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)

1. myininaya

What does HCF mean?

2. myininaya

And Xth CBSE?

3. myininaya

Oh CBSE is some kind of exam,

4. anonymous

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

5. myininaya

oh so hcf is gcf

6. myininaya

or also known as gcd

7. anonymous

yes it is

8. 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

9. anonymous

i guess you've put it right

10. myininaya

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

11. myininaya

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

12. myininaya

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

13. 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..

14. myininaya

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

15. anonymous

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

16. 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)

17. 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)

18. anonymous

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

19. myininaya

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

20. myininaya

so I think we are actually done

21. 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

22. myininaya

anyways I know you are gone now :p

23. ganeshie8

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

24. ganeshie8

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

25. 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

26. ganeshie8

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

27. myininaya

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

28. myininaya

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

29. myininaya

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

30. ganeshie8

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

31. ganeshie8

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

32. 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 $$p'=2$$

33. anonymous

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

34. 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 $$3^4\not|\ pqr$$

35. 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 $$r=9$$

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

37. 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 $$(4,18,1),(2,36,1)$$

38. anonymous

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