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I'm aware that there are already a few questions like this but unfortunately I wasn't able to find an answer yet.

$$ (14^{2014)^{2014}} \pmod {60} $$

So I started off by putting the modular in : $$ (14^{2014}\pmod{60})^{2014} $$

So I looked at the inner break first and decided to calculate the inside. Since $gcd(14,60) \not= 1$ I can't use Euler/Fermat and have to do it via prime factorization and if Euler/Fermat works than this, otherwise I would have used CRT.

$$ 60 = 5*3*2^2$$ $$14^{2014} \equiv 1 \pmod 5$$ $$14^{2014} \equiv 1 \pmod 3$$ $$14^{2014} \equiv 0 \pmod 2$$

So $$14^{2014} \equiv 2 \pmod {60}$$

I now have left: $$2^{2014} \pmod {60}$$

Then I used CRT again, well I wanted to but saw that was actually easier again to get the values using Euler/Fermat after prime factorization like above. $$2^{2014} \equiv 4 \pmod {5}$$ $$2^{2014} \equiv 1 \pmod {3}$$ $$2^{2014} \equiv 0 \pmod {2}$$

So I assumed that it is 4 and $$2^{2014} \equiv 0 \pmod {4}$$

So $$2^{2014} \equiv 4 \pmod {60}$$

Unfortunately 4 is not the answer our tutor told us, he meant that the correct result is 16. Can anyone find where I miscalculated or where I was thinking wrong?

Edit: Actually I found one answer if I start off in another way:

$$14^{2014*2014} \pmod {60}$$

If I then do prime factorization and look at the remainders of the exponent I found that $$ 2014*2014 \equiv 2 \pmod {60}$$ and therefore $$14^2 = 196$$ $$196 \pmod{60} = 16$$

Great! Still, if anyone knows why the other approach doesn't work please tell me.

Edit 2: I thought about it and I'm not sure if I'm allowed to just look at the exponents instanced. Maybe it's just coincidence. Edit 3: To the post with the similar question: The solution is not fully the same and I think in this post it is easier to understand. Furthermore it was nice if this question stays because I wanted to show it to my peers.

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The conclusion

$14^{2014}≡2 (60)$

is not correct. It should be

$14^{2014}≡16 (60)$.

Here is my computation:

$14^{2014}≡1(5)$

$14^{2014}≡1(3)$

$14^{2014}≡0(4)$

And hence also:

$(14^{2014})^{2014}≡1(5)$

$(14^{2014})^{2014}≡1(3)$

$(14^{2014})^{2014}≡0(4)$

Therefore:

$(14^{2014})^{2014}≡16(60)$

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  • $\begingroup$ Ah I see I have to take 4 if primal factorization = 5*3*2*2 and not just 2. I see why $14^{2014} \pmod {60} $ is cong. 16. Thank you very much for taking the time! Appreciate. $\endgroup$ – Somebody Jun 28 '15 at 20:57

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