■1次合同式(その4)

[Q]2^6754=?  (mod1155)

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 1155=3・5・7・11

フェルマーの小定理より

  2^2=1  (mod3)

  2^4=1  (mod5)

  2^6=1  (mod7)

  2^10=1  (mod11)

 したがって,

  2^6754=1  (mod3)

  2^6754=4  (mod5)

  2^6754=2  (mod7)

  2^6754=5  (mod11)

 こうして,連立合同式

[1]x=1  (mod3)

[2]x=4  (mod5)

[3]x=2  (mod7)

[4]x=5  (mod11)

が得られる.

[1]x=1+3y

[2]1+3y=4  (mod5)→y=1  (mod5)

→y=1+5z→x=1+3y=4+15z

[3]4+15z=2  (mod7)→z=5  (mod7)

→z=5+7t→x=79+105t

[4]79+105t=5  (mod11)→t=6  (mod11)

→t=6+11u→x=709+1155u

[A]2^6754=709  (mod1155)

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