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 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>.
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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.
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== Solution 2 ==
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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 ==
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 22:51, 23 November 2020

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 at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$. This means that each square is in the form $(12c)^2$, where $12 c$ is a positive integer less than $\sqrt{10^6}$. There are $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}$ solutions.

Solution 2

After some work you will eventually find its just $\left\lfloor \frac{1000}{12}\right\rfloor = \boxed{083}$ 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

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