How Does Fermat's Little Theorem Apply to Calculating 3^302 (mod 5)?

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Discussion Overview

The discussion revolves around the application of Fermat's Little Theorem to calculate \(3^{302} \mod 5\). Participants explore the theorem's implications and how it can be used to simplify the calculation.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant states the theorem as \(a^{p-1} \equiv 1 \mod p\) and questions how it applies to \(3^{302} \mod 5\).
  • Another participant reiterates that according to Fermat's Little Theorem, \(3^{4} \equiv 1 \mod 5\).
  • A third participant provides a breakdown of the calculation, suggesting that \(3^{302} \mod 5\) can be simplified using the theorem, leading to the expression \(3 \cdot 3 \cdot 1 \mod 5\).
  • A later reply expresses gratitude for the insights, indicating that the discussion has helped in understanding related problems.

Areas of Agreement / Disagreement

Participants appear to agree on the application of Fermat's Little Theorem to the problem, but there is no explicit consensus on the final calculation or its implications.

Contextual Notes

The discussion does not resolve the final value of \(3^{302} \mod 5\) explicitly, and some assumptions about the application of the theorem remain unexamined.

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I have a question regarding mods and Fermat's Little Theorem. I know Fermat's little theorem states that a^p-1 congruent to 1 (mod p). Also, i know that for every interger a we have that a^p congruent to a (mod p). So, my question is: What is the answer for 3^302 (mod 5)? Would it be 3^301 congruent 1 (mod 5)? I am having a bit of difficulty understanding this concept. Any help?
 
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Well, Fermat's little theorem says 3^(5-1) = 1 (mod 5)...
 
Consider 300 = 4*75 = (5-1)*75. So:

So 3^(302) mod 5 = 3*3*[3^(300)] mod 5 = 3*3*[(3^75)^(5 - 1)] mod 5 = 3*3*1 mod 5

I think you can do the rest :wink:
 
thank you thank you all. by using this i can figure out the other five problems.
 

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