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. $123968$
• B. $133068$
• C. $133268$
• D. $187332$

Answer 1

The correct option is C $133268$

$D\left(8\right)=$ numbers divisible by $8=200,208,216,224,232,240,..,592,600.$

Total $D\left(8\right)$ numbers $=51$

Sum of $D\left(8\right)$ numbers $=[\frac{51}{2}\times (200+600\left)\right]=(51\times 400)=20400$

$D\left(12\right)=$ numbers divisible by $12=204,216,228,240,252,264,..,588,600.$

Total $D\left(12\right)$ numbers $=34$

Sum of $D\left(12\right)$ numbers $=[\frac{34}{2}\times (204+600\left)\right]=(17\times 804)=13668$

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

Total $D(8\cap 12)$ numbers $=17$

Sum of $D(8$ intersect $12)$ numbers $=[\frac{17}{2}\times (216+600\left)\right]=(17\times 408)=6936$

So, $D(8\cup 12)=$ numbers divisible by either $8$ or $12$

$=D\left(8\right)+D\left(12\right)-D(8\cap 12).$

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

Now, sum of all natural numbers ranging from $200$ to $600=[\frac{401}{2}\times (200+600\left)\right]=(401\times 400)=160400.$

Sum of all natural numbers from $200$ to $600$ which are neither divisible by $8$nor by $12=(160400-27132)$

$=133268.$