Question

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)

Answer 1

What does HCF mean?

Answer 2

And Xth CBSE?

Answer 3

Oh CBSE is some kind of exam,

Answer 4

HCF means Highest Common Factor
and
Xth CBSE means it is for Xth standard syllabus of India's Central Board of Secondary Education

Answer 5

oh so hcf is gcf

Answer 6

or also known as gcd

Answer 7

yes it is

Answer 8

\[pqr=72 \\ p,q,r \in \mathbb{Z}^+ \\ \gcd(p,q)=2 \\ \gcd(p,r)=1 \\ \text{ find } lcm(p,q,r)\]
Ok just want to state it like this because it is easier for me to read

Answer 9

i guess you've put it right

Answer 10

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

Answer 11

\[lcm(p,q)=\frac{p q}{\gcd(p,q)}=\frac{pq}{2} \\ lcm(p,r)=\frac{pr}{\gcd(p,r)}=\frac{pr}{1}=pr\]

Answer 12

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

Answer 13

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..

Answer 14

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

Answer 15

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

Answer 16

\[\gcd(p,q,r)=\gcd(\gcd(p,q),r)=\gcd(2,r) \\ \gcd(\gcd(p,r),q)=\gcd(1,q)=1 \\ \text{ so } \gcd(2,r)=1 \text{ which means } r \text{ can't be even }\]
but anyways we now have
\[\gcd(p,q,r)=1 \]

we still need to find gcd(q,r)

Answer 17

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)

Answer 18

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

Answer 19

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

Answer 20

so I think we are actually done

Answer 21

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

Answer 22

anyways I know you are gone now :p

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

$$pqr=72=2^3\cdot3^2\\\operatorname{gcd}(p,q)=2\\\operatorname{gcd}(p,r)=1$$... 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 \(p'=2\)

Answer 33

so \(p=4,q=6,r=3\) and thus \(\operatorname{lcm}(p,q,r)=\operatorname{lcm}(2^2,2\cdot3,3)=2^2\cdot3=12\)

Answer 34

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 \(3^4\not|\ pqr\)

Answer 35

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 \(r=9\)

Answer 36

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 \(\operatorname{lcm}(4,18,1)=36\)

Answer 37

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 \((4,18,1),(2,36,1)\)

Answer 38

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