Question

The sum of all natural number from 200 to 600 (both inclusive ) which are neither divisible by 8 nor by 12 .

Answer 1

Dear Student,

D(8) = numbers divisible by 8 = 200, 208, 216, 224, 232, 240,………….., 592, 600.

Let a =200
d=8
t_n = a+(n-1)d
=>600=200+(n-1)8
=>(600-200)/8 =n-1
=>50=n-1
=>n=51

Total D(8) numbers = 51

Sum of D(8) numbers = [(51/2)*(200 + 600)] = (51*400) = 20400

D(12) = numbers divisible by 12 = 204, 216, 228, 240, 252, 264,………….., 588, 600.

Let a=200
d=12
t_n​​​​​​​ =600
t_n=a+(n-1)d
=> n ={(t_n -a)/d} +1
=> n={(600-200)/12} +1 =34

Total D(12) numbers = 34

Sum of D(12) numbers = [(34/2)*(204 + 600)] = (17*804) = 13668

Now, D(8 intersect 12) = numbers divisible by both 8 and 12 = 216, 240, 264,…., 576, 600.

Total D(8 intersect 12) numbers = 17

Sum of D(8 intersect 12) numbers = [(17/2)*(216 + 600)] = (17*408) = 6936

So, D(8 union 12) = numbers divisible by either 8 or 12

= D(8) + D(12) - D(8 intersection 12).

So, sum of D(8 union 12) numbers = 20400 + 13668 - 6936 = 27132.

Now, sum of all natural numbers ranging from 200 to 600 = [(401/2)*(200 + 600)] = (401*400) = 160400.

So, sum of all natural numbers from 200 to 600 (both inclusive) which are neither divisible by 8 nor by 12 = (160400 - 27132) = 133268.



Regards