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Question

The sum of all the natural numbers from 200 to 600(both inclusive) which are neither divisible by 8 nor by 12 is?

A
123968
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B
133068
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C
133268
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D
187332
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Solution

The correct option is C 133268
D(8)= numbers divisible by 8=200,208,216,224,232,240,..,592,600.
Total D(8) numbers =51
Sum of D(8) numbers =[512×(200+600)]=(51×400)=20400
D(12)= numbers divisible by 12=204,216,228,240,252,264,..,588,600.
Total D(12) numbers =34
Sum of D(12) numbers =[342×(204+600)]=(17×804)=13668

Now, D(812)= numbers divisible by both 8 and 12=216,240,264,...,576,600.

Total D(812) numbers =17

Sum of D(8 intersect 12) numbers =[172×(216+600)]=(17×408)=6936

So, D(812)= numbers divisible by either 8 or 12

=D(8)+D(12)D(812).

So, sum of D(812) numbers =20400+136686936=27132.

Now, sum of all natural numbers ranging from 200 to 600=[4012×(200+600)]=(401×400)=160400.

Sum of all natural numbers from 200 to 600 which are neither divisible by 8nor by 12=(16040027132)
=133268.

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