Question

Find the sum of all three-digit numbers which give a remainder of $4$ when they are divided by $5$.

Answer 1

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 $999$

$\Rightarrow n=\frac{(999-99)}{5}=\frac{900}{5}=180$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{180}{2}[2\times 104+(180-1\left)5\right]$

$=90[208+179\times 5]$

$=90[208+895]$

$=90\times 1103$

$=99270$

Thus, the sum of all $3$-digit numbers $104$ to $999$ which leave a remainder of $4$ when divided by $5$ is $99270$.