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

(Solution)
(Solution (Basic Casework and Combinations))
 
(2 intermediate revisions by 2 users not shown)
Line 2: Line 2:
 
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?
  
== Solution ==
+
== Solution (Basic Casework and Combinations) ==
 
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>.  
 
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>.  
  
Line 13: Line 13:
  
 
~lpieleanu (Minor editing)
 
~lpieleanu (Minor editing)
 +
~ rollover2020 (extremely minor editing)
  
 
== See also ==
 
== See also ==

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png