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if A={1,2,...,600} includes all natural numbers between 1 to 600. I want to find number of elements of A that not divisible by 3 or 5 or 7?

any hint or idea?

Asaf Karagila
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Davied Zuhraph
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  • The inclusion-exclusion principle gives that they are roughly $$600\cdot \left(1-\frac{1}{3}\right)\cdot \left(1-\frac{1}{5}\right) \cdot \left(1-\frac{1}{7}\right).$$ – Jack D'Aurizio Sep 26 '14 at 11:16

1 Answers1

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Let $A$ be the set of integers in $I=[1,600]$ that are divisible by three, $B$ the set of integers in $I$ that are divisible by five and $C$ the set of integers that are divisible by seven. The inclusion-exclusion principle gives that the answer is: $$600-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$$ or: $$ 600-200-120-85+40+28+17-5 = \color{red}{275.}$$

Jack D'Aurizio
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    If you notice, the cardinality of $A\cap B$ is just $\left\lfloor\frac{600}{3\cdot 5}\right\rfloor$ and so on, so what is written in the first line is roughly $$600 \cdot \left(1-\frac{1}{3}\right) \cdot \left(1-\frac{1}{5}\right) \cdot \left(1-\frac{1}{7}\right)$$ as stated in the above comment. – Jack D'Aurizio Sep 26 '14 at 11:37