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)

- anonymous

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- myininaya

What does HCF mean?

- myininaya

And Xth CBSE?

- myininaya

Oh CBSE is some kind of exam,

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## More answers

- 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

oh so hcf is gcf

- myininaya

or also known as gcd

- anonymous

yes it is

- 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

i guess you've put it right

- myininaya

we might be able to use this
\[lcm(a,b)=\frac{| a \cdot b|}{\gcd(a,b) }\]

- 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

drop the absolute value because you know p and q and r are positive so there product will be too

- 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

I guess we aren't given the gcd(q,r)

- anonymous

No, I've copied the question faithfully from the worksheet I received.

- 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

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

Thanks for trying, maybe I'll tackle it with fresh mind tomorrow...
Right now I'm very tired......and its past midnight here......

- myininaya

you know I think q might be even since gcd(p,q)=2

- myininaya

so I think we are actually done

- 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

anyways I know you are gone now :p

- ganeshie8

do the integers `4, 6, 3` satisfy the given conditions

- ganeshie8

was just cooking up based on prime factorization of 72..

- 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

yeah looks there is no unique answer, lcm could be either 24 or 36

- myininaya

I'm going to see if I can find where I went wrong with my so called logic

- myininaya

guess it was in assuming that q was even with the given that r was not even that gcd(q,r)=1

- myininaya

because like you said r=3 is not even
and q =6 is even
and gcd(6,3) is not 1

- ganeshie8

thats it i guess, you have worked the case when gcd(q, r)=1
the other case is gcd(q, r)=3

- ganeshie8

seems 4, 18, 1 also satisfy
my batery is dying brb. .

- 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

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

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

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

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

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

in fact \((4,6,3)\) also gives rise to \((2,12,3)\) if you redistribute your \(2\)s

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