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? | |
<math> | <math> | ||
| − | (\mathrm {A}) \ | + | (\mathrm {A}) \ 100 \qquad (\mathrm {B}) \ 102 \qquad (\mathrm {C})\ 104 \qquad (\mathrm {D}) \ 106 \qquad (\mathrm {E})\ 108 |
</math> | </math> | ||
== Solution == | == Solution == | ||
| − | <math> | + | 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>. |
| + | |||
| + | 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| | + | {{AMC12 box|year=2005|num-b=17|num-a=19|ab=A}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 10:03, 23 September 2007
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?
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: 
. Hence, the number of prime-looking number is 
.
We can calculate 
using the Principle of Inclusion-Exclusion: (the values of 
and their intersections can be found quite easily)
Substituting, we find that our answer is 
.
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 | |
