homework help > math > other math q&a board > topic: find (a^5)(mod7) for a = 1,2,3,4,5,6 ...

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posted by  diogenes on 11/19/2009 6:45:26 PM  |  status: Live  |  Earned Karma: 25

find (a^5)(mod7) for a = 1,2,3,4,5,6

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N/A N/A N/A N/A 11/20/2009 at 6:00:00 AM
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not sure what this question is asking to be honest if someone could do say for a = 2 and 3, I'm thinking I could figure it out
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posted by CC Skier on 11/19/2009 7:54:48 PM  |  status: Live
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The answer is the remainder when that number is divided by 7.  For example 10 is equal to 3 mod 7 since 10 divided by 7 is 1 remainder 3. The traditional way to write that is 10 = 1*7 + 3
I've highlighted the remainders in red below

15(mod 7) = 1(mod 7) since 1=0*7 + 1

25(mod 7) = 32(mod 7) = 4(mod 7)  since 32 = 4*7 + 4

35(mod 7) = 243(mod 7) = 5(mod 7) since 243 = 34*7 + 5

45(mod 7) = 1024(mod 7) = 2(mod 7) since 1024 = 146*7 + 2

55(mod 7) = 3125(mod 7) = 3(mod 7) since 3125 = 446*7 + 3

65(mod 7) = 7776(mod 7) = 6(mod 7) since 7776 = 1110*7 + 6
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posted by diogenes on 11/19/2009 8:05:24 PM  |  status: Live
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since 5 has an order of 6, wouldn't 5^5 be the most you can reduce it in mod 7?, whereas 6^5 is congruent to 6(mod7) since its order is 2
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