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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^(p1)
== 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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View Full Document 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.
 Spring '10
 Obayo

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