MHB When Is \( (k+1)^n+(k+2)^n+(k+3)^n+(k+4)^n+(k+5)^n \) Divisible by 5?

  • Thread starter Thread starter Ackbach
  • Start date Start date
Ackbach
Gold Member
MHB
Messages
4,148
Reaction score
93
Here is this week's POTW:

-----

For which nonnegative integers $n$ and $k$ is
$$ (k+1)^n+(k+2)^n+(k+3)^n+(k+4)^n+(k+5)^n $$
divisible by $5?$

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
Congratulations to kaliprasad, kiwi, and castor28 for their correct solutions to this week's POTW, which was Problem 109 in the MAA archives. kaliprasad's solution follows:

[sp]
Let us define$f(k,n) = (k+1)^n +(k +2)^n +(k +3)^n + (k+4)^n +(k +5)^n$Hence $f(k+1,n) - f(k,n ) = (x+6)^n - (k+1)^n$ and it is dvisible by $(k+6) - (k+1)$ or 5. this is independent of nSo $f(k+1,n)$ is divisible by 5 iff $f(k,n)$ is divisible by 5Further as 5 is prime we have as per Fermat's Little theorem $x^5=x$ for all x .Hence if $f(k,p)$ is divisible by 5 then $f(k,p+5)$ is divisible by 5So we need to check for $f(0,0),f(0,1),f(0,2),f(0,3),f(0,4)$ and we get

$f(0,0)=5,f(0,1) = 15, f(0,2) = 55, f(0,3) = 225, f(0,4) = 979$All except $f(0,4)$ is divisible by 5So it is divisible by 5 for n which is not of the form 5p+ 4 and for all k
for non negative k and p
[/sp]
 
Back
Top