# 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

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 $10^6$.

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