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 3 factors of <math>2</math> and 1 factor of <math>3</math>.
 
The [[prime factorization]] of <math>24</math> is <math>2^3\cdot3</math>. Thus, each square must have 3 factors of <math>2</math> and 1 factor of <math>3</math>.
  
This means that the square is in the form <math>(12c)^2</math>, where c is a positive integer. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions.
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This means that the square is in the form <math>(12c)^2</math>, where <math>c</math> is a positive integer. There are <math>\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}</math> solutions.
  
 
== See also ==
 
== See also ==

Revision as of 14:33, 19 April 2008

Problem

How many positive perfect squares less than $10^6$ are multiples of $24$?

Solution

The prime factorization of $24$ is $2^3\cdot3$. Thus, each square must have 3 factors of $2$ and 1 factor of $3$.

This means that the square is in the form $(12c)^2$, where $c$ is a positive integer. There are $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}$ solutions.

See also

2007 AIME I (ProblemsAnswer KeyResources)
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