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Calculate how many positive natural numbers $≤ 200$ are divisible by one and only one between $12$ and $15$.

My attempt:

$B_1=[200/12]=16$

$B_2=[200/15]=13$

$B_1∩B_2=3$

$B_1+B_2+2|B1∩B2|=16+13+6=35$

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  • $\begingroup$ Do you know the inclusion-exclusion principle? Are you trying to use it? $\endgroup$ – ajotatxe Jan 21 '18 at 16:48
  • $\begingroup$ How do you arrive at $B_1\cap B_2=3$? $\endgroup$ – Hagen von Eitzen Jan 21 '18 at 16:50
  • $\begingroup$ @HagenvonEitzen Because $60,120,180$ $\endgroup$ – Jonsa Jan 21 '18 at 16:54
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Let $A$ be the number of positive integers $\leq 200$ divisible by $12$ and let $B$ be the number of positive integers $\leq 200$ divisible by $15$. Then we are after $A\Delta B=(A\setminus B)\cup (B\setminus A)$ (the symmetric difference). But $$ |A\Delta B|=|A|+|B|-2|A\cap B|\tag{1}. $$ Note that $|A|=[200/12]$ and $B=[200/15]$ where $[\cdot]$ is the floor. An integer is divisible by $12$ and $15$ iff it is divisible by the least common multiple, namely, $60$. Hence $|A\cap B|=[200/60]$.

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  • $\begingroup$ The answer is $23$? $\endgroup$ – Jonsa Jan 21 '18 at 17:01
  • $\begingroup$ Yes, $16+13-2(3)=23$. $\endgroup$ – Sri-Amirthan Theivendran Jan 21 '18 at 17:03

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