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. 1,23,968
• B. 1,33,068
• C. 1,33,268
• D. 1,87,332

Hint:

Use inclusion-exclusion for numbers divisible by multiple conditions. Always adjust for overlap using LCM.

Answer 1

The Correct Option is C

Total numbers from 200 to 600: \[ \text{Count} = 600 - 200 + 1 = 401 \] Sum of first \(n\) natural numbers: \[ \text{Sum} = \frac{n}{2}(a + l) \Rightarrow \frac{401}{2}(200 + 600) = \frac{401 \times 800}{2} = 401 \times 400 = 160400 \] Now remove numbers divisible by 8 or 12. Step 1: Numbers divisible by 8 from 200 to 600 First = 200, Last = 600, Common difference = 8 \[ n = \frac{600 - 200}{8} + 1 = 51
\text{Sum}_{8} = \frac{51}{2}(200 + 600) = \frac{51 \times 800}{2} = 20400 \] Step 2: Numbers divisible by 12 from 204 to 600 First = 204, Last = 600, Common difference = 12 \[ n = \frac{600 - 204}{12} + 1 = 34
\text{Sum}_{12} = \frac{34}{2}(204 + 600) = 17 \times 804 = 13668 \] Step 3: Numbers divisible by LCM(8,12) = 24 First = 216, Last = 600 \[ n = \frac{600 - 216}{24} + 1 = 17
\text{Sum}_{24} = \frac{17}{2}(216 + 600) = \frac{17 \times 816}{2} = 6936 \] Now apply inclusion-exclusion: \[ \text{Sum}_{8 \cup 12} = \text{Sum}_{8} + \text{Sum}_{12} - \text{Sum}_{24}
= 20400 + 13668 - 6936 = 27132 \] Final required sum: \[ 160400 - 27132 = \boxed{133268} \] \fbox{Final Answer: (C) 1,33,268}