It is an AP whose first term is \\\\(104\\\\), common difference is \\\\(5\\\\) and the last term is \\\\(999\\\\)\\\\(a = 104, d = 5\\\\)Last term is \\\\(999\\\\)\\\\(\\Rightarrow n=\\dfrac{(999-99)}{5} = \\dfrac{900}{5} =180\\\\)\\\\(S_n = \\dfrac{n}{2}[2a +(n-1)d]\\\\)\\\\(= \\dfrac{180}{2}[2\ imes 104 + (180-1)5]\\\\)\\\\(= 90[208 + 179\ imes 5]\\\\)\\\\(=90[208+895]\\\\)\\\\(= 90\ imes 1103\\\\)\\\\(= 99270\\\\)Thus, the sum of all \\\\(3\\\\)-digit numbers \\\\(104\\\\) to \\\\(999\\\\) which leave a remainder of \\\\(4\\\\) when divided by \\\\(5\\\\) is \\\\(99270\\\\).