Using Fermat - Using Fermat's theorem mod p and/or Euler's...

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Using Fermat's theorem mod p and/or Euler's Totien theorem, find the congruence of 2749^633605 mod 5281. Fermats theorem: Here: Let “ == “ denote congruence a^(p-1) == 1 mod p then; 633605/5280 = 120 remainder 5 2749^633605 = (2749^120)^5280 * 2749 ^5 So, (2749^120)^5280 == 1 mod 5281 Solving this; 633605 's binary representation is 10011010101100000101 2749 to the power of (2^19) (mod 5281) = 2888 2749 to the power of (2^18) (mod 5281) = 4270 2749 to the power of (2^17) (mod 5281) = 538 2749 to the power of (2^16) (mod 5281) = 4427 2749 to the power of (2^15) (mod 5281) = 3389 2749 to the power of (2^14) (mod 5281) = 1739 2749 to the power of (2^13) (mod 5281) = 444 2749 to the power of (2^12) (mod 5281) = 1227 2749 to the power of (2^11) (mod 5281) = 3061 2749 to the power of (2^10) (mod 5281) = 1845 2749 to the power of (2^9) (mod 5281) = 2888 2749 to the power of (2^8) (mod 5281) = 4270 2749 to the power of (2^7) (mod 5281) = 538
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2749 to the power of (2^6) (mod 5281) = 4427 2749 to the power of (2^5) (mod 5281) = 3389
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This note was uploaded on 06/20/2010 for the course UNKNOWN TECH taught by Professor Obayo during the Spring '10 term at University of Arkansas for Medical Sciences.

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Using Fermat - Using Fermat's theorem mod p and/or Euler's...

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