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

## 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.