Question

Cho số A = \(2017^{2018}+2018^{2017}\). Tìm số dư khi chia A cho 10 ?

Answer 1

Ta có:
2017² = 9 ( theo mod 10 ) ; 2017⁸=(2017²)⁴=9⁴=1 ( theo mod 10 )
2017¹⁰ = (2017²)^{5 =} 9⁵=9 ( theo mod 10 )
2017¹⁰⁰=( 2017¹⁰)¹⁰=9¹⁰=1 ( theo mod 10 )
2017¹⁰⁰⁰=( 2017¹⁰⁰)¹⁰=1¹⁰=1 ( theo mod 10 )
2017²⁰⁰⁰=( 2017¹⁰⁰⁰)²=1²= 1( theo mod 10 )
2017²⁰¹⁸=2017²⁰⁰⁰.2017¹⁰.2017⁸=1.9.1=9( theo mod 10 )

2018=8 ( theo mod 10 ) ;2018²=4 ( theo mod 10 )
2018⁷=8⁷=2 ( theo mod 10 )
2018¹⁰=(2018²)⁵=4⁵=4 ( theo mod 10 )
2018¹⁰⁰=(2018¹⁰)¹⁰=4¹⁰=6 ( theo mod 10 )
2018¹⁰⁰⁰= (2018¹⁰⁰)¹⁰=6¹⁰=6 ( theo mod 10 )
2018²⁰⁰⁰=(2018¹⁰⁰⁰)²=6²=6( theo mod 10 )
2018²⁰¹⁷=2018²⁰⁰⁰.2018¹⁰.2018⁷=6.4.2=8

⇒ A = 2017²⁰¹⁸+2018²⁰¹⁷=9+8=7 ( theo mod 10 )

⇒ Số dư khi chia A = 2017²⁰¹⁸+2018²⁰¹⁷ cho 10 là 7

Answer 2

Ta có:
20172 = 9 ( theo mod 10 ) ; 20178=(20172)4=94=1 ( theo mod 10 )
201710 = (20172)5 = 95=9 ( theo mod 10 )
2017100=( 201710)10=910=1 ( theo mod 10 )
20171000=( 2017100)10=110=1 ( theo mod 10 )
20172000=( 20171000)2=12= 1( theo mod 10 )
20172018=20172000.201710.20178=1.9.1=9( theo mod 10 )

2018=8 ( theo mod 10 ) ;20182=4 ( theo mod 10 )
20187=87=2 ( theo mod 10 )
201810=(20182)5=45=4 ( theo mod 10 )
2018100=(201810)10=410=6 ( theo mod 10 )
20181000= (2018100)10=610=6 ( theo mod 10 )
20182000=(20181000)2=62=6( theo mod 10 )
20182017=20182000.201810.20187=6.4.2=8

⇒ A = 20172018+20182017=9+8=7 ( theo mod 10 )

⇒ Số dư khi chia A = 20172018+20182017 cho 10 là 7