Results 1 to 5 of 5

Math Help - need a easy way to solve it

  1. #1
    Senior Member
    Joined
    Sep 2012
    From
    Sweden
    Posts
    250
    Thanks
    6

    need a easy way to solve it

    For which integers are x^5+4x^4+3x+1 divisible by 5?
    i have solved this prob and there is none but i did it on a "hard way"
    i did make it to x(x^3(x+4)+3)+1 and try put like x to 1,2,3 etc.. is there any easy way to solve it?


    (sorry for bad english:/)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,110
    Thanks
    68

    Re: need a easy way to solve it

    Wolfram gives this as solution:
    solve x^5+4*x^4+3*x+1-5*k=0 for x - Wolfram|Alpha

    What d'heck does it mean...Plato?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    From
    Melbourne Australia
    Posts
    214
    Thanks
    28

    Re: need a easy way to solve it

    Let's plase all of the integers into one of 5 different classes:

    x = 5a + 0, or
    x = 5a + 1, or
    x = 5a + 2, or
    x = 5a + 3, or
    x = 5a + 4

    for each type of number I would try substituting for x in your original equation and determining if it is possible for the result to be divisible by 5. For example:

    Let X = 5a + 0. Yor equation becomes:

    (5a)^5+4(5a)^4+3(5a)+1 = 5(5^4a^5+4.5^3.a^4+3a)+1 and this cannot be divisible by 5 so the enture class of numbers x=5a+0 cannot give a result that is divisible by 5.

    Let X = 5a + 1. Yor equation becomes:

    (5a+1)^5+4(5a+1)^4+3(5a+1)+1

    When you multiply out the first bracket only the last term is not divisible by 5, the last term is 1
    When you multiply out the second bracket only the last term is not divisible by 5, the last term is 4
    When you multiply out the third bracket only the last term is not divisible by 5, the last term is 3

    so we can say that for some k

    (5a+1)^5+4(5a+1)^4+3(5a+1)+1 = 5k + 1+4+3+1=5k'+4 and this cannot be divisible by 5 so the enture class of numbers x=5a+1 cannot give a result that is divisible by 5.


    Now you can examine the remaining three classes of integers in the same way.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member MaxJasper's Avatar
    Joined
    Aug 2012
    From
    Canada
    Posts
    482
    Thanks
    54

    Lightbulb Re: need a easy way to solve it

    f({x})\text{=}x^5+4x^4+3x+1

    for k=1,2,3,....n

    \text{Mod}(f(k), 5) = \{4,3,2,1,1, 4,3,2,1,1, 4,3,2,1,1, ....\}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: need a easy way to solve it

    Just check x = 0,1,2,3,4. If x = k works, then any x that is congruent to k (mod 5) also works.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: February 9th 2013, 08:59 PM
  2. Easy-to-solve doubt! Please help me!
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 7th 2010, 11:12 AM
  3. Solve for x (easy)!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 31st 2010, 12:19 AM
  4. should be easy to solve for x radical
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 2nd 2010, 01:52 PM
  5. I can't solve this easy question ?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 8th 2009, 01:57 AM

Search Tags


/mathhelpforum @mathhelpforum