Difference between revisions of "2007 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
How many [[positive]] [[perfect squares]] less than <math>10^6</math> are [[multiple]]s of <math>24</math>?  
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How many [[positive]] [[perfect square]]s less than <math>10^6</math> are [[multiple]]s of <math>24</math>?  
  
 
== Solution ==
 
== Solution ==
The [[prime factorization]] of <math>24 = 2^33</math>; thus each square must be the square of a number which has 2 factors of <math>2</math> and 1 factor of <math>3</math> (must be a multiple of <math>2^23 = 12</math>). There are <math>\frac{996-12}{12} + 1 = 083</math> solutions.
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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>. 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</math> solutions.
  
 
== See also ==
 
== See also ==

Revision as of 14:57, 15 March 2007

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 $\left\lfloor \frac{1000}{12}\right\rfloor$ 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