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

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== Problem ==
 
== Problem ==
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Call a number ''prime-looking'' if it is [[composite]] but not divisible by <math>2, 3,</math> or 5. The three smallest prime-looking numbers are <math>49, 77</math>, and <math>91</math>. There are <math>168</math> prime numbers less than <math>1000</math>. How many prime-looking numbers are there less than <math>1000</math>?
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Call a number ''prime-looking'' if it is [[composite]] but not divisible by <math>2, 3,</math> or <math>5.</math> The three smallest prime-looking numbers are <math>49, 77</math>, and <math>91</math>. There are <math>168</math> prime numbers less than <math>1000</math>. How many prime-looking numbers are there less than <math>1000</math>?
  
 
<math>
 
<math>

Revision as of 23:13, 11 October 2014

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 $1000$?

$(\mathrm {A}) \ 100 \qquad (\mathrm {B}) \ 102 \qquad (\mathrm {C})\ 104 \qquad (\mathrm {D}) \ 106 \qquad (\mathrm {E})\ 108$

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 complementary counting. We can split the numbers from $1$ to $1000$ into several groups: $\{1\},$ $\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},$ $\{\mathrm{numbers\ divisible\ by\ 3 = S_3}\},$ $\{\mathrm{numbers\ divisible\ by\ 5 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},$ $\{\mathrm{prime-looking}\}$. Hence, the number of prime-looking numbers is $1000 - 165 - 1 - |S_2 \cup S_3 \cup S_5|$ (note that $2,3,5$ are primes).

We can calculate $S_2 \cup S_3 \cup S_5$ using the Principle of Inclusion-Exclusion: (the values of $|S_2| \ldots$ and their intersections can be found quite easily)

$|S_2 \cup S_3 \cup S_5| = |S_2| + |S_3| + |S_5| - |S_2 \cap S_3| - |S_3 \cap S_5| - |S_2 \cap S_5| + |S_2 \cap S_3 \cap S_5|$
$= 500 + 333 + 200 - 166 - 66 - 100 + 33 = 734$

Substituting, we find that our answer is $1000 - 165 - 1 - 734 = 100 \Longrightarrow \mathrm{(A)}$.

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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