The integers from $1$ to $100,$ which are divisible by $2,$ are $2,4,6 \ldots \ldots 100$

This forms an $A.P.$ with both the first term and common difference equal to $2.$

$\Rightarrow 100=2+(n-1) 2$

$\Rightarrow n=50$

$\therefore 2+4+6+\ldots \ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]$

$=\frac{50}{2}[4+98]$

$=(25)(102)$

$=2550$

The integers from $1$ to $100 ,$ which are divisible by $5,10 \ldots . .100$

This forms an $A.P.$ with both the first term and common difference equal to $5 .$

$\therefore 100=5+(n-1) 5$

$\Rightarrow 5 n=100$

$\Rightarrow n=20$

$\therefore 5+10+\ldots .+100=\frac{20}{2}[2(5)+(20-1) 5]$

$=10[10+(19) 5]$

$=10[10+95]=10 \times 105$

$=1050$

The integers, which are divisible by both $2$ and $5,$ are $10,20, \ldots \ldots 100$

This also forms an $A.P.$ with both the first term and common difference equal to $10.$

$\therefore 100=10+(n-1)(10)$

$\Rightarrow 100=10 n$

$\Rightarrow n=10$

$\therefore 10+20+\ldots .+100=\frac{10}{2}[2(10)+(10-1)(10)]$

$=5[20+90]=5(110)=550$

$\therefore$ Required sum $=2550+1050-550=3050$

Thus, the sum of the integers from $1$ to $100,$ which are divisible by $2$ or $5,$ is $3050$