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發問:
456^789mod123 點樣可以簡化至可以計算? 請詳細列出步驟! thx!
最佳解答:
1. 123=41*3, 456=3*152 2. 456^789=0 (mod 3) 456^789= 5^789 (mod 41) =125^263 (mod 41) = 2^263=(2^7)^37*2^4= 5^37*16 (mod 41) = 125^12*5*16 = 2^12*(-2) (mod 41) =2^7* 2^5*(-2)= 5*32*(-2)= 8 (mod 41) 3. 456^789=8 (mod 41) so 456^789= 41*a+8 then 456^789=41*(3b+r)+8 (r=0,1,2) while 456^789=0 (mod 3) so 41(3b+r)+8=0 (mod 3) then 2r-1=0 (mod 3), thus r=2 456^789=41(3b+2)+8= 123b+ 90 thus 456^789=90 (mod 123) (other method by Fermat's little thm) Fermat: 5^40=1 (mod 41), then 5^760=1 (mod 41) so 456^789= 5^789=5^29 (mod 41) = (125^9)*25 =2^9*25 (mod 41) = (2^7)*(2^2)*25 = 5*4*25=8 (mod 41)
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想問一條mod數發問:
456^789mod123 點樣可以簡化至可以計算? 請詳細列出步驟! thx!
最佳解答:
1. 123=41*3, 456=3*152 2. 456^789=0 (mod 3) 456^789= 5^789 (mod 41) =125^263 (mod 41) = 2^263=(2^7)^37*2^4= 5^37*16 (mod 41) = 125^12*5*16 = 2^12*(-2) (mod 41) =2^7* 2^5*(-2)= 5*32*(-2)= 8 (mod 41) 3. 456^789=8 (mod 41) so 456^789= 41*a+8 then 456^789=41*(3b+r)+8 (r=0,1,2) while 456^789=0 (mod 3) so 41(3b+r)+8=0 (mod 3) then 2r-1=0 (mod 3), thus r=2 456^789=41(3b+2)+8= 123b+ 90 thus 456^789=90 (mod 123) (other method by Fermat's little thm) Fermat: 5^40=1 (mod 41), then 5^760=1 (mod 41) so 456^789= 5^789=5^29 (mod 41) = (125^9)*25 =2^9*25 (mod 41) = (2^7)*(2^2)*25 = 5*4*25=8 (mod 41)
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