[ASK] About Power Again

Apr 2018
182
38
Sukoharjo
The sum of ten integers is 0. Show that the sum of the fifth powers of these numbers is divisible by 5.

For this one I don't know what I have to do at all other than brute-forcing which may even be impossible.
 

romsek

MHF Helper
Nov 2013
6,758
3,041
California
Where are you getting these problems? I'm beginning to suspect you are just throwing random stuff out to waste people's time.
 
Apr 2018
182
38
Sukoharjo
I saved them to use later at my job (making students' worksheet) since they seemed relatively easy at a first glance. However, those questions turned to be difficult, so I need some assistance.

If you feel that this is wasting your time, why bother commenting in the first place? I'll just wait for the other people who don't think negatively to help.
 
Jun 2013
1,134
605
Lebanon
\(\displaystyle x^5\equiv x \pmod 5\)
 

romsek

MHF Helper
Nov 2013
6,758
3,041
California
I saved them to use later at my job (making students' worksheet) since they seemed relatively easy at a first glance. However, those questions turned to be difficult, so I need some assistance.

If you feel that this is wasting your time, why bother commenting in the first place? I'll just wait for the other people who don't think negatively to help.
well let's see

first question had a fairly simple answer not in your list of answer choices

second question had an absurdly complicated answer that would take a human hours if not days to figure out

pardon me if I suspect your 3rd question

I will in fact ignore you from now on. Thank you.
 
Apr 2018
182
38
Sukoharjo
\(\displaystyle x^5\equiv x \pmod 5\)
Ummm... Care to elaborate?

well let's see

first question had a fairly simple answer not in your list of answer choices

second question had an absurdly complicated answer that would take a human hours if not days to figure out

pardon me if I suspect your 3rd question

I will in fact ignore you from now on. Thank you.
You know, you can ignore people without suspecting them wasting your time on purpose.
 
Jun 2013
1,134
605
Lebanon
By Fermat's Theorem or by some other more elementary method we have

\(\displaystyle x^5\equiv x \pmod 5\) meaning \(\displaystyle x^5- x\) is divisible by \(\displaystyle 5\) for all integers \(\displaystyle x\)

so that

\(\displaystyle \sum _i a_i=0\)

implies

\(\displaystyle \sum _i a_i^5\equiv \sum _i a_i=0\)
 
Apr 2018
182
38
Sukoharjo
I think I kinda have a grasp now. Thanks.