A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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
 one year ago
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)

This Question is Closed

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Oh CBSE is some kind of exam,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0HCF means Highest Common Factor and Xth CBSE means it is for Xth standard syllabus of India's Central Board of Secondary Education

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2or also known as gcd

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[pqr=72 \\ p,q,r \in \mathbb{Z}^+ \\ \gcd(p,q)=2 \\ \gcd(p,r)=1 \\ \text{ find } lcm(p,q,r)\] Ok just want to state it like this because it is easier for me to read

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess you've put it right

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2we might be able to use this \[lcm(a,b)=\frac{ a \cdot b}{\gcd(a,b) }\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[lcm(p,q)=\frac{p q}{\gcd(p,q)}=\frac{pq}{2} \\ lcm(p,r)=\frac{pr}{\gcd(p,r)}=\frac{pr}{1}=pr\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2drop the absolute value because you know p and q and r are positive so there product will be too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2I guess we aren't given the gcd(q,r)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I've copied the question faithfully from the worksheet I received.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\gcd(p,q,r)=\gcd(\gcd(p,q),r)=\gcd(2,r) \\ \gcd(\gcd(p,r),q)=\gcd(1,q)=1 \\ \text{ so } \gcd(2,r)=1 \text{ which means } r \text{ can't be even }\] but anyways we now have \[\gcd(p,q,r)=1 \] we still need to find gcd(q,r)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for trying, maybe I'll tackle it with fresh mind tomorrow... Right now I'm very tired......and its past midnight here......

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you know I think q might be even since gcd(p,q)=2

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so I think we are actually done

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2Like 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
 one year ago
Best ResponseYou've already chosen the best response.2anyways I know you are gone now :p

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1do the integers `4, 6, 3` satisfy the given conditions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1was just cooking up based on prime factorization of 72..

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2p=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
 one year ago
Best ResponseYou've already chosen the best response.1yeah looks there is no unique answer, lcm could be either 24 or 36

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I'm going to see if I can find where I went wrong with my so called logic

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2guess it was in assuming that q was even with the given that r was not even that gcd(q,r)=1

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2because like you said r=3 is not even and q =6 is even and gcd(6,3) is not 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1thats it i guess, you have worked the case when gcd(q, r)=1 the other case is gcd(q, r)=3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1seems 4, 18, 1 also satisfy my batery is dying brb. .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$pqr=72=2^3\cdot3^2\\\operatorname{gcd}(p,q)=2\\\operatorname{gcd}(p,r)=1$$... 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
 one year ago
Best ResponseYou've already chosen the best response.0so \(p=4,q=6,r=3\) and thus \(\operatorname{lcm}(p,q,r)=\operatorname{lcm}(2^2,2\cdot3,3)=2^2\cdot3=12\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 \(3^4qr\) but \(3^4\not\ pqr\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0if 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
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0in fact \((4,6,3)\) also gives rise to \((2,12,3)\) if you redistribute your \(2\)s
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.