Question

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

Answer 1

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

Answer 2

Solution -:

let us is 1 + 2 + 3 + ...+100

$= \frac{100}{2} (1 + 100)$

[ n =100 a =1 ]

50× 101 =5050

let's S1 = 3+6+9....+99

$= 3(1 + 2 + 3... + 33) \\ 3 \times \frac{33}{2} (1 + 33) \\ = 99 \times 17 = 1683$

$let \: s2 \: = 5 + 10 + 15 + ... + 100 \\ 5(1 + 2 + 3 + .. + 20 \\ 5 \times \frac{20}{2} 1 + 20$

= 50×21 = 1050

$lets \: s3 \: = 15 + 30 + 45 + ... + 90$

$15(1 + 2 + 3.. + 6) \\ 15 \times \frac{6}{2} (1 + 6) \\ 45 \times 7 = 315$

Required sum = S - S1 - S2+ S3

= 5050-1683-1050+315

= 2632

Answer is 2632