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Reverse LCM/HCF 
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#1
May2414, 11:50 AM

P: 49

Hi all
I was hoping someone could help answer the following question: I think of two numbers A & B The HCF of A & B is 8 The LCM of A& B is 192 What are the numbers A&B? I do the following: A*B = HCF & LCM A*B = 8 * 192 A*B = 1536. Prime factor (PF) both numbers to get: PF of 8 = 2 x 2 x 2 PF of 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3 That is as far i can get, can someone help? Thanks 


#2
May2414, 12:02 PM

P: 424

What makes you think there is a unique solution?



#3
May2414, 12:11 PM

P: 49

i know that there could be several solutions but i am struggling to understand the process.
alos what could be the possbile numbers and how do you get to them? 


#4
May2414, 12:13 PM

P: 424

Reverse LCM/HCF
If you have no idea how to approach the problem I would suggest the following process:
Start listing the A's that could work in this problem, i.e. the A has 8 as a divisor and which itself is a divisor of 192. For instance your list might start: A=8 A=16 ... For every such A ask yourself what B should be for their LCM and HCF to be what you wanted. This problem is small enough that you can do it for every A, but you will probably see the pattern before you are done. EDIT: And remember to always look at prime decomposition. Never think of 16 or 32 think of 2^4 and 2^5. EDIT2: The process I outlined is not an efficient process and there is an easy way to solve problems such as these if you know the right mathematical results about prime decomposition and have a little experience, but I think working it out manually is a good way to get a feel for the problem and I don't want to just give you the solution. 


#5
May2414, 03:03 PM

Math
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Thanks
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#6
May2414, 03:17 PM

HW Helper
P: 2,264

from
HCF(A,B)=8 LCM(A,B)=192 next consider A/8 and B/8 what are HCF(A/8,B/8) LCM(A/8,B/8) again problems of this type can have multiple solutions in general without further restrictions 


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