Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
 
Line 7: Line 7:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
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>.
+
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>, etc... 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>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:56, 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 $12$: $12^2$, $24^2$, $36^2$, $48^2$, etc... 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_1&oldid=188795"