Question

find the sum of three digit numbers which leave remainder 2 when divided by 5

Answer 1

first number a = 102

last number = 997

997 = 102 + (n-1)5

997 - 102 = (n-1)5

895/5 = n -1

n = 179 - 1 = 178

Sum = 178 (102+997)/2  = 178 x 1099/2 = 97811