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#### Monoxdifly

##### Well-known member

- Aug 6, 2015

- 271

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

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- Thread starter
- #1

- Aug 6, 2015

- 271

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

- Oct 18, 2017

- 245

Hi Monoxdifly ,

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

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.

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- #3

- Aug 6, 2015

- 271

So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?Hi Monoxdifly ,

The number of integers does not matter. This would be a great time to use Fermat's Little Theorem : for any integer $a$, $a^5\equiv a\pmod5$, since $5$ is prime.

- Oct 18, 2017

- 245

Hi Monoxdifly ,So you mean that it applies to any number? That "The sum of ten integers is 0" is just a distraction, then?

The distraction is the fact that there are ten integers. The fact that the sum is $0$ is essential.

More precisely, if the integers are $a, b, c, \ldots$, Fermat's theorem tells us that $a^5\equiv a, b^5\equiv b\ldots\pmod5$.

We have therefore $a^5 + b^5 + \cdots\equiv a + b + \cdots\pmod 5$. As we are told that $a + b +\cdots=0\equiv0\pmod5$, we conclude that $a^5 + b^5 + \cdots\equiv0\pmod5$, which means that the sum is divisible by $5$.

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- #5

- Aug 6, 2015

- 271

I think I kinda get a grasp here. Thanks for your help.