<p>In the figure above, circular region A represents all integers from 10 to 100, inclusive; circular region B represents all integers that are multiples of 3; and circular region C represents all squares of integers. How many numbers are represented by the shaded region?</p>
<p>(A) 24
(B) 25
(C) 26
(D) 27
(E) 28</p>
<p>This is from the PSAT we all took this year.</p>
<p>It completely tricked me.</p>
<p>Well, I actually wrote out all the multiples of three, then crossed out the ones that were squares - 36 and 81. There’s probably a faster, more computational way to do it, but eh, I like writing things out.</p>
<p>I was left with 28 multiples, or E.</p>
<p>How the heck do you know NOT to include squares?</p>
<p>That’s what got me. I assumed it was just multiples of 3 within 10 and 100 :|</p>
<p>btw you’re correct</p>
<p>Sorry but I got 12 to 99 = 3 x 4 to 3 x 33.
There are 29 multiples in between those two.
29 - 2 values (81 and 36) = 27?</p>
<p>EDIT: Nevermind, I realized that the set should be inclusive so there are actually 30 multiples.</p>
<p>There are 33 multiples of 3 from 1-100. Cross out the single digits (3, 6, 9) and you’re left with 30 multiples. Now, multiples of 3, squared, namely 6 squared and 9 squared (36, 81), can be crossed out, and so you’re finally left with 28 integers.</p>
<p>The shaded region consists of all integers in A and B but NOT in C. Thus you need to exclude perfect squares.</p>
<p>@teteate you’re wrong…</p>
<p>Gah I still don’t get this</p>
<p>The Venn diagram is understood as such:</p>
<p>A is all integers from 10 to 100, inclusive
B is all multiples of 3
C is the perfect squares.</p>
<p>The shaded area is contained by circles A and B, but does not contain C. Therefore the numbers you’re looking for are: integers from 10 to 100 (inclusive) and are also multiples of three, but are NOT perfect squares.</p>
<p>The numbers that apply to both A and B are the multiples of 3 from 12 to 99. There are 30 of these. Because C is not included in the shaded region, you have to take out the squares 36 and 81, therefore, giving you the answer 28.</p>
<p>For some reason I can’t load the question up. But taking thesmiter’s post #8, the number of integers from 10 - 100, multiples of 3, but not perfect squares is easy to count.</p>
<p>First note that the multiples of 3 are 12, 15, 18, …, 99. There are 30 such numbers. The perfect squares are 6^2 and 9^2 (36 and 81) so we remove these from the set. Therefore C has 28 elements.</p>
<p>*Easy way to count number of elements in {12, 15, …, 99}: Divide each element by 3, subtract 3 from each remaining element, leaving {1, 2, …, 30}. You don’t have to worry about the inclusive set.</p>
<p>^Wouldn’t it be easier to just divide the highest multiple by 3? 99/3 = 33. That counts {3,6,9}, so subtract 3 to get rid of those, bam you have 30 elements. Minus 36 and 81 because they’re perfect squares and bang, 28, there’s your answer.</p>
<p>@AKShockwave That works as well…but if you have a sequence like {112, 125, 138, …, 775}, then finding the cardinality is a little trickier using that method.</p>
<p>In that case it’s just (775 - 112)/13, add 1 for inclusiveness.</p>
<p>Do 15/3. You get 5, and notice there are 5 multiples in between, namely 3, 6, 9, 12, and 15 itself. So if you did 99/3 (99 because it’s the highest multiple of 3 before 100), you get 33, meaning there are thirty-three 3s before 100. Because the question requires you to look for multiple of 3s within 10 and 100, you must immediately take away the 3, 6, and 9 from the total amount that represents the shaded region. You get 30, but you must remember that some square of integers have results that fall both in between 10 and 100 and are multiples of 3. By testing, you realize that those numbers are 36 and 81, for 6^2 and 9^2, respectively. So, we have 5 numbers to take away (3, 6, 9, 36, 81) from our total of 33, giving us 28, or E.</p>
