JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
Let $A=\{n\in[100,700]\cap\mathbb{N}:n$ is neither a multiple of $3$ nor a multiple of $4\}$. Then the number of elements in $A$ is
Permutation and Combination
Solution
Multiples of 3 in $[100,700]$: $102,\dots,699$: $(699-102)/3+1=200$. Multiples of 4: $100,\dots,700$: $151$. Multiples of 12: $108,\dots,696$: $50$. Multiple of 3 or 4: $200+151-50=301$. Total numbers $=601$. Neither: $601-301=300$. Option C.