my bad!! i read the question wrong.

The question is:

1^1+2^2+3^3+...+10^10

We can write the integers having power greater than 4 in the form of

5 ^(4n+1) +6^(4n+2)+7^(4n+3)+….+10(4n+2)

Any number whose power will be greater than 4 the units digit will start repeating after 4

E.g.

2^1=2, 2^2=4, 2^3=8 , 2^4=16 (just the units digit)

2^5=32 2^6=64 ,2^7=128

Similarly, we find that units digit for

1=>1

2=>4

3=>7

4=>6

5=>5^(4n+1)=>5^1 (units digit)=>5

6=>6^(4n+1)=>6^2 (units digit)=>6

7=>3

8=>6

9=>9

10=>0

Therefore,

Remainder=(1+4+7+6+5+6+3+6+9+0)/5=2

Answer is C