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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

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TheRealFakeNews
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2 Answers2

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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 $1,2,3,4,6, 9, 12, 18, 36$.

André Nicolas
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  • Why is it that you chose not to make use of lcm? Also, what do you mean by "must 'have'"? – TheRealFakeNews Jan 28 '13 at 04:48
  • 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 05:00
  • Details of the converse: note $,n = 36k = 2^2\cdot 3^2 k\Rightarrow ,n^2 = 2^4\cdot 3^4 k^2,$ is divisible by both $,24 = 2^3\cdot 3,$ and $,108 = 2^2\cdot 3^3,,$ so $,n\in S.,$ So we conclude $,n\in S\iff 36\mid n,,$ i.e. $,S = 36!:\Bbb N\ \ $ – Bill Dubuque Mar 19 '21 at 13:47
  • @BillDubuque This answer doesn't answer my question since he is not using LCM based solution. My goal with my question was not to just get the solution but to understand the logic behind why multiples of 36 are members of S and in doing so get a better grasp on using LCM... – GRANZER Mar 19 '21 at 15:29
  • ..As lulu had mentioned (https://math.stackexchange.com/questions/4068208/need-help-understanding-this-tricky-lcm-problem?noredirect=1#comment8405397_4068208), it needs a separate argument to tell that 'all' multiples of 36 are members of S and I am trying to understand that argument so it can be used for other problems too... – GRANZER Mar 19 '21 at 15:30
  • ...This post: math.stackexchange.com/a/1527553/502336 gives an explanation but it's not clear to me. I know 'only some' of the integer multiples of the square root of LCM(24,108) are members of S(and also so squirt(LCM(24,108))=squirt(216) is not a member of S as its not a perfect square). But 'all' the integer multiples of squrt((2^4)(3^4))= 36 are a member of S. Now ((2^4)(3^4)) is not the LCM(24, 108) it's an integer multiple of the LCM, (2^3)*(3^3). Why are all the multiples of 36 members of S? Can you please explain this? Thank you. – GRANZER Mar 19 '21 at 15:30
  • @BillDubuque I am writing down the precise point in the solution at which I have doubt. Can you please explain it to me?. Let us take this statement as RULE1:Every common integer multiple of 2 numbers is an integer multiple of LCM (Ref: math.stackexchange.com/questions/878639/…). Also, Let set B be the set of squares of members of set S. Every member of set B is an integer multiple (only some of the multiple) of LCM(24, 108) = (2^3)(3^3)= 216, by Rule1.. – GRANZER Mar 21 '21 at 15:30
  • ...But, Squrt(216) is not a member of set S as its not a perfect square. So, the lowest value member of set B is (2^4)(3^4)=1296 and so the lowest member of S is squrt(1296)=36. Now the solutions I have read goes on to state that since 1296 is the lowest member of set B all the members of set B are (only some of the) integer multiples of 1296. I understand how the members of set B are multiples of 216 (by Rule1), but 1296 is not the LCM of 36 and 108... – GRANZER Mar 21 '21 at 15:30
  • ...So, now how is it that we say that the next possible member of B is 12962 = 2592 and completely leap over values like 2167 OR 2168? I am asking this because the statement that possible values in B are multiples of 216 made sense to me (as this is the LCM) but 1296 is not the LCM its just the least multiple that is a perfect square. How is it obvious that there are no other perfect square between 1296 and 12962= 2592 like maybe 216*7 = 1512?. I may be missing some fundamental and obvious math here (some thing intutive like Rule1) but cant put my finger on it. Can you please help me? – GRANZER Mar 21 '21 at 15:31
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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.

lab bhattacharjee
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