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The number of numbers lying between 1 and 200 which are divisible by either of two , three or five?

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A)numbers divisible by 2: $\frac{200}{2} = 100$

B)numbers divisible by 3: $\frac{200}{3} = 66$

C)numbers divisible by 5: $\frac{200}{5} = 40$

counting twice

AB)numbers divisible by 6: $\frac{200}{6} = 33$

AC)numbers divisible by 10: $\frac{200}{10} = 20$

BC)numbers divisible by 15: $\frac{200}{15} = 13$

counting 3 times

ABC)numbers divisible by 30: $\frac{200}{30} = 6$

Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 - 13 + 6 = 146

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  • $\begingroup$ why you don't include the numbers divisible by 7 ? $\endgroup$ – zeraoulia rafik Jul 6 '15 at 0:49
  • $\begingroup$ Nice example of inclusion-exclusion $\endgroup$ – Simon S Jul 6 '15 at 0:50
  • $\begingroup$ @zeraouliarafik The question didn't ask about multiples of $7$. $\endgroup$ – Thomas Andrews Jul 6 '15 at 2:18
  • $\begingroup$ @Conrado costa sir the answer in the book is 145 ? That is why i am getting confused how to approach ? $\endgroup$ – sushmitha Jul 7 '15 at 5:42

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