Results 1 to 4 of 4

Math Help - some problems

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    12

    some problems

    please I need some help on these qustion



    1) Use the theory of congruences to show that [89|(2^44) − 1].


    2) For any n 2 N find the remainder when the number
    1^5 + 2^5 + 3^5 + + (n − 1)^5 + n^5
    is divided by 4.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    I don't know how far you've studied these things... Nevermind...

    We want to show that 2^{44}\equiv 1 [89]

    The cardinal of (Z/89Z)* is 88 because 89 is a prime integer.
    The elements of (Z/89Z)* are the positive integers <89 and coprime with 89. Because 89 is prime, any integer <89 will be coprime with 89. Hence, 2 belongs to (Z/89Z)*.

    We know that the order m of an element belonging to (Z/89Z)*, here 2, divides its cardinal, 88.

    Hence, m \in \{1;2;4;8;11;22;44;88\}

    m is such as :
    2^m \equiv 1 [89]

    2^m is > 89.

    \Longrightarrow m \neq 1;2;4

    Now, 2^8=256=3*89-11 \equiv -11 [89]

    2^{11}=2^8 \cdot 2^3=2^8 \cdot 8 \equiv (-11) \cdot 8 [89]

    ----> 2^{11} \equiv -88 [89] \Longleftrightarrow 2^{11} \equiv 1 [89]


    \boxed{2^{44}=(2^{11})^4 \equiv 1^4 [89]}

    QED.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Maha View Post
    please I need some help on these qustion



    1) Use the theory of congruences to show that [89|(2^44) − 1].


    2) For any n 2 N find the remainder when the number
    1^5 + 2^5 + 3^5 + + (n − 1)^5 + n^5
    is divided by 4.
    For the second one :
    We're looking for c : (1^5+2^5+3^5+\dots+(n-1)^5+n^5) \equiv c [4]

    1^5 \equiv 1 [4]
    2^5=2^2 \cdot 2^3=4 \cdot 2^3 \equiv 0 [4]
    3^5=3^2 \cdot 3^2 \cdot 3 \equiv 1 \cdot 1 \cdot 3 [4] \Longleftrightarrow 3^5 \equiv -1 [4]
    4^5 \equiv 0 [4]
    5^5 \equiv 1 [4]

    \dots

    Now, do you see any induction to this ?

    Hint : the possible values for the remainder is either 0 either 1, all depends on the congruence of n to 4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Note that a number a can be: <br />
a \equiv -1,0,1,2\left( {\bmod .4} \right)<br />

    We won't consider the even terms, since they're all multiple of 4.


    Case n = 2m - 1<br />

    \sum_{k=1}^{m}{(2k-1)^5}\equiv{\sum_{k=1}^{m}{(-1)^{k-1}}<br />
} (\bmod.4)

    Thus: : \sum_{k=1}^{m}{(2k-1)^5}\equiv{<br />
\frac{{1 - \left( { - 1} \right)^m }}{2} } (\bmod.4)

    Case n = 2m<br />

    We get the same result, since the term we have added is even

    Therefore the general result is: \sum_{k=1}^n{k^5}\equiv{<br />
\frac{{1 - \left( { - 1} \right)^{\left\lceil {\tfrac{n}<br />
{2}} \right\rceil } }}<br />
{2}}(\bmod.4)<br />
where <br />
\left\lceil x \right\rceil <br />
is the ceiling function
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 3rd 2011, 05:35 AM
  2. binomial problems/normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 19th 2010, 11:46 PM
  3. binomial problems as normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 19th 2010, 11:41 PM
  4. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 05:39 PM
  5. Help thease problems problems
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 1st 2008, 11:03 AM

Search Tags


/mathhelpforum @mathhelpforum