Question 1148455
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            The solution to part (a) given by @boreal is WRONG.


            I came to bring the correct solution.



<pre>
The number of integers from 1 to 201 (inclusive) that are multiple of 6 is  33 
 (notice that {{{201/6}}} = 33.5, approximately).


The number of integers from 1 to 201 (inclusive) that are multiple of 11 is  18 
 (notice that {{{201/11}}} = 18.3, approximately).


The number of integers from 1 to 201 (inclusive) that are multiple of 6 and of 11 simultaneously is  3:  66, 132 and 198).


From it, the number of integers from 1 to 201 (inclusive) that are multiple of 6 or  11  is  33 + 18 - 3 = 48.    <U>ANSWER</U>



    We must first add  33 and 18  and then subtract the "3" from this sum (!)

    It is because when we add 33 and 18, we count TWICE the integers in the intersection of these subsets (!)
</pre>

Solved.


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See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Counting elements in sub-sets of a given finite set</A>

in this site, where you can find other similar solved problems.



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