Difference between revisions of "2005 AMC 12A Problems/Problem 18"

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== Problem ==
 
== Problem ==
Call a number ''prime-looking'' if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?
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Two is <math>10 \%</math> of <math>x</math> and <math>20 \%</math> of <math>y</math>. What is <math>x - y</math>?
  
 
<math>
 
<math>
(\mathrm {A}) \ 100 \qquad (\mathrm {B}) \ 102 \qquad (\mathrm {C})\ 104 \qquad (\mathrm {D}) \ 106 \qquad (\mathrm {E})\ 108
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(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 20
 
</math>
 
</math>
  
 
== Solution ==
 
== Solution ==
The given states that there are 168 prime numbers less than 1000, which is a fact we must somehow utilize. Since there seems to be no easy way to directly calculate the number of "prime-looking" numbers, we can apply the [[not principle]]. We can split the numbers from 1 to 1000 into several groups: <math>\{1\},</math> <math>\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 2 = S_3}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 2 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},</math> <math> \{\mathrm{prime-looking}\}</math>. Hence, the number of prime-looking number is <math>1000 - 165 - 1 - |S_2 \cup S_3 \cup S_5|</math>.
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<math>2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10  \mathrm{(D)}</math>.
 
 
We can calculate <math>S_2 \cup S_3 \cup S_5</math> using the [[Principle of Inclusion-Exclusion]]: (the values of <math>|S_2| \ldots</math> and their intersections can be found quite easily)
 
 
 
<div style="text-align:center;"><math>|S_2 \cup S_3 \cup S_5| = |S_2| + |S_3| + |S_5| - |S_2 \cap S_3| - |S_3 \cap S_5| - |S_2 \cap S_5| + |S_2 \cap S_3 \cap S_5|</math><br /><math>= 500 + 333 + 200 - 166 - 66 - 100 + 33 = 734</math></div>
 
 
 
Substituting, we find that our answer is <math>1000 - 165 - 1 - 734 = 100 \Longrightarrow \mathrm{(A)}</math>.
 
  
 
== See also ==
 
== See also ==
{{AMC12 box|year=2005|num-b=17|num-a=19|ab=A}}
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{{AMC12 box|year=2005|before=First question|num-a=2|ab=A}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 10:02, 23 September 2007

Problem

Two is $10 \%$ of $x$ and $20 \%$ of $y$. What is $x - y$?

$(\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 20$

Solution

$2 = \frac {1}{10}x \Longrightarrow x = 20,\quad 2 = \frac{1}{5}y \Longrightarrow y = 10,\quad x-y = 20 - 10=10  \mathrm{(D)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First question
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Problem 2
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All AMC 12 Problems and Solutions
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