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Difference between revisions of "2005 AIME I Problems/Problem 3"

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== Solution ==
 
== Solution ==
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Suppose <math>n</math> is such an [[integer]]. Then <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math>.  
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Suppose <math>n</math> is such an [[integer]]. Then one of its factors is <math>1</math>, so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math>.  
  
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In the first case, the three proper divisors of <math>n</math> are 1, <math>p</math> and <math>q</math>.  Thus, we need to pick two prime numbers less than 50. There are fifteen of these (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47) so there are <math> {15 \choose 2} =105</math> numbers of the first type.
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In the first case, the three proper divisors of <math>n</math> are <math>1</math>, <math>p</math> and <math>q</math>.  Thus, we need to pick two prime numbers less than <math>50</math>. There are fifteen of these (<math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43</math> and <math>47</math>) so there are <math> {15 \choose 2} =105</math> numbers of the first type.
  
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In the second case, the three proper divisors of <math>n</math> are 1, <math>p</math> and <math>p^2</math>.  Thus we need to pick a prime number whose square is less than 50.  There are four of these (2, 3, 5 and 7) and so four numbers of the second type.  
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In the second case, the three proper divisors of <math>n</math> are 1, <math>p</math> and <math>p^2</math>.  Thus we need to pick a prime number whose square is less than <math>50</math>.  There are four of these (<math>2, 3, 5</math> and <math>7</math>) and so four numbers of the second type.  
  
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Thus there are <math>105+4=109</math> integers that meet the given conditions.
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Thus there are <math>105+4=\boxed{109}</math> integers that meet the given conditions.
  
 
== See also ==
 
== See also ==

Revision as of 17:35, 26 April 2008

Problem

How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?

Solution

Suppose $n$ is such an integer. Then one of its factors is $1$, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$.

In the first case, the three proper divisors of $n$ are $1$, $p$ and $q$. Thus, we need to pick two prime numbers less than $50$. There are fifteen of these ($2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ and $47$) so there are ${15 \choose 2} =105$ numbers of the first type.

In the second case, the three proper divisors of $n$ are 1, $p$ and $p^2$. Thus we need to pick a prime number whose square is less than $50$. There are four of these ($2, 3, 5$ and $7$) and so four numbers of the second type.

Thus there are $105+4=\boxed{109}$ integers that meet the given conditions.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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