Difference between revisions of "2005 AIME I Problems/Problem 3"
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== Problem == | == Problem == | ||
− | How many [[positive integer]]s have exactly three [[proper divisor]]s, 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 == | ||
− | <math>n=p\cdot q</math> or <math>n=p^3</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>. |
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
== See also == | == See also == | ||
− | * [[ | + | * [[Divisor_function#Demonstration | Counting divisors of positive integers]] |
− | + | {{AIME box|year=2005|n=I|num-b=2|num-a=4}} | |
− | |||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:26, 7 November 2019
Problem
How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50?
Solution
Suppose is such an integer. Because has proper divisors, it must have divisors,, so must be in the form or for distinct prime numbers and .
In the first case, the three proper divisors of are , and . Thus, we need to pick two prime numbers less than . There are fifteen of these ( and ) so there are numbers of the first type.
In the second case, the three proper divisors of are 1, and . Thus we need to pick a prime number whose square is less than . There are four of these ( and ) and so four numbers of the second type.
Thus there are 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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