Using Turbo Pascal, we have this following code:

uses crt;

var a,b,c,u,i:longint;

begin

clrscr;

i:=0;

u:=0;

for a:=1 to 9 do

for b:=0 to 9 do

for c:=0 to 9 do

begin

i:=100*a+10b+c;

if (i mod 5)=0 then u:=u+i;

end;

write(u);

readln;

end.

Gives $u=98550$

Java code:

int sum = 0; for(int i = 100; i < 1000; i += 5) { sum += i; } System.out.println(sum);

Python Code

        for n in range(99, 1000):
                 if (n % 10 == 0 or n % 5==0):
                        sum=sum+n

        print  sum

The 3-digit numbers which are divisible by 5 are $100,105,110,......,995$ . These numbers are in AP with first term $(a)=100$ and common difference $(d)=5$ . Let $a_{n}$ be last term. So,

$a_{n}=995 \implies a+(n-1)d=995 \implies 100+(n-1)5=995 \implies n-1=\frac{1095}{5}=179 \implies n=180$

Thus, there are 180 terms in the AP, so we use here the formula of sum of AP, $S=\frac{n}{2}(a+l)$ where $l=$ Last term

So, $S=\frac{180}{2}(100+995)=90\times 1095 = \boxed{98550}$