# LCM. What am I missing?

Problem statement:

Let $S$ be the set of all positive integers $n$ such that $n^2$ is a multiple of both $24$ and $108$. Which of the following integers are divisors of every integer $n$ in $S$?

The choices are: $12, 24, 36, 72$. The solution is $12$ and $36$.

My solution was $S = \{36, 36* 6^3, 36*6^5, ...\}$

I can't figure out this problem. I started by finding the $\text{lcm}(24,108)$ and solved for $n$. I got some of the numbers in the set $S$, but not all of them, and I don't see what I'm missing.

Update: added the choices and what my results

-
Could you post your results for the set $S$? –  anorton Jan 28 '13 at 2:36

We have $n^2$ a multiple of $24=2^3\cdot 3$. If we think of the prime power factorization of $n$, we see it must have at least a $3^1$, and at least a $2^2$.

Similarly, since $108=2^2\cdot 3^3$, the number $n$ must "have" at least one $2^1$, and at least a $3^2$.

So $n$ must be divisible by $2^2\cdot 3^2$.

Conversely, anything divisible by $2^2\cdot 3^2$ is in $S$.

So the numbers that divide every element of $S$ are precisely the divisors of $36$. It is now straightforward to make a complete list: the positive ones are $1,2,3,4,6, 9, 12, 18, 36$.

-
Why is it that you chose not to make use of lcm? Also, what do you mean by "must 'have'"? –  AlanH Jan 28 '13 at 4:48
I chose not to make use of lcm to make the argument more basic, more elementary. If the prime power factorization of $n$ had highest power of $2$ equal to $2^1$, then the highest power of $2$ that would divide $n^2$ would be $2^2$. But we are told that $24$ divides $n^2$, so $2^3$ divides $n^2$. –  André Nicolas Jan 28 '13 at 5:00

So, $n^2$ is multiple of lcm$(24,108)=216=2^3\cdot3^3$

Clearly, $n$ is of the form $2^a\cdot 3^b\cdot f$ where $a,b,f$ are positive integers and $(2\cdot3,f)=1$.

$\implies n^2$ is of the form $2^{2a}\cdot3^{2b}\cdot f^2$

As $216\mid n^2\implies 3\le 2a$ or $2a\ge3\implies a\ge2, a=c+2$(say,) where integer $c\ge0$

Similarly, $b\ge 2, a=d+2$(say,) where integer $d\ge0$

So, $n=2^{2+c}3^{2+d}\cdot f=36\cdot 2^c3^d\cdot f$

So, $n$ must be divisible by $36$, hence by its divisors.

-