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

    Common Factor Question

    Module 5 Day 9 Challenge Part 3
    2
    2
    24
    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.
    • S
      spaceblastxy1428 M1★ M3★ M4 M5★
      last edited by

      Problem:
      Why isn't there an "earlier multiple" for the \(N=3\) case?
      i.) \(3\) is a prime
      ii.) \(3\) and \(10\) have no common factors
      iii.) \(10\) is not prime
      iv.) \(3\) is not a perfect square

      Explanation:
      For there to be an "earlier multiple" of \(3\) that makes a multiple of \(10,\) we would need a number less than \(30\) that is divisible by both \(3\) and \(10.\)
      Unfortunately, \(3\) and \(10\) are relatively prime, so their least common multiple is \(30;\) hence, there is no such "earlier multiple." Thus, the correct answer is ii.).

      Don't \(3\) and \(10\) have a common factor of \(1?\)

      1 Reply Last reply Reply Quote 2
      • N
        nastya MOD M0 M1 M2 M3 M4 M5
        last edited by debbie

        Hi @spaceblastxy1428!

        Actually, that's a good question. Logically, you are right, but \(1\) is a special number that is a factor of all of integer numbers, since any number is divisible by \(1.\) So any two random integer numbers have a common factor of \(1.\) That is why when mathematicians say that "two numbers doesn't have common factors", they mean that those two numbers don't have any common factors except for \(1.\) 🙂

        1 Reply Last reply Reply Quote 3

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