Difference between revisions of "2005 AMC 12A Problems/Problem 18"
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| − | The given states that there are <math>168</math> prime numbers less than <math>1000</math>, 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 <math>1</math> to <math>1000</math> into several groups: <math>\{1\},</math> <math>\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 3 = S_3}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 5 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},</math> <math> \{\mathrm{prime-looking}\}</math>. Hence, the number of prime-looking numbers is <math>1000 - | + | The given states that there are <math>168</math> prime numbers less than <math>1000</math>, 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 <math>1</math> to <math>1000</math> into several groups: <math>\{1\},</math> <math>\{\mathrm{numbers\ divisible\ by\ 2 = S_2}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 3 = S_3}\},</math> <math> \{\mathrm{numbers\ divisible\ by\ 5 = S_5}\}, \{\mathrm{primes\ not\ including\ 2,3,5}\},</math> <math> \{\mathrm{prime-looking}\}</math>. Hence, the number of prime-looking numbers is <math>1000 - (168-3) - 1 - |S_2 \cup S_3 \cup S_5|</math> (note that <math>2,3,5</math> are primes). |
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) | 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) | ||
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{{AMC12 box|year=2005|num-b=17|num-a=19|ab=A}} | {{AMC12 box|year=2005|num-b=17|num-a=19|ab=A}} | ||
| − | [[Category:Introductory | + | [[Category:Introductory Number Theory Problems]] |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:15, 25 November 2020
Problem
Call a number prime-looking if it is composite but not divisible by 
or 
The three smallest prime-looking numbers are 
, and 
. There are 
prime numbers less than 
. How many prime-looking numbers are there less than ?
Solution
The given states that there are prime numbers less than , 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 
to into several groups: 
. Hence, the number of prime-looking numbers is 
(note that 
are primes).
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
