The product of three integer p,q,r is 72 where p, q, r are positive integers. HCF of p and q is 2. HCF of p and r is 1. Find the LCM of p,q,r (This question is for Xth CBSE)

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The product of three integer p,q,r is 72 where p, q, r are positive integers. HCF of p and q is 2. HCF of p and r is 1. Find the LCM of p,q,r (This question is for Xth CBSE)

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What does HCF mean?
And Xth CBSE?
Oh CBSE is some kind of exam,

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HCF means Highest Common Factor and Xth CBSE means it is for Xth standard syllabus of India's Central Board of Secondary Education
oh so hcf is gcf
or also known as gcd
yes it is
\[pqr=72 \\ p,q,r \in \mathbb{Z}^+ \\ \gcd(p,q)=2 \\ \gcd(p,r)=1 \\ \text{ find } lcm(p,q,r)\] Ok just want to state it like this because it is easier for me to read
i guess you've put it right
we might be able to use this \[lcm(a,b)=\frac{| a \cdot b|}{\gcd(a,b) }\]
\[lcm(p,q)=\frac{p q}{\gcd(p,q)}=\frac{pq}{2} \\ lcm(p,r)=\frac{pr}{\gcd(p,r)}=\frac{pr}{1}=pr\]
drop the absolute value because you know p and q and r are positive so there product will be too
I also went about it this way but how to find the LCM of all three??? I feel Euclid's Division Lemma has to be used in some way..
I guess we aren't given the gcd(q,r)
No, I've copied the question faithfully from the worksheet I received.
\[\gcd(p,q,r)=\gcd(\gcd(p,q),r)=\gcd(2,r) \\ \gcd(\gcd(p,r),q)=\gcd(1,q)=1 \\ \text{ so } \gcd(2,r)=1 \text{ which means } r \text{ can't be even }\] but anyways we now have \[\gcd(p,q,r)=1 \] we still need to find gcd(q,r)
so we know p or q is even since the product of p,q, and r is even hmmm if we assume q is even then gcd(q,r)=1 \[lcm(p,q,r)=\frac{p qr \gcd(p,q,r)}{\gcd(p,q) \cdot \gcd(p,r) \cdot \gcd(q,r)} \frac{72 (1)}{2 \cdot 1 \cdot 1}=\frac{72}{2}=36\] but I don't know if q is even now what if we assume q is odd gcd(q,r) is harder to determine but I think we can somehow use gcd(1,q)=1 and gcd(2,r)=1 to find gcd(q,r)
Thanks for trying, maybe I'll tackle it with fresh mind tomorrow... Right now I'm very tired......and its past midnight here......
you know I think q might be even since gcd(p,q)=2
so I think we are actually done
Like I'm saying that we are given actually that both p and q are even since gcd(p,q)=2 which means 2 must be a factor of both p and q
anyways I know you are gone now :p
do the integers `4, 6, 3` satisfy the given conditions
was just cooking up based on prime factorization of 72..
p=4,q=6,r=3 yep \[\gcd(4,6)=2 \\ \gcd(4,3)=1 \] --- now given this case we have also \[\gcd(6,3)=3\] oops
yeah looks there is no unique answer, lcm could be either 24 or 36
I'm going to see if I can find where I went wrong with my so called logic
guess it was in assuming that q was even with the given that r was not even that gcd(q,r)=1
because like you said r=3 is not even and q =6 is even and gcd(6,3) is not 1
thats it i guess, you have worked the case when gcd(q, r)=1 the other case is gcd(q, r)=3
seems 4, 18, 1 also satisfy my batery is dying brb. .
$$pqr=72=2^3\cdot3^2\\\operatorname{gcd}(p,q)=2\\\operatorname{gcd}(p,r)=1$$... so we're told that \(p=2p',q=2q'\) where \(p',q'\) and \(p',r\) are coprime pairs where \(p'q'r=2\cdot3^2\) so it follows that we can pick \(q'=r=3\) trivially and \(p'=2\)
so \(p=4,q=6,r=3\) and thus \(\operatorname{lcm}(p,q,r)=\operatorname{lcm}(2^2,2\cdot3,3)=2^2\cdot3=12\)
the reason there is not a unique solution is because we are not told what \(\gcd(q,r)\) can be, so we're free to choose \(\gcd(q,r)\in\{1,3\}\) -- these are the only two permissible possibilities as \(\gcd(q,r)\) cannot be even otherwise \(\gcd(p,r)\ne 1\) and \(\gcd(q,r)\) must divide \(pqr=72\), and \(3^2\) is too big as then \(3^4|qr\) but \(3^4\not|\ pqr\)
the solution above is the unique one that yields \(\gcd(q,r)=3\); if we instead presume \(\gcd(q,r)=1\) then we force \(r=1\) or \(r=9\)
if we take \(r=9\) then \(p=4,q=2\) or \(p=2,q=4\) hence \((2,4,9),(4,2,9)\) are also solutions that they give \(\operatorname{lcm}(2,4,9)=36\); if we take \(r=1\) then \(p=18,q=4\) or \(p=4,q=18\) so \(\operatorname{lcm}(4,18,1)=36\)
and notice those solutions are obvious since in the case of \(r=9=3^2\) your \(3\)s in \(pqr=2^33^2\) are taken care of while you're free to apportion the three \(2\)s between \(p,q\) as either \(3=2+1=1+2\); similarly, with \(r=1\) it's clear that we need to put our \(3\)s in either \(p,q\) alone as otherwise \(3|\gcd(p,q)\) but \(\gcd(p,q)=2\), so we either put them in \(p\) and get \((18,4,1)\) or \((36,2,1)\) we put them in \(q\) and get \((4,18,1),(2,36,1)\)
in fact \((4,6,3)\) also gives rise to \((2,12,3)\) if you redistribute your \(2\)s

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