Let is neither a multiple of 3 nor a multiple of 4. Then the number of elements in A is

Question

Let $A = {n \in [100, 700] \cap N : n$ is neither a multiple of 3 nor a multiple of 4}. Then the number of elements in A is [2024]

Smart Achievers · Accepted Answer

(2)

$Multiples of 3 are 102, 105, \dots, 699$

$a_{n} = 699 = 102 + (n - 1) \cdot 3$

$\Rightarrow 597 = 3 (n - 1) \Rightarrow n = 200$

$Now, multiples of 4 are 100, 104, \dots, 700$

$a_{m} = 700 = 100 + (m - 1) \cdot 4$

$\Rightarrow m = 151$

$Multiples of 3 and 4 are 108, 120, \dots, 696$

$a_{p} = 696 = 108 + (p - 1) \cdot 12$

$\Rightarrow p = 50$

$n (3 \cup 4) = n (3) + n (4) - n (3 \cap 4)$

$= n + m - p = 200 + 151 - 50 = 301$

$∴ The number of elements in A = 601 - 301 = 300$

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