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 2749 to the power of (2^4) (mod 5281) = 1739 2749 to the power of (2^3) (mod 5281) = 4837 2749 to the power of (2^2) (mod 5281) = 1538 2749 to the power of (2^1) (mod 5281) = 5171 2749 to the power of (2^0) (mod 5281) = 2749 Hence, Hence 2749 to the power of 633605 (mod 5281) = 2888 X 1 X 1 X 4427 X 3389 X 1 X 444 X 1 X 3061 X 1 X 2888 X 4270 X 1 X 1 X 1 X 1 X 1 X 1538
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