Question

Find the sum of all $3$ digit natural numbers, which are divisible by $8$

Answer 1

The three digit natural numbers divisible by $8$ are $104,112,120,....,992$.
Let $S_n$ denote their sum. That is, $S_n=104+112+120+128+,....+992$.
Now, the sequence $104,112,120,...,992$ forms an A.P.
Here, $a=104,d=8$ and $l=992$.
$\therefore n=\frac{l-a}{d}+1=\frac{992-104}{8}+1$
$=\frac{888}{8}+1=112$.
Thus, $S_{112}=\frac{n}{2}[a+l]=\frac{112}{2}[104+992]=56\left(1096\right)=61376$.
Hence, the sum of all three digits numbers, which are divisible by $8$ is equal to $61376$.