Question

find the sum of all three-digit numbers that give a remainder of 4 when they are divided by 5.

Answer 1

First 3 digit number with given condition is 104 and the last such number is 999
Number of terms, n =
$\frac{999 - 104}{5} + 1 = 180$
$sum = \frac{180}{2} \times (104 + 999) = 99270$

Answer 2

Note that the numbers which when divided by $\sf{5}$, leaves a remainder of $\sf{4}$ must be of the form $\sf{5n+4}$

So, the first three digit number of the form $\sf{5n+4}$ will be $\sf{104}$, starting from $\sf{n=20}$

Now, we create an $\sf{A.P.}$ of common difference $\sf{5}$ till the last three digit number of the form $\sf{5n+4}$ which is $\sf{999}$, ending with $\sf{n=199}$, which tells us we have created an $\sf{A.P.}$ of $\sf{199-20+1=180}$ terms.

$\sf{A.P. \to 104, \: 109, \: 114, \: ..., \: 999}$

Hence, The sum of these numbers is

$\sf{\dfrac{n}{2} (a + l)}$

$= \sf{ \dfrac{180}{2} (104 + 999) = 90(1103)}$

$\sf{ = \boxed{ \sf{99270}}}$