how to apply the concept of "divisibility by 37" for the following question.

4 is written 3400 times side by side to give a big number i.e. 44444.........444(3400 digits). What is the remainder when divided by 37?

the remainder will be 4.
444 is completely divisible by 37 hence 44444......4444(3n digits) will also be completely divisible by 37
444444.........44444((3n+1) times) will give a remainder of 4 upon dividing by 37
4444.......4444((3n+2)times) will give a remainder of 7 upon dividing by 37. since 44 gives a remainder of 7 upon dividing by 37.
since 3400 can be represented as 3n +1 i.e, 331133 +1 hence the remainder will be 4

37*3 is 111
Given is 4444...(3400 times ) so write it as
4*11111...(3400 times )
now let us find a relation
I.e., 111*1001 is 111111( 6 1's) 111*1001001 is 111111111( 9 1's)
so we are getting no of 1's as 3 multiples.
so for 111....(3399) I.e. 4*(11111..1110+1)..
and 40*11111....3399 times+4
therefore the remainder will be 4.

4444.....444(3400 times) can be written as 4 *1111.....1111(3400 times)

37*3=111

111.....111(34000 times)= 111*(1001001001........) (we can generate upto 1111.....11111(3399 times)

so 4444....(3400)=4*(111*(1001001001...001)*1+1)

so the remainder should be 4