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The number of integers between $1$ and $500$ (both inclusive) that are divisible by $3$ or $5$ or $7$ is ____________ .
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Here, we can apply the property of set. Let $D_n$ denote divisibility by $n,$ $D_{n_1,n_2}$ denote divisibility by both $n_1$ and $n_2$ and so on.

$N(D_3 \cup D_5 \cup D_7)=N(D_3)+N(D_5)+N(D_7) -N(D_{3,5})-N(D_{ 5,7})-N(D_{3,7})+N(D_{3,5,7})$
$\quad \quad =166+100+71-33-14-23+4$
$\quad \quad =271$
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71 votes

EASY way to solve using GATE interface CALCULATOR;

$P(3\cup5\cup7=P(3)+P(5)+P(7)-P(3\times5)-P(5\times7)-P(3\times7)+-P(3\times7\times5)$

  • $P(3)=\frac{500}{3}=\lfloor166.66\rfloor=166$
  • $P(5)=\frac{500}{5}=100$
  • $P(7)=\frac{500}{7}=\lfloor71.42\rfloor=71$
  • $P(3\times5)=\frac{500}{15}=\lfloor33.33\rfloor=33$
  • $P(7\times5)=\frac{500}{35}=\lfloor14.28\rfloor=14$
  • $P(3\times7)=\frac{500}{21}=\lfloor23.8\rfloor=23$
  • $P(3\times5\times 7)=\frac{500}{105}=\lfloor4.76\rfloor=4$

$\therefore P(3\cup5\cup7)= 166+100+71-33-14-23+4 = 271$

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| AUBUC |= | A | + | B |+| C | - | A ⋂ B | - | A ⋂ C | -| B ⋂ C | + | A ⋂ B ⋂ C |
166 + 100 + 71 - 33 - 14 -23 + 4  = 271

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