Results 1 to 5 of 5

Math Help - help would be appreciated, thanks!

  1. #1
    Newbie
    Joined
    Jul 2007
    Posts
    15

    help would be appreciated, thanks!

    as n takes each positive integer value in turn (that is n=1, n=2, n=4 and so on) how many different values are obtained for the remainder when n(squared) is divede by n+4?

    a)1 b)8 c)9 d)16 e) infinitely many


    help would be appreciated, thanks!

    xxx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by mich13 View Post
    as n takes each positive integer value in turn (that is n=1, n=2, n=4 and so on) how many different values are obtained for the remainder when n(squared) is divede by n+4?

    a)1 b)8 c)9 d)16 e) infinitely many


    help would be appreciated, thanks!

    xxx
    So for n being an integer what are the possible remainders for:
    \frac{n^2}{n + 4}

    Start by looking at n = 1:
    \frac{1^2}{1 + 4} = \frac{1}{5}
    so the remainder is 1.

    n= 2: \frac{2^2}{2 + 4} = \frac{4}{6} so the remainder is 4. (No simplifying the fraction!)

    n= 3: \frac{3^2}{3 + 4} = \frac{9}{7} so the remainder is 2.

    n= 4: \frac{4^2}{4 + 4} = \frac{16}{8} so the remainder is 0.

    n= 5: \frac{5^2}{5 + 4} = \frac{25}{9} so the remainder is 7.

    Continue by doing the long division:
    \frac{n^2}{n + 4} = n - 4 + \frac{16}{n + 4}

    Does this help?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2007
    Posts
    15

    ?

    noo it ddnt help
    im 13 i dnt get a WORD of that lol
    but thanks anywways.. i appreciate your time xxx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by topsquark View Post
    Continue by doing the long division:
    \frac{n^2}{n + 4} = n - 4 + \frac{16}{n + 4}
    Well, you can also do this without long division

    \frac{{n^2 }}<br />
{{n + 4}} = \frac{{(n + 4)(n - 4) + 16}}<br />
{{n + 4}} = n - 4 + \frac{{16}}<br />
{{n + 4}}

    Though I don't think you used long division for this

    --

    mich13, I suggest you appreciate the help, read carefully. Our help it depends of the time.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by mich13 View Post
    noo it ddnt help
    im 13 i dnt get a WORD of that lol
    but thanks anywways.. i appreciate your time xxx
    I simply considered the cases for n = 1 to n = 5. This is just division. My point is, what is the remainder when n is larger than 12? Try a few of these.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set help appreciated!
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 5th 2010, 12:22 AM
  2. Set help appreciated
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 26th 2010, 09:17 PM
  3. Appreciated help.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 27th 2009, 01:07 PM
  4. help would be very much appreciated please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 18th 2008, 01:50 AM
  5. Help would be appreciated..........
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 19th 2006, 06:41 AM

Search Tags


/mathhelpforum @mathhelpforum