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?

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.

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