Find Remainder with Fermat's Theorem

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This discussion focuses on using Fermat's Little Theorem to find the remainder of 52005 when divided by 4010. The problem highlights the decomposition of 4010 into its prime factors: 2, 5, and 401. By applying Fermat's Little Theorem, participants are guided to compute \(5^{2005} \mod 2\), \(5^{2005} \mod 5\), and \(5^{2005} \mod 401\). The results from these calculations can then be combined using the Chinese Remainder Theorem to determine the overall remainder.

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  • Fermat's Little Theorem
  • Chinese Remainder Theorem
  • Modular arithmetic
  • Prime factorization
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jeedoubts
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How to use fermit's thereom in finding remainder of a number when divided by another number ?

(eg remainder of 52005 when divided by 4010 ?)
 
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How did you arrive at such a problem? Fermat's (little) theorem deals prime powers.
 
I'm assuming you don't know or don't want to use Euler's theorem.

Note 4010 = 2*5*401.

Can you find integers a,b,c such that
[tex]\begin{align*}<br /> 5^{2005} &\equiv a \pmod 2 \\<br /> 5^{2005} &\equiv b \pmod 5 \\<br /> 5^{2005} &\equiv c \pmod {401}<br /> \end{align*}[/tex]
? (perhaps using Fermat's little theorem)

If you can, then you can use these results and the Chinese remainder theorem to find 5^2005 modulo 2*5*401.
 

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