
(19^(99)+1999)^(19999)÷20 的余数是多少,过程?
2个回答

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19 mod 20 = -1
1999 mod 20 = -1
2^6 mod 20 = 4
运用余数定理
(19^(99)+1999)^(19999) mod 20
= [ (-1)^99+(-1)^1999 ]^19999 mod 20
= (-2)^19999 mod 20
= [ - 2^19999 ] mod 20
= [ - (2^6 mod 20)^3333*2 ] mod 20
= [ - (4^3333*2) mod 20 ] mod 20
= [ -(2^6)^1111*2 mod 20 ] mod 20
= [ -(4^1111*2) mod 20 ] mod 20
= [ - (2^6)^370 mod 20)*8 ] mod 20
= [ -4^370*8 mod 20 ] mod 20
= [ -(2^6)^123*2^5 mod 20 ] mod 20
= [ - 4^123*2^5 mod 20 ] mod 20
= [ - (2^6)^41*2^5 mod 20 ] mod 20
= [ - (4^41*2^5) mod 20 ] mod 20
= [ -(2^6)^14*8 mod 20 ] mod 20
= [ - (4^14*8) mod 20 ] mod 20
= [ -(2^6)^5*2 mod 20 ] mod 20
= [ -(4^5*2) mod 20 ] mod 20
= [ -(2^6)*2^5 mod 20 ] mod 20
= [ - (4*2^5) mod 20 ] mod 20
= [ -(2^6*2) mod 20 ] mod 20
= [ - (4*2) mod 20 ] mod 20
= - 8 mod 20
= -8
余数等于 20+(-8) = 12
1999 mod 20 = -1
2^6 mod 20 = 4
运用余数定理
(19^(99)+1999)^(19999) mod 20
= [ (-1)^99+(-1)^1999 ]^19999 mod 20
= (-2)^19999 mod 20
= [ - 2^19999 ] mod 20
= [ - (2^6 mod 20)^3333*2 ] mod 20
= [ - (4^3333*2) mod 20 ] mod 20
= [ -(2^6)^1111*2 mod 20 ] mod 20
= [ -(4^1111*2) mod 20 ] mod 20
= [ - (2^6)^370 mod 20)*8 ] mod 20
= [ -4^370*8 mod 20 ] mod 20
= [ -(2^6)^123*2^5 mod 20 ] mod 20
= [ - 4^123*2^5 mod 20 ] mod 20
= [ - (2^6)^41*2^5 mod 20 ] mod 20
= [ - (4^41*2^5) mod 20 ] mod 20
= [ -(2^6)^14*8 mod 20 ] mod 20
= [ - (4^14*8) mod 20 ] mod 20
= [ -(2^6)^5*2 mod 20 ] mod 20
= [ -(4^5*2) mod 20 ] mod 20
= [ -(2^6)*2^5 mod 20 ] mod 20
= [ - (4*2^5) mod 20 ] mod 20
= [ -(2^6*2) mod 20 ] mod 20
= [ - (4*2) mod 20 ] mod 20
= - 8 mod 20
= -8
余数等于 20+(-8) = 12
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