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ZIML Division M Oct. 2018 #12

 
 
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Re: ZIML Division M Oct. 2018 #12
LensmireJohn - 2021年12月16日 Thursday 13:39
 

Hi Virginia! I don't want to spoil the whole question, but let's review the problem and give some of the first steps.

Key information from the problem (plus prime factorizations):

  1. $\text{gcd}(a,b,c) = 10 = 2\times 5$
  2. $\text{lcm}(a,b,c) = 600 = 2^3\times 3\times 5^2$
  3. $\text{lcm}(a,b) = 40 = 2^3\times 5$

We want to find the smallest value of $c$ that works with the above information. Let's start by summarizing what the above tell us about $c$ specifically:

  1. $c$ is a multiple of $2\times 5 = 10$.
  2. $c$ is a factor of $2^3\times 3\times 5^2 = 600$.

This gives us a good start, but how can we us 3? Remember that if we take the LCM of three numbers, it's allowed to first take the LCM of two of them. Thus, we must have (combining 2 and 3):$$600=2^3\times 3\times 5^2 = \text{lcm}(a,b,c) = \text{lcm}(2^3\times 5, c),$$since $\text{lcm}(a,b) = 40 = 2^3\times 5$.

Now we have 3 statements that give us restrictions on $c$, try to use them to find the smallest possible value for $c$. Let us know if you have more questions!