Question

Chứng minh:

4^2018 - 1 chia hết cho 3

5^2019 - 1 chia hết cho 4

4^2019 + 1 chia hết cho 5

5^2017 + 1 chia hết cho 6

giúp mk với nha mn

Answer 1

a, Ta có: \(4\equiv1\left(mod3\right)\)

\(\Rightarrow4^{2018}\equiv1\left(mod3\right)\)

\(\Rightarrow4^{2018}-1⋮3\)

b, Ta có: \(5\equiv1\left(mod4\right)\)

\(\Rightarrow5^{2019}\equiv1\left(mod4\right)\)

\(\Rightarrow5^{2019}-1⋮4\)

c, \(4\equiv-1\left(mod5\right)\)

\(\Rightarrow4^{2019}\equiv-1\left(mod5\right)\)

\(\Rightarrow4^{2019}+1⋮5\)

d, \(5\equiv-1\left(mod6\right)\)

\(\Rightarrow5^{2017}\equiv-1\left(mod6\right)\)

\(\Rightarrow5^{2017}+1⋮6\)

Answer 2

1. Vì \(4\) chia \(3\) dư \(1\)

\(\Rightarrow4^{2018}\) chia \(3\) dư \(1^{2018}=1.\)

\(\Rightarrow4^{2018}-1\) chia hết cho \(3.\)

Answer 3

a, Ta có: 

⇒42018≡1(mod3)⇒42018≡1(���3)

⇒42018−1⋮3⇒42018−1⋮3

b, Ta có: 

⇒52019≡1(mod4)⇒52019≡1(���4)

⇒52019−1⋮4⇒52019−1⋮4

c, 

⇒42019≡−1(mod5)⇒42019≡−1(���5)

⇒42019+1⋮5⇒42019+1⋮5

d, 

⇒52017≡−1(mod6)⇒52017≡−1(���6)

⇒52017+1⋮6