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    Module 5 Day 7 Your Turn Part 3
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      amusingminnow M0 M1 M2 M3 M4 M5
      last edited by

      Let a=b^2 and b=b, then the LCM = b^2 and the GCD = b.
      This means that LCM/GCD=b, which fulfills the requirements.
      Why did you mark this as wrong?

      quacker88Q 1 Reply Last reply Reply Quote 1
      • quacker88Q
        quacker88 MOD @amusingminnow
        last edited by

        @amusingminnow These types of problems are tricky -- it's asking that given \(\frac{\text{lcm}(a,b)}{\gcd(a,b)}=b \), which of the options must be true. So the only thing we know that's true for sure is that \(\frac{\text{lcm}(a,b)}{\gcd(a,b)}=b \).
        We can't just assume that \(a=b^2\), because the problem never told us that! It just so happens that if \(a=b^2\), then the requirements are fulfilled. But, \(a\) doesn't have to be equal to \(b^2\). For example, \(a=9,b=6\) works.

        Make sure that when you're doing these problems, start with the given, and then find what must be true using only the given. Using that, you get that the fifth option is the right answer. If you want to look at the thread for the solution you can find it here: https://forum.poshenloh.com/topic/724/confusing-solution/2
        Hope this helps!

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