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I assume I should find $n \le 100$ such that $$2^{10000}\equiv n\! \pmod {100}$$ I can see that $$2^{10000}= \left(2^{2}\right)^{5000}\equiv 0 \! \pmod {4}$$ and since $\phi (25)=20$ and $gcd(2,25)=1$ we also get $$2^{10000}=\left(2^{20}\right)^{500}\equiv 1\! \pmod {25}$$

But since i get different remainders i dont know how to proceed

Nasal
  • 798
  • See here. You obtain $n=76$. – Dietrich Burde Feb 28 '20 at 12:53
  • I looked all over the web and I have seen CRT mentioned everywhere but not sure how to apply it in this situation – Nasal Feb 28 '20 at 12:53
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    Or since this is such a small case (applying the Chinese Remainder Theorem here is overkill if you only want to find the solution), simply look at numbers that are 1 more than a multiple of 25 (i.e. 1, 26,...), until you find one that is also a multiple of 4. – Jaap Scherphuis Feb 28 '20 at 12:55
  • Note that the Chinese remainder theorem tells you you have a solution (and gives a guaranteed method for finding it). But knowing there is a solution you can look for a short-cut computation with confidence. – Mark Bennet Feb 28 '20 at 13:01
  • $\bmod 25!:,\ n = 4k \equiv 1\equiv -24\iff k\equiv -6\iff n=4k\equiv -24\equiv 76\pmod{!100}\ \ $ – Bill Dubuque Feb 28 '20 at 15:19

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