1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Non calc paper question

  1. Mar 31, 2007 #1
    Hi, I have a question

    tThe number 40 can be written as [tex]2^m[/tex] x n, where m and n are prime numbers

    I managed to guess that m = 3 and n = 5 by trials and error by using low prime numbers. Is this the proper/best way to work out the answer. If not can someone show me a proper way

  2. jcsd
  3. Mar 31, 2007 #2
    Your way works, because it got you the right answer, but consider what if the number was larger. The way I would do it, which is just one way of many, would be to write out the factors of 40 = {1x40, 2x20, ... , 40x1}. Since you know that 2^m can at most be 40, and also must be prime (meaning the first factor can only be 2^2, 2^3, 2^5 (which is not in the set of factors so it can be discarded)), that will narrow down the options of m, and from there you can find which of the 2 options of n will give the right factors. 4x10 (no!), or 8x5 (yes!)

    My way would only work for numbers relatively small though because you don't to do a bunch of factors for numbers in tens of thousands. There could be something better out there with logarithms or systems of equations.
    Last edited: Mar 31, 2007
  4. Mar 31, 2007 #3

    Gib Z

    User Avatar
    Homework Helper

    Yes there are better ways using Modular Arithmetic i think, but this is a GSCE or something paper, and considering as don't learn logs, I think they'll be fine with the guess and check for now, which is probably why they've been given a small number.
  5. Mar 31, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    Scratch last comment, I found an easier, yet systematic method that doesn't require guess and check.

    [tex]40 = 2^m \cdot n[/tex].

    Repeated divide by 2.
    [tex]20=2^{m-1}\cdot n[/tex]
    [tex]10=2^{m-2} \cdot n[/tex]
    [tex]5=2^{m-3}\cdot n[/tex]

    Since 5 can not be divided by 2 any further, this is the lowest you can reduce it. Since 2 is obviously not a factor, m-3 must be zero, or m=3. And its also obvious to see n then must be 5. HAPPY SUNSHINE :D:D:D
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook