Difference between revisions of "2007 AIME I Problems/Problem 1"
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The [[prime factorization]] of <math>24</math> is <math>2^3\cdot3</math>. Thus, each square must have at least <math>3</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math> and its square root must have <math>2</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math>. | The [[prime factorization]] of <math>24</math> is <math>2^3\cdot3</math>. Thus, each square must have at least <math>3</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math> and its square root must have <math>2</math> factors of <math>2</math> and <math>1</math> factor of <math>3</math>. | ||
This means that each square is in the form <math>(12c)^2</math>, where <math>12 c</math> is a positive integer less than <math>\sqrt{10^6}</math>. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. | This means that each square is in the form <math>(12c)^2</math>, where <math>12 c</math> is a positive integer less than <math>\sqrt{10^6}</math>. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. | ||
| + | == Solution 2 == | ||
| + | After some work you will eventually find its just <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions. | ||
== See also == | == See also == | ||
Revision as of 21:51, 23 November 2020
Contents
Problem
How many positive perfect squares less than 
are multiples of 
?
Solution
The prime factorization of is 
. Thus, each square must have at least 
factors of 
and 
factor of and its square root must have factors of and factor of .
This means that each square is in the form 
, where 
is a positive integer less than 
. There are 
solutions.
Solution 2
After some work you will eventually find its just solutions.
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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. 
