6 years, 11 months ago
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?
6 years, 11 months ago
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
6 years, 11 months ago
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.
6 years, 11 months ago
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