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Confusing solution

Module 5 Day 7 Your Turn Part 3
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  • I
    InTeReStInG M5
    last edited by debbie Jan 21, 2021, 10:10 PM Jan 20, 2021, 9:26 PM

    Module 5 Day 7 Your Turn Part 3 Mini-Question Solution

    Let the GCD of a and b be g; then we can express a as g × A and b as g × B, where A and B don't have any factors in common. Then the LCM of a and b is g × A × B, and the LCM divided by the GCD would be A × B. Since this is equal to b, that means that g × B = A × B, which means that g = A. Then a = A^2, and b is a multiple of the square root of a.
    Bold part is confusing

    Q 1 Reply Last reply Jan 20, 2021, 10:55 PM Reply Quote 5
    • Q
      quacker88 MOD @InTeReStInG
      last edited by quacker88 Jan 20, 2021, 10:58 PM Jan 20, 2021, 10:55 PM

      @neatlobster (I'm gonna copy over the question first):

      Suppose that for integers a and b,lcm(a,b)gcd⁡(a,b)=b\text{Suppose that for integers } a \text{ and } b, \frac{\text{lcm}(a,b)}{\gcd(a,b)} =b Suppose that for integers a and b,gcd(a,b)lcm(a,b)​=b
      Which of the following must be true?\text{Which of the following must be true?}Which of the following must be true?

      a=b2gcd⁡(a,b)=1One of the numbers is b times the othera=1a is a square, and b is a multiple of aa=b^2 \\ \gcd(a,b)=1 \\ \text{One of the numbers is } b \text{ times the other} \\ a=1 \\ a \text{ is a square, and } b \text{ is a multiple of }\sqrt{a} a=b2gcd(a,b)=1One of the numbers is b times the othera=1a is a square, and b is a multiple of a​

      Following the explanation, we have the following:
      gcd⁡(a,b)=g \gcd(a,b)=g gcd(a,b)=g
      a=g⋅Aa = g \cdot A a=g⋅A
      b=g⋅Bb = g \cdot B b=g⋅B
      lcm(a,b)=g⋅A⋅B\text{lcm}(a,b)=g \cdot A \cdot B lcm(a,b)=g⋅A⋅B

      Remember, the problem gave us that lcm(a,b)gcd⁡(a,b)=b\frac{\text{lcm}(a,b)}{\gcd(a,b)}=b gcd(a,b)lcm(a,b)​=b. Using our substitutions for the lcm & gcd, we have that

      g⋅A⋅Bg=b\frac{g \cdot A \cdot B}{g} = b gg⋅A⋅B​=b

      Now, here's the clever substitution: recall that b=g⋅Bb = g \cdot Bb=g⋅B. So,
      g⋅A⋅Bg=b  ⟹  b⋅Ag=b  ⟹  Ag=1  ⟹  A=g\frac{g \cdot A \cdot B}{g} = b \implies \frac{b \cdot A }{g} = b \implies \frac{A}{g} = 1 \implies A=g gg⋅A⋅B​=b⟹gb⋅A​=b⟹gA​=1⟹A=g

      Since a=g⋅A,a=A2a = g \cdot A , a = A^2 a=g⋅A,a=A2, so aaa is a perfect square.

      And, since a=A2  ⟹  a=A=ga=A^2 \implies \sqrt{a} = A = ga=A2⟹a​=A=g

      b=g⋅B  ⟹  b=a⋅Bb = g \cdot B \implies b= \sqrt{a} \cdot Bb=g⋅B⟹b=a​⋅B , so bbb is a multiple of a\sqrt{a}a​.
      Hence, the answer is the fifth one.

      I know this is a lot, let me know if you have any more questions. I'd be glad to help!

      P.S. Here's a hint: try using gcd⁡(a,b)⋅lcm(a,b)=a⋅b\gcd(a,b) \cdot \text{lcm}(a,b) = a \cdot bgcd(a,b)⋅lcm(a,b)=a⋅b. It's a very useful fact!

      1 Reply Last reply Reply Quote 7
      • I
        InTeReStInG M5
        last edited by Jan 21, 2021, 2:39 PM

        Thanks so much!

        D 1 Reply Last reply Jan 23, 2021, 5:12 PM Reply Quote 4
        • D
          debbie ADMIN M0★ M1 M5 @InTeReStInG
          last edited by Jan 23, 2021, 5:12 PM

          @neatlobster @quacker88 Thank you!! 🙂

          1 Reply Last reply Reply Quote 3

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