Question

find the sum of all 3 digit numbers which leave reaminder 3 when divided by 5.

Answer 1

hey..

The series of A.P is 103,108,113.....

a = 103
d = 108 - 103 = 5
an = 998
n = ?

an = a + (n-1)d
998 = 103 + (n-1)5
998-103 = 5n-5
895+5 = 5n
900/5 = n
n = 180

therefore 180 three digit numbers leave a reminder 3 when divided with 5

s180 = 180/2 (103+998)
= 90 (1101)
= 99090

therefore the sum of all the three digit numbers which leave a reminder 3 when divided by 5 is 99090