Given integers are: 1,2,3,......................,100

Number that are divisible by2 are: 2,4,6,............100

It forms an AP. Let n is the number of term in this AP

      l = a + (n-1)*d

=> 100 = 2 + 2(n-1)

=> 100-2 = 2(n-1)

=> 98 = 2(n-1)

=> n-1 = 98/2

=> n-1 = 49

=> n = 49+1

=> n = 50

Now Sum = (n/2) * (a+l)

               = (50/2)*(2+100)

               = 25*102

               = 2550 

Number that are divisible by 5 are: 5, 10,15,................,100

It forms an AP. Let n is the number of term in this AP

      l = a + (n-1)*d

=> 100 = 5 +5(n-1)

=> 100-5 = 5(n-1)

=> 95= 5(n-1)

=> n-1 = 95/5

=> n-1 = 19

=> n = 19+1

=> n = 20

Now Sum = (n/2) * (a+l)

               = (20/2)*(5+100)

               = 10*105

               = 1050 

Number that are divisible by both 2 and 5 are: 10, 20,35,................,100

It forms an AP. Let n is the number of term in this AP

      l = a + (n-1)*d

=> 100 = 10 +10(n-1)

=> 100-10 = 10(n-1)

=> 90= 10(n-1)

=> n-1 = 90/10

=> n-1 = 9

=> n = 9+1

=> n = 10

Now Sum = (n/2) * (a+l)

               = (10/2)*(10+100)

               = (10*110)/2

               = 10*55

              = 550

So sum of integers that are divisible by 2 or 5 is = 2550 + 1050 - 550 = 3600 - 605 = 3050