Question 1126619
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Let's do it in more simple way.


<pre>
From the elementary set theory,  {(A U B) n C}}} = {(A n C) U (B n C)}.


    {(A n C)} is the set of all prime numbers between 1 and 100 intersected with the set of all integer numbers multiple 3 in this domain.

    So, {{A n C)} consists of one single element {3}.


    {(B n C)} is the set of all even numbers between 1 and 100 intersected with the set of all integer numbers multiple of 3 in this domain.


    So, {(B n C)} is the set of all multiple of 6 between 1 and 100 inclusively.



Therefore,  

    {(A U B) n C}}} = {(A n C) U (B n C)} = <U>the set of all multiple of 6 between 1 and 100 inclusively</U>  PLUS  <U>the number 3</U>.


    The amount of all integers multiple of 6 between 1 and 100 inclusively is  16   (because  {{{100/6}}} = 16.66. . . ).


    Plus that single "3" gives you the answer of 17.
</pre>

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The tutor &nbsp;@MathLover1  &nbsp;missed the number &nbsp;"54" &nbsp;in her final list.