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

(Removed Solution 2 because it's literally just the answer, with no details.)
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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.
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== Solution 2 ==
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The perfect squares divisible by <math>24</math> are all multiples of <math>12</math>: <math>12^2</math>, <math>24^2</math>, <math>36^2</math>, <math>48^2</math>... Since they all have to be less than <math>10^6</math>, or <math>1000^2</math>, the closest multiple of <math>12</math> to <math>1000</math> is <math>996</math> (<math>12*83</math>), so we know that this is the last term in the sequence. Therefore, we know that there are <math>\boxed{083}</math> perfect squares divisible by <math>24</math> that are less than <math>10^6</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2007|n=I|before=First Question|num-a=2}}

Revision as of 12:54, 7 February 2023

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

The perfect squares divisible by $24$ are all multiples of $12$: $12^2$, $24^2$, $36^2$, $48^2$... Since they all have to be less than $10^6$, or $1000^2$, the closest multiple of $12$ to $1000$ is $996$ ($12*83$), so we know that this is the last term in the sequence. Therefore, we know that there are $\boxed{083}$ perfect squares divisible by $24$ that are less than $10^6$.

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

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