Question

the sum of the integers from 1 to 100 which are not divisible by 3 or 5is .​

Answer 1

Answer:

Solution:

Sum of integers which are not divisible by 3 or 5 = sum of first 100 natural numbers – sum of multiples of 3 – sum of multiples of 5 + sum of multiples of both 3 and 5

Sum of first 100 natural numbers = n(n+1)/2

= 100(101)/2

= 5050

Sum of multiples of 3 = 3 + 6 + 9+..99

Here a = 3 and d = 3

an = a+(n-1)d = 99

3+(n-1)3 = 99

3n = 99

n = 99/3= 33

Sn = (n/2)(2a+(n-1)d)

= (33/2)(2×3+(33-1)3)

= (33/2)(6+32×3)

= (33/2)×102

= 1683

Sum of multiples of 5 = 5 + 10 + 15+…100

Here a = 5 and d = 5

an = a+(n-1)d = 100

5+(n-1)5 = 100

5n = 100

n = 100/5= 20

Sn = (n/2)(2a+(n-1)d)

= (20/2)(2×5+19×5)

= 10×105

= 1050

Sum of multiples of both 3 and 5 = 15 + 30 + 45 +…90

Here n = 6

a = 15, d = 15

Sum = (6/2)(15+90)

= 315

Required sum = 5050 – 1683- 1050 +315

= 2632

Hence option (4) is the answer.

Step-by-step explanation:

$i \: hope \: its \: helpful \: to \: you$

Answer 2

Answer:

Here is the answer

Step-by-step explanation:

Divided by 3

6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54, 57,60,63,66,69,72,75,78,81,84,87,90,93,96,99

By 5 are

5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100

Hope it helps you❤