Tìm số dư khi chia B= $11^{2019}$ +$11^{2020}$ cho 19
2 câu trả lời
$11^3≡1(mod 19)$
$=>11^{2019}≡1(mod 19)$
$11≡11(mod 19)$
$=> 11^{2020}≡11(mod 19)$
$=>11^{2019}+11^{2020}≡12(mod 19)$
$=>11^{2019}+11^{2020}$ chia $19$ dư $12$
$=>B$ chia $19$ dư $12$
`B=11^2019 +11^2020`
`=11^2019 .12`
Có `11^3 =1331≡1(mod 19)`
`⇒(11^3 )^673 ≡1^673 =1(mod 19)`
`⇒ (11^3 )^673 .12 ≡1.12(mod 19)`
`⇒11^2019 .12≡12(mod 19)`
Vậy ` B:19` dư `12`
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