Question

Find the sum of all three digit number which when divided by 5 leaves a remainder 3

Answer 1

the sum of three digit least no whose divisible by 5 remaining 3=103
the sum of three digit highest no whose divisible 5 remaining 3=998
prove it
[5)103 (2
=_10___
remainder= 3
5)998(199
=_ 5___
=49
__45___
=48
__45___
remainder=3

Answer 2

Hello...

A.T.Q

smallest 3 digit number =103

largest 3 digit number =998

3 digit number after 103=108

ap is formed 103,108........998

first term=103

common difference=5

Total no. of three digit numbers which leave the remainder 3 when divided by 5=n

998=103+(n-1)5

180=n

sum of ap=180/2 *(103+998)=90*1101
=99090