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

[physics kiddy]

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

 

[RamaWolf]

We have;

1^{5} = 1 mod 4
2^{5} = 0 mod 4
3^{5} = 81 * 3 = 3 mod 4
4^{5} = 0 mod 4

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

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

Regards

 

[ramsey2879]

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

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.

 

[physics kiddy]

Please explain what is mod functions and equations. That's urgent ...

 

[icystrike]

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

 

[physics kiddy]

Thank you all very very much ...

 

[Xitami]

\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}