Forum — Daily Challenge
    • Categories
    • Recent
    • Tags
    • Popular
    • Users
    • Groups
    • Login

    Why does brute force not work in this question?

    Module 4 Day 7 Your Turn Part 2
    5
    5
    63
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • The Blade DancerT
      The Blade Dancer M0★ M1★ M2★ M3★ M4 M5
      last edited by debbie

      Why does brute force not work in this question? (I got i., which was apparently wrong). Is there some sort of unseen problem with just multiplying to get the 9th power of 3, and the 9th power of 2, and comparing them?

      The Blade Dancer
      League of Legends, Valorant: Harlem Charades (#NA1)
      Discord: Change nickname if gay#7585

      debbieD 1 Reply Last reply Reply Quote 3
      • debbieD
        debbie ADMIN M0★ M1 M5 @The Blade Dancer
        last edited by debbie

        @The-Blade-Dancer Hello! 🙂 Nice to hear from you again. I think you had the right idea with this question! From what you said, you put \(\frac{6561}{256}\) as your answer, so you knew that the answer is in the form of a power of three-halves, \( \left( \frac{3}{2} \right)^n.\)

        In fact, from what you say below, you actually correctly thought that the answer is \( \left( \frac{3}{2} \right)^9.\) 🙂 👍 But since you got choice i.) \( \left( \frac{6561}{256} \right) \) as your answer, you accidentally calculated \( \left( \frac{3}{2} \right)^8 \) instead of \( \left( \frac{3}{2} \right)^9.\) 🤦 It's 🆗!

        @The-Blade-Dancer said in Brute force

        Is there some sort of unseen problem with just multiplying to get the 9th power of 3, and the 9th power of 2, and comparing them?

        Actually, \( 3^9 = 19683\) and \(2^9 = 512,\) so the answer should be \( \boxed{\frac{19683}{512}}.\)

        The reason for the ratio \( \left( \frac{3}{2} \right) \) is this: at each step, the ratio of the growth of the first sequence compared with the growth of the second sequence is \( 2: 3,\) which can be simplified to \( 1: \frac{3}{2}.\) It's similar to the idea of a frame of reference, or relative speed, which we see in physics. We can pretend that the first sequence isn't changing, that its terms are all equal to \(1,\) and that the terms of the second sequence are changing relative to the terms of the first sequence.

        The terms of the second sequence are exactly the ratios of the \(n^{\text{th}}\) terms of the second sequence compared with the first, and so are the powers of \( \left( \frac{3}{2} \right).\)

        $$\begin{aligned} \text{ 1st sequence} \text{ } \text{ } & \text{ } \text{ } \text{ 2nd sequence} \\ \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right) \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^2 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^3 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^4 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^5 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^6 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^7 \\ 1 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \left( \frac{3}{2} \right)^8 \\ \textcolor{red}{1} \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ }\boxed{ \textcolor{red}{\left( \frac{3}{2} \right)^9} }\\ \ldots \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } & \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \ldots \\ \end{aligned} $$

        We're allowed to do this because we care only about the ratio of the \(9^{\text{th}}\) terms, not their actual values.

        🙂

        T F 2 Replies Last reply Reply Quote 2
        • T
          Tylenol M2 M3★ M4 M5 @debbie
          last edited by

          @debbie Why do we multiply by (2/3)^9 instead of (2/3)^8?
          I thought that the first term is 1 so we have to subtract 1 from n. does 1 not count as a term?

          I 1 Reply Last reply Reply Quote 2
          • I
            Iamnotawhale M0★ M1★ M2★ M3★ M4★ M5★ @Tylenol
            last edited by

            I have the same problem. Shouldn’t the ratio be (2/3)^8? In the problem, both sequences start with 1A48EF776-11FE-480A-8794-7E0CCFA03A04.jpeg
            In the solution, however it is clear that the sequences start with 2 and 3 respectively09DAAA46-3697-4691-B2C8-E9B2BC84405E.jpeg

            1 Reply Last reply Reply Quote 1
            • F
              forgivingsidewinder M0★ M1★ M2 M3 M4 M5 @debbie
              last edited by

              @debbie You only needed to do 2^9 since they both are prime, and their are know common denominators in the choices

              1 Reply Last reply Reply Quote 0

              • 1 / 1
              • First post
                Last post
              Daily Challenge | Terms | COPPA