $5$. – fleablood Jan 04 '20 at 17:27
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So in terms of $6$ and not five we get $604^{20}\equiv 4^{20}\equiv (-2)^{20}\equiv 2^{20}\equiv 4^{10} \equiv 2^{10} \equiv 4^{5} =(-2)^5 = -32 \equiv -2\equiv 4\pmod 6$. – fleablood Jan 04 '20 at 17:31
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So $2851^{604^{20}}\equiv 5^{604^{20}}\equiv 5^4\equiv 25^2 \equiv 11^2 \equiv (-3)^2 \equiv 9\pmod {14}$. But, yes, that was the correct idea. – fleablood Jan 04 '20 at 17:33
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@fleablood, Thank you for correction, I wouldn't notice at the moment. – Matcha Latte Jan 04 '20 at 17:37
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2That is the correct way of doing it. – fleablood Jan 04 '20 at 17:43
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2BTW.... The only time $\phi(M)$ is odd is if $M = 2$ and $\phi(M) = 1$. – fleablood Jan 04 '20 at 17:51
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@fleablood I appreciate your attention to the question. Thank you very much! – Matcha Latte Jan 04 '20 at 17:53
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Close....
but in calculating $\phi(14)$ you forgot to include $1$ as one of the values relative prime to $14$.
It's a good idea to get used to the idea of calculating $\phi$ using it's multiple distributive property over primes.
$\phi(14) =\phi(2*7) =\phi(2)\phi(7) = (2-1)(7-1) = 6$.
Then we have $604^{20}\equiv 4^{20}\pmod 6$.
Calculating $4^{20}\pmod 6$ can be done are a variety of ways. But if you had a really difficult case you could use the Chinese Remainder theorem $4^{20}\equiv 0 \pmod 2$ and $4^{20}\equiv 1^{20\%\phi(3)=0}\equiv 1\pmod 3$ so $4^20\equiv 0\pmod 2$ and $1\pmod 3$ and the only option for that is $4^{20}\equiv 4 \pmod 6$.
.... or we could have simply done a few successiv squarings. $4^2\equiv 16\equiv 4\pmod 6$ so $4^k\equiv 4 \pmod 6$.
So we have $ 2851^{604^{20}}\equiv 5^4\pmod {14}$
And $5^4\equiv 25^2 \equiv 11^2 \equiv (-3)^2 \equiv 9\pmod {14}$.
fleablood
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See here for much more on why $!\bmod 12!:\ 4^{\large 2}\equiv 4,\Rightarrow, 4^N\equiv 4\ $ [modular idempotents] – Bill Dubuque Feb 19 '20 at 17:39