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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 $S$ ?

Indicate all such integers:

$A:12$

$B:24$

$C:36$

$D:72$

The answers are $A$ and $C$

First I took the lcm of $24$ and $108$ which is $2^3\times3^3$ but then it says that "the prime factorization of a square number must contain only even exponents. Thus, the least multiple of $(2^3)(3^3)$ that is a square is $(2^4)(3^4)$"

Can somebody explain why that is true?

What if the lcm was $2^3\times3^4$ ? Would I just make it $2^4\times3^4$ ?

Help!

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O.rka
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  • This is what I noticed, but I am not confident enough to post it as an answer: ${n \in \mathbb{Z},|, \exists k \in \mathbb{Z}: n^2 = 2^4\times3^4\times2k} \equiv {n \in \mathbb{Z},|, \exists j \in \mathbb{Z} : n = 4\times9\times2j ,, , \text{where $\sqrt{2j} \in \mathbb{Z}$}}$ – A.E Jul 28 '13 at 23:52
  • Well suppose if $n^2 \geq 4$ and suppose that atleast one of the exponents is odd. Let $n^2=p_1^{a_1}...p_n^{a_n}$ Suppose that $a_1$ is odd let $n_1=2k+1$ for some $k\in\mathbb{z}$ Taking the square root of $n^2$ we see that $2\nmid 2k+1$ Thus a square of a number must contain even exponents. – user60887 Jul 29 '13 at 00:37

2 Answers2

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Let $n=p_1^{a_1}\cdots p_r^{a_r}$ where the $p_i$ are primes, so $n^2=p_1^{2a_1}\cdots p_r^{2a_r}$. As you observed, $n^2$ must be a multiple of $LCM(24, 108)=2^{3} 3^{3}$, so $2a_1\ge 3$ and $2a_2\ge 3$ with $p_1=2$ and $p_2=3$. Therefore $a_1\ge 2$ and $a_2\ge 2$, so n is a multiple of $2^{2} 3^{2}=36$. Thus S consists of all positive multiples of 36, so the integers which divide every integer in S are simply the divisors of 36.

user84413
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We know that solution set is multiple of LCM . But here n^2 must be multiple of LCM . For that to happen it can be the number must be a perfect square . That can happen only if the power is even . Or else we will end up with square root in n .

In your case $ 2^3 * 3^4 $ as $n^2$. n will have $\sqrt{2}$ which makes n a non integer . so the numbers must be 216*6 , 216*6^3 ,...

Harish Kayarohanam
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  • so would i need to recognize that 2^3*3^3 = 216 which is not a perfect square and then find the smallest multiple of 216 that is a perfect square? – O.rka Jul 28 '13 at 23:36
  • yes. Or else it will result in n being a non integer. IN question it is said n is integer – Harish Kayarohanam Jul 28 '13 at 23:38