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Re: Congruence Question
Posted:
Oct 24, 2012 1:05 PM
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If p and q are prime, then p^q=p(mod q). 3 and 509 are prime.
Since 3^509=3(mod 509), 3^508=(3^509)/3=1 (mod 509) since inverses are unique modulo a prime and 3^512=3^3*(3*509)=81(mod 509).
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From: Edward <discussions@mathforum.org>
To: discretemath@mathforum.org
Sent: Wednesday, 24 October 2012, 1:52
Subject: Congruence Question
So now I'm moving on and learning about congruence for the first time. Trying to take it all in and did some example problems to sharpen my skills, but I have two questions.
List all congruence classes for congruence mod 9, giving the most usual and one other name for each. Is this right?
-27 -18 -9 0 9 18 27 36 45 54 63 72 81 90
-26 -17 -8 1 10 19 28 37 46 55 64 73 82 91
-25 -16 -7 2 11 20 29 38 47 56 65 74 83 92
-24 -15 -6 3 12 21 30 39 48 57 66 75 84 93
-23 -14 -5 4 13 22 31 40 49 58 67 76 85 94
-22 -13 -4 5 14 23 32 41 50 59 68 77 86 95
-21 -12 -3 6 15 24 33 42 51 60 69 78 87 96
-20 -11 -2 7 16 25 34 43 52 61 70 79 88 97
-19 -10 -1 8 17 26 35 44 53 62 71 80 89 98
2.Use Fermat's Little Theorem to compute
3^508 (mod 509), 3^509(mod 509) and 3^512 (mod 509)
Any ideas for question 2? I'm lost
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