anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
What does HCF mean?
myininaya
  • myininaya
And Xth CBSE?
myininaya
  • myininaya
Oh CBSE is some kind of exam,

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anonymous
  • anonymous
HCF means Highest Common Factor and Xth CBSE means it is for Xth standard syllabus of India's Central Board of Secondary Education
myininaya
  • myininaya
oh so hcf is gcf
myininaya
  • myininaya
or also known as gcd
anonymous
  • anonymous
yes it is
myininaya
  • myininaya
\[pqr=72 \\ p,q,r \in \mathbb{Z}^+ \\ \gcd(p,q)=2 \\ \gcd(p,r)=1 \\ \text{ find } lcm(p,q,r)\] Ok just want to state it like this because it is easier for me to read
anonymous
  • anonymous
i guess you've put it right
myininaya
  • myininaya
we might be able to use this \[lcm(a,b)=\frac{| a \cdot b|}{\gcd(a,b) }\]
myininaya
  • myininaya
\[lcm(p,q)=\frac{p q}{\gcd(p,q)}=\frac{pq}{2} \\ lcm(p,r)=\frac{pr}{\gcd(p,r)}=\frac{pr}{1}=pr\]
myininaya
  • myininaya
drop the absolute value because you know p and q and r are positive so there product will be too
anonymous
  • anonymous
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..
myininaya
  • myininaya
I guess we aren't given the gcd(q,r)
anonymous
  • anonymous
No, I've copied the question faithfully from the worksheet I received.
myininaya
  • myininaya
\[\gcd(p,q,r)=\gcd(\gcd(p,q),r)=\gcd(2,r) \\ \gcd(\gcd(p,r),q)=\gcd(1,q)=1 \\ \text{ so } \gcd(2,r)=1 \text{ which means } r \text{ can't be even }\] but anyways we now have \[\gcd(p,q,r)=1 \] we still need to find gcd(q,r)
myininaya
  • myininaya
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)
anonymous
  • anonymous
Thanks for trying, maybe I'll tackle it with fresh mind tomorrow... Right now I'm very tired......and its past midnight here......
myininaya
  • myininaya
you know I think q might be even since gcd(p,q)=2
myininaya
  • myininaya
so I think we are actually done
myininaya
  • myininaya
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
myininaya
  • myininaya
anyways I know you are gone now :p
ganeshie8
  • ganeshie8
do the integers `4, 6, 3` satisfy the given conditions
ganeshie8
  • ganeshie8
was just cooking up based on prime factorization of 72..
myininaya
  • myininaya
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
ganeshie8
  • ganeshie8
yeah looks there is no unique answer, lcm could be either 24 or 36
myininaya
  • myininaya
I'm going to see if I can find where I went wrong with my so called logic
myininaya
  • myininaya
guess it was in assuming that q was even with the given that r was not even that gcd(q,r)=1
myininaya
  • myininaya
because like you said r=3 is not even and q =6 is even and gcd(6,3) is not 1
ganeshie8
  • ganeshie8
thats it i guess, you have worked the case when gcd(q, r)=1 the other case is gcd(q, r)=3
ganeshie8
  • ganeshie8
seems 4, 18, 1 also satisfy my batery is dying brb. .
anonymous
  • anonymous
$$pqr=72=2^3\cdot3^2\\\operatorname{gcd}(p,q)=2\\\operatorname{gcd}(p,r)=1$$... 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 \(p'=2\)
anonymous
  • anonymous
so \(p=4,q=6,r=3\) and thus \(\operatorname{lcm}(p,q,r)=\operatorname{lcm}(2^2,2\cdot3,3)=2^2\cdot3=12\)
anonymous
  • anonymous
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 \(3^4\not|\ pqr\)
anonymous
  • anonymous
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 \(r=9\)
anonymous
  • anonymous
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 \(\operatorname{lcm}(4,18,1)=36\)
anonymous
  • anonymous
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 \((4,18,1),(2,36,1)\)
anonymous
  • anonymous
in fact \((4,6,3)\) also gives rise to \((2,12,3)\) if you redistribute your \(2\)s

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