Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$ x = 5^{1345}\bmod 58$.

I wrote a program that finds and period of residues and builds a table. This table consists of $k$ lines where $k$ is a number of residues in one repeating block, as residues repeat periodically. In other words, I represent this equation as $x = 5^{n \cdot k+m} \bmod 58$, where $m$ is # of residue in table, and that residue is the answer.

But how to solve this equation mathematically? This algorithm is too complicated to be done on paper. I know that it's possible to use Fermat's little theorem here, but can't understand how. Hope someone will help me to understand this.

share|improve this question
1  
Hint: Use Chinese remainder theorem. –  tetori Jan 14 '13 at 14:19
add comment

2 Answers

up vote 5 down vote accepted

$5^{1345} = 5 \cdot 5^{1344} = 5 \cdot (5^{28})^{48} = 5 \cdot (5^{\phi(58)})^{48} = 5 \cdot 1^{48} = 5 \mod 58$.

share|improve this answer
    
Thanks, I got it! –  Shemhamforasch Jan 14 '13 at 14:28
add comment

Use Euler's theorem to get that: $$5^{28}\bmod{58} =1$$ And use: $$5^{1345}\bmod{58} = 5^{48\cdot28+1}\bmod{58}=5\cdot 1^{48}=5$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.