Question

What is the last digit of 4^n,where n is the natural number​

Answer 1

ANSWER

3

4n

=(3

4

)

n

=81

n

=(1+80)

n

=1+[

n

C

1

80+

n

C

2

80

2

+

n

C

3

80

3

...+

n

C

n

80

n

]

=1+λ

Taking 80 common from λ we get

=1+80k where k is a constant...(i)

3

1+80k

=3(3

80k

)

=3(9

40k

)

=3(81

20k

)

=3([1+80]

20k

)

=3(1+α80) ...similar to (i)

=3+240α

Hence 3

3

4n

+1

=3+240α+1

=240α+4

=10(24α)+4

=10β+4

Answer 2

Last digit of4ⁿ if n is even then Last digit will be 6 and if it's odd then Last digit will be 4