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I am a 9th grade student and having problems solving this question

  1. Oct 2, 2011 #1
    I am a 9th grade student and having problems solving this question :

    What is the remainder when 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 6^5 .... 100^5 is divided by 4
     
  2. jcsd
  3. Oct 2, 2011 #2
    Re: quotient

    We have;

    [itex]1^{5} [/itex] = 1 mod 4
    [itex]2^{5} [/itex] = 0 mod 4
    [itex]3^{5} [/itex] = 81 * 3 = 3 mod 4
    [itex]4^{5} [/itex] = 0 mod 4

    and for this partial sum we get 1+0+3+0 = 0 mod 4

    Now: what happens with [itex]5^{5}[/itex],[itex]6^{5}[/itex],[itex]7^{5}[/itex],[itex]8^{5}[/itex] compared with [itex]1^{5}[/itex],[itex]2^{5}[/itex],[itex]3^{5}[/itex],[itex]4^{5}[/itex] ?

    Regards
     
  4. Oct 2, 2011 #3
    Re: quotient

    I think from your title, you somehow feel that the quotient is material to this problem, but all that is needed is what would be the remainder of the sum of all those powers when divided by 4. Hint, it will be either 0,1,2 or 3. Rama Wolf show that 1^5 + 2^5 + 3^5 +4^5 will give a remainder of zero when divided by 4, i.e. the sum equals zero mod 4.

    An important fact about mod equations may be useful to consider here. (a+m)*(b+m) = a*b + (a+b)*m + m^2 = = a*b mod m since the remainder of m^2 + (a+b)*m = 0 when divided by m. Therefore 1^5 == 5^5 mod 4, etc.
     
    Last edited: Oct 2, 2011
  5. Oct 2, 2011 #4
    Re: quotient

    Please explain what is mod functions and equations. That's urgent ...
     
  6. Oct 2, 2011 #5
    Re: quotient

    Hi kiddy. Since i think we probably have to be a little tactful in explaining since u a just a ninth grader. Mod function is basically is giving you the remainder. When we say 5==1 mod 4 it means 5 has a remainder of 1 when divided by 4
     
  7. Oct 2, 2011 #6
    Re: quotient

    Thank you all very very much ....
     
  8. Oct 3, 2011 #7
    Re: quotient

    [tex]\sum_{i=1}^{n}i^5= \frac{2 n^6+6 n^5+5 n^4-n^2}{12}=\frac{n^2(n+1)(2n^2+2n-1)}{12}=\frac{(((2n+6)n+5)n^2-1)n^2}{12}[/tex]
     
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