SOLUTION: If A = {all prime numbers between 1 and 100}, B = {2,4,6,...,98,100}, and C = {3,6,9,...,96,99}, how many elements does (A union B) intersection C have?

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Question 1126619: If A = {all prime numbers between 1 and 100}, B = {2,4,6,...,98,100}, and C = {3,6,9,...,96,99}, how many elements does (A union B) intersection C have?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

all prime numbers between 1 and 100
A={2 ,3 , 5 ,7 , 11,13 , 17 , 19 , 23 , 29, 31 , 37 , 41, 43 ,47 ,53 ,59 , 61 ,67,71 }... (25 primes)
B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100}

( A U B ) The union of two sets is the set of elements contained in both sets.

( A U B )={2, 3,4,5, 6,7, 8, 10,11, 12,13, 14, 16,17, 18,19, 20, 22, 23,24, 26, 28,29, 30,31, 32, 34, 36, 37,38, 40, 41,42,43, 44, 46,47, 48, 50, 52, 53,54, 56, 58, 59,60,61, 62, 64, 66, 68, 70, 71,72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100}


C = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99}

( A U B ) ∩ C
look for the elements you can find in both (A U B) and C: the intersection of two sets is the set of elements that are common to both sets

I will highlight elements that are common

A U B={2, ,4,5, ,7, 8, 10,11, ,13, 14, 16,17, ,19, 20, 22, 23,, 26, 28,29, ,31, 32, 34, , 37,38, 40, 41,,43, 44, 46,47, , 50, 52, 53,54, 56, 58, 59,,61, 62, 64, , 68, 70, 71,, 74, 76, , 80, 82, , 86, 88, , 92, 94, , 98, 100}

C= { , , 9, , 15, , 21,, 27, , 33,, 39, , 45, , 51, 54, 57, , 63, , 69, , 75, , 81, , 87, , 93,, 99}
so,
( A U B ) ∩ C={ , ,, ,,, ,,,,,, , , , }

( A U B ) ∩ C have elements


Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.
Let's do it in more simple way. 

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)} = the set of all multiple of 6 between 1 and 100 inclusively  PLUS  the number 3.


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


    Plus that single "3" gives you the answer of 17.

-------------

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


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