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

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How many [[positive integer]]s have exactly three [[proper divisor]]s (positive integral [[divisor]]s excluding itself), each of which is less than 50?
 
How many [[positive integer]]s have exactly three [[proper divisor]]s (positive integral [[divisor]]s excluding itself), each of which is less than 50?
  
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== Solution ==
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== Solution (Basic Casework and Combinations) ==
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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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Suppose <math>n</math> is such an [[integer]]. Because <math>n</math> has <math>3</math> proper divisors, it must have <math>4</math> divisors,, 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 <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 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> ways to choose a pair of primes from the list and thus <math>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 <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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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.  
  
 
Thus there are <math>105+4=\boxed{109}</math> integers that meet the given conditions.
 
Thus there are <math>105+4=\boxed{109}</math> integers that meet the given conditions.
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~lpieleanu (Minor editing)
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~ rollover2020 (extremely minor editing)
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 08:37, 23 January 2024

Problem

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

Solution (Basic Casework and Combinations)

Suppose $n$ is such an integer. Because $n$ has $3$ proper divisors, it must have $4$ divisors,, 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 ${15 \choose 2} =105$ ways to choose a pair of primes from the list and thus $105$ 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.


~lpieleanu (Minor editing) ~ rollover2020 (extremely minor editing)

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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