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[Moderators please note this is NOT a duplicate of

GRE problem involving LCD, prime factorization, and sets.

LCM. What am I missing?

GRE test prep question [LCM and divisors]

These posts only answer how 36 is the lowest value memeber of S, but not HOW and WHY '36 is the lowest member of S' means 'all the integer multiples of 36 are also members of S'. ]

Here is a tricky math problem. I could understand half of its solution. But need help understanding the last bit.

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?

a)12 b)24 c)36 d)72

Solution: Since n^2 is said to be multiple of 24 and 108 it implies that n2 is also multiple of LCM of 24 and 108. The LCM of 24 and 108 is (2^3)(3^3). But this doesn't mean that every multiple of (2^3)(3^3) will be a member of {n^2}, and also since (2^3)(3^3) is not a perfect square its also not a member of {n^2}. The smallest member of {n^2} will be (2^3)(3^3)(6) i.e. (2^4)(3^4) and so the first member of set S is squrt((2^4)(3^4)) = (2^2)(3^2) = 36.

I have understood how to find the first member of set S and this in itself is enough to get the solution as (c) and (d).

The solution goes on the state that

Other members of {n^2} are all the positive integer multiples of (2^4) (3^4) and so the other members of set S are positive integer >multiples of 36.

I am not able to understand how it was concluded that the rest of the members of {n^2} are all the integer multiples of (2^4)(3^4) and no other (other than multiples of (2^4)(3^4)) numbers. Can anyone explain how just because (2^4)(3^4) is a member of {n^2} means all its multiples are also members of {n^2}?

Thank you.

NOTE: I get how 36 is the lowest member of S, as has be answered in other questions...what I am not able to understand is how does 36 being the lowest member of S means all its integer multiples are also part of S. Also, how is it that only the multiples of 36 are members of S?

Solution from book: enter image description here enter image description here

GRANZER
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    The duplicates are very clear and straight forward...I don't see any other way to phrase the answer. – lulu Mar 19 '21 at 13:03
  • Your specific question is very unclear. What on earth is a "member" of $n^2$? Natural numbers do not have members. – lulu Mar 19 '21 at 13:04
  • @lulu Here {n^2} is the set whose members are squares of the member of S (Since in the question 'n' is give as any of the member belonging to S .) – GRANZER Mar 19 '21 at 13:09
  • That doesn't make a lot of sense. I expect you just mean that ${n^2}=S$, but why make your readers guess what you mean? In any case, the answers to the duplicates are very clear and straight forward. I suggest studying them. – lulu Mar 19 '21 at 13:12
  • @lulu I get how 36 is the lowest member of set S. But why does that mean all the integer multiples of 36 are also members of S? – GRANZER Mar 19 '21 at 13:12
  • Who says it does? You need to make a separate argument to show that multiples of $36$ are all in $S$. Happily, there is no difficulty doing that...as the answers to the duplicates make clear. – lulu Mar 19 '21 at 13:13
  • Andre's answer in the dupe shows that $,n\in S\iff 36\mid n,,$ i.e. $,S = 36\Bbb N,,$ so the divisors of $S$ are precisely the divisors of $36.,$ Is that not clear? – Bill Dubuque Mar 19 '21 at 13:23
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    Please don't duplicate your prior question(s). This violates site policy. – Bill Dubuque Mar 19 '21 at 13:28
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    I added a comment to Andre's answer giving the details of the converse. If anything remains unclear please explain precisely which step in the answer you cannot follow. – Bill Dubuque Mar 19 '21 at 13:43
  • @BillDubuque Andre's answer doesn't answer my question since he is not using LCM based solution. My goal with this question is not to get the solution but to understand the logic behind why multiples of 36 are members of S. As lulu mentioned above,it needs a seperate argument to tell that 'all' multiples of 36 are members of S. This post:https://math.stackexchange.com/a/1527553/502336 gives a explanation but it's not clear to me. – GRANZER Mar 19 '21 at 13:50
  • ...I know 'only some' of the integer multiples of squre root of LCM(24,108) are member of S(and so squirt(LCM(24,108))=squirt(216) is not a memeber of S). But 'all' the integer multiples of squrt((2^4)(3^4))= 36 are member of S. Now ((2^4)(3^4)) is not the LCM(24, 108) it's a integer multiple of the LCM. Why are all the multiples of 36 members of S? – GRANZER Mar 19 '21 at 13:50
  • A proof follows immediately by FTA = Fund. Theorem of Arithmetic (existence and uniqueness of prime factorizations) as below

    $$\large\begin{align} n\in S!!! \overset{\rm def!!}\iff, & 2^3\cdot 3, 2^2\cdot 3^3\mid n^2\ \iff, & 2^3\cdot 3^3\mid n^2\ \iff, & 2^2\cdot 3^2\mid n \end{align}\qquad$$

    For which step(s) above do you need further details?

     – Bill Dubuque Mar 19 '21 at 16:13
  • Or, equivalently, doing expt halving $,e\to \lceil e/2\rceil,$ before we do the lcm, as in Andre's answer:

    $$\large\begin{align} n\in S!!! \overset{\rm def!!}\iff, & 2^3\cdot 3, 2^2\cdot 3^3\mid n^2\ \iff, & 2^2\cdot 3,,\ 2\cdot 3^2\mid n\ \iff, & 2^2\cdot 3^2\mid n \end{align}\qquad$$

     – Bill Dubuque Mar 19 '21 at 16:30
  • @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: https://math.stackexchange.com/questions/878639/proving-that-any-common-multiple-of-two-numbers-is-a-multiple-of-their-least-com). 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 12:04
  • ...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 12:05
  • ...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 feel I am missing some fundamental and obvious math understanding here. – GRANZER Mar 21 '21 at 12:06
  • Your quoted argument is so imprecise that it is impossible to uniquely infer what is intended. The argument is either incomplete or incorrect, and it is impossible to know which is the case without the author filling in the many missing details. You already have links to complete rigorous proofs. These are what you should concentrate on. – Bill Dubuque Mar 21 '21 at 16:25
  • @BillDubuque I am really sorry about the imprecise nature of the argument. I have added pics of the solution from the book. I have also, marked the line in red that has stumped me. The line states "since every value of n^2 is a multiple of (2^4)(3^4)". I understand why every value of n^2 is a multiple of (2^3)(3^3). How is it obvious here that values of n^2 has to be multiples of (2^4)(3^4)? – GRANZER Mar 22 '21 at 03:52

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