SOLUTION: The L.C.M of two numbers is 120 and their G.C.F is 6. one of the numbers is 30, what is the other number?

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Question 593989: The L.C.M of two numbers is 120 and their G.C.F is 6. one of the numbers is 30, what is the other number?

Answer by Edwin McCravy(20086)   (Show Source): You can put this solution on YOUR website!
The L.C.M of two numbers is 120 and their G.C.F is 6. one of the numbers is 30, what is the other number?
Easy way:

The product of the GCF and the LCM of two numbers is the product of the two numbers

Let the other number be N.

Then GCF×LCM = 30N or

      6(120) = 30N
         720 = 30N
          24 = N

----------------------------------
HARDER WAY but involves thinking not a memorized formula:
    

 30 = 2    ×3×5
  ? = ?  
---------------
      2    ×3   = 6
      2×2×2×3×5 = 120



Rule for L.C.M
The L.C.M. of two positive integers must have a prime factor 
the MOST number of times which ONE of them has it as a factor. 

Ruke for G.C.F
The G.C.F. of two positive integers must have a prime factor 
the LEAST number of times which BOTH of them have it as a factor.

L.C.M. = 120 has 2 as a factor three times, so the MOST number 
of times 30 or the other number has it as a factor is three times.
30 only has 2 as a factor 1 time, so the other number must have 2
as a factor 3 times.  So the other number is at least 2×2×2


G.C.F. = 6 has 2 and 3 once each as its prime factors.  30 has both these
factors once so the other number must have both of them once, so the other
number must have factor 2 and 3.

We have already determined that it must have 2 as a factor, (in fact it
must have it three times); therefore the other number has 2 as a factor 
3 times and 3 as a factor once.

So the other number is 2×2×2×3 = 24.
Infact you can fill it in from this chart:

 30 = 2    ×3×5
  ? = ?  
---------------
      2    ×3   = 6
      2×2×2×3×5 = 120

Bring up the factor not represented by one of the numbers 


 30 = 2    ×3×5
 24 = 2×2×2×3  
---------------
      2    ×3   = 6
      2×2×2×3×5 = 120


Edwin

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